Engineering Mathematics-I (M101) Fourth Edition WBUT–2015 9789339222055, 9339222059

Table of contents :
Title
Contents
1 Matrix I
2 Matrix II
3 Successive Differentiation
4 Mean Value Theorems and Expansion of Functions
5 Reduction Formula
6 Calculus of Functions of Several Variables
7 Line Integral, Double Integral and Triple Integral
8 Infinite Series
9 Vector Analysis
Question Papers

Citation preview

Engineering Mathematics-I (M101) Fourth Edition

WBUT–2015

About the Authors Sourav Kar is presently Assistant Professor of Mathematics at Siliguri Institute of Technology, Siliguri, West Bengal. He did his postgraduation as well as PhD in Mathematics from the University of North Bengal, West Bengal, and also obtained a B.Ed. degree from SRK B.Ed. College, Darjeeling. He has been involved in teaching and research for more than eleven years and has published twelve research papers in various national and international journals. He has also presented several research papers in national and international conferences. Subrata Karmakar is presently Assistant Professor of Mathematics at the Siliguri Institute of Technology, Siliguri, West Bengal. He is a postgraduate in Mathematics from Utkal University, Bhubaneswar, and holds a B.Ed. degree from Regional Institute of Education (NCERT), Bhubaneswar, Odisha. He has qualified different national-level examinations like GATE and SLET. He has been involved in teaching and research for more than eleven years in applied mathematics. During his service at Siliguri Institute of Technology, as an active academician, he has participated in different workshops/ conferences/seminars conducted by DST, AICTE and UGC. He is pursuing research in the University of North Bengal. Dr Kar and Karmakar have jointly published two other books, Engineering Mathematics II and Engineering Mathematics III, for WBUT with McGraw Hill Education (India).

Engineering Mathematics-I (M101) Fourth Edition

WBUT–2015 Sourav Kar Assistant Professor Department of Mathematics Siliguri Institute of Technology Siliguri, West Bengal

Subrata Karmakar Assistant Professor Department of Mathematics Siliguri Institute of Technology Siliguri, West Bengal

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

Published by McGraw Hill Education (India) Private Limited, P-24, Green Park Extension, New Delhi 110 016 Engineering Mathematics-I, 4/e (WBUT–2015) Copyright © 2015, 2013, 2011 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. Print Edition ISBN-13: 978-93-392-2205-5 ISBN-10: 93-392-2205-9 Managing Director: Kaushik Bellani Head—Products (Higher Education & Professional): Vibha Mahajan Assistant Sponsoring Editor: Koyel Ghosh Editorial Executive: Piyali Chatterjee Manager—Production Systems: Satinder S Baveja Assistant Manager—Editorial Services: Sohini Mukherjee Senior Production Manager: P L Pandita Senior Graphic Designer—Cover: Meenu Raghav Senior Publishing Manager (SEM & Tech. Ed.): Shalini Jha Assistant Product Manager: Tina Jajoriya General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Script Makers, 19, A1-B, DDA Market, Paschim Vihar, New Delhi 110063 and printed at Adarsh Printers, B-47, Jhilmil Industrial, D.S.I.D.C., G.T. Road, Shahdara, Delhi - 110 095 Cover Printer: SDR Printers Pvt. Ltd. RZXYYRCVRBDQX Visit us at: www.mheducation.co.in

Dedicated To my teacher Dr Sanjib Kr Datta and My beloved family members Sourav Kar

To the Holy Mother Maa Sarada Subrata Karmakar

Contents Preface Roadmap to the Syllabus 1.

Matrix I 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18

2.

xiii xvii 1.1–1.65

Introduction 1.1 Different Type of Matrices 1.2 Algebraic Operations on Matrices 1.3 Transpose of a Matrix 1.7 Symmetric and Skew-Symmetric Matrices 1.9 Some Special Types of Matrices 1.11 Determinant of a Square Matrix 1.13 Properties of the Determinants 1.13 Minors and Cofactors of a Determinant 1.15 Expansion of a Determinant 1.17 Laplace Method of Expansion of a Determinant 1.20 Product of Determinants 1.22 Adjoint of a Determinant 1.24 Singular and Non-Singular Matrices 1.25 Adjoint of a Matrix 1.25 Inverse of a Nonsingular Matrix 1.26 Orthogonal Matrix 1.28 Trace of a Matrix 1.30 Worked-Out Examples 1.31 Short and Long Answer Type Questions 1.54 Multiple-Choice Questions 1.62

Matrix II 2.1 Introduction 2.1 2.2 Rank of a Matrix 2.1 2.3 Elementary Row and Column Operations 2.2

2.1–2.64

Contents

viii

2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14

3.

Successive Differentiation 3.1 3.2 3.3 3.4

4.

3.1–3.28

Introduction 3.1 Successive Differentiation 3.1 n-th Derivative of Some Important Functions 3.2 n-th Order Derivative of Product of Two Functions of Same Variable 3.7 Worked-Out Examples 3.9 Short and Long Answer Type Questions 3.22 Multiple-Choice Questions 3.26

Mean Value Theorems and Expansion of Functions 4.1 4.2 4.3 4.4 4.5 4.6 4.7

5.

Row Equivalent and Column Equivalent Matrices 2.3 Row-Reduced Echelon Matrix 2.4 Determination of Rank of Matrix by Elimentary Operations 2.4 Solution of Simultaneous Linear Equations by Matrix Inversion Method 2.5 Solution of Simultaneous Linear Equations by Cramer’s Rule 2.7 System of Homogeneous and Nonhomogeneous Linear Equations 2.8 Consistency and Inconsistecy of the System of Linear Equations 2.10 Existence of the Solution of Homogeneous System 2.10 Existence of the Solution of a Non-Homogeneous System 2.13 Eigen Values and Eigen Vectors 2.18 Diagonalisation of a Square Matrix 2.25 Worked-Out Examples 2.29 Short and Long Answer Type Questions 2.55 Multiple-Choice Questions 2.62

4.1–4.52

Introduction 4.1 Rolle’s Theorem 4.1 Lagrange’s Mean Value Theorem (Lagrange’s MVT) 4.6 Cauchy’s Mean-Value Theorem (Cauchy’s MVT) 4.15 Taylor’s Theorem (Generalised Mean-Value Theorem) 4.18 Maclaurin’s Theorem 4.21 Infinite Series Expansion of Functions 4.22 Worked-Out Examples 4.32 Short and Long Answer Type Questions 4.47 Multiple-Choice Questions 4.50

Reduction Formula 5.1 Introduction 5.1 5.2 Reduction Formula for

5.1–5.42

Contents

(a ) ( b)

òsin

n

ix

xdx, Where n(>1) is a Positive Integer

p 2 sin n xdx, 0

ò

Where n(>1) is a Positive Integer

5.1

5.3 Reduction Formula for (a ) ( b)

òcos

n

xdx, Where n(>1) is a Positive Integer

p 2 cos n xdx, 0

ò

Where n(>1) is a Positive Integer

5.6

5.4 Reduction Formula for (a )

òsin

m

xcosn xdx, Where m (>1) and n(>1) are Positive

Integers ( b)

p 2 sin m xcos n xdx, 0

ò

Where m (>1) and n(>1) are Positive

Integers 5.10 5.5 Reduction Formula for (a ) ( b)

òcos

m

xsin nxdx, Where m and n are Positive Integers

p 2 cos m xsin nxdx, 0

ò

Where m and n are Positive Integers

5.14

5.6 Reduction Formula for dx (a ) ò , Where n(>1) is a Positive Integer 2 2 n x +a

(

( b)

)

dx

¥

ò0

2 n

(x + a ) 2

, Where n(>1) is a Positive Integer

Worked-Out Examples 5.21 Short and Long Answer Type Questions Multiple-Choice Questions 5.40

6.

5.36

Calculus of Functions of Several Variables 6.1 6.2 6.3 6.4 6.5 6.6 6.7

5.17

Introduction 6.1 Functions of Several Variables 6.1 Limit and Continuity 6.2 Partial Derivatives 6.7 Composite Functions 6.13 Homogeneous Function and Euler’s Theorem 6.17 Differentiation of Implicit Functions 6.21

6.1–6.64

Contents

x

6.8 Total Differentials 6.22 6.9 Jacobians and Their Properties 6.23 6.10 Maxima and Minima 6.25 Worked-Out Examples 6.30 Short and Long Answer Type Questions Multiple-Choice Questions 6.62

7.

6.57

Line Integral, Double Integral and Triple Integral 7.1 Definition of Line Integrals 7.1 7.2 Double Integrals 7.4 7.3 Triple Integrals 7.12 Worked-Out Examples 7.17 Short and Long Answer Type Questions Multiple-Choice Questions 7.25

7.23

8.  Infinite Series  8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15

7.1–7.26

8.1–8.54

Introduction 8.1 Preliminary Ideas of Sequences 8.1 Different Types of Sequences 8.2 Bounded Sequence 8.2 Monotone Sequence 8.3 Limit of a Sequence 8.4 Convergent Sequence 8.4 Divergent Sequence 8.5 Infinite Series 8.5 Convergence and Divergence of Infinite Series 8.6 Properties of Convergence of Infinite Series 8.8 Different Tests of Convergence of Infinite Series 8.9 Alternating Series 8.16 Absolute Convergence 8.17 Conditional Convergence 8.18 Worked-Out Examples 8.20 Short and Long Answer Type Questions 8.46 Multiple-Choice Questions 8.51

9. Vector Analysis 9.1 Introduction 9.1 9.2 Scalars and Vectors 9.1 9.3 Scalar Product or Dot Product of Vectors 9.6

9.1–9.74

Contents

9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 9.20

xi

Vector or Cross Products of Vectors 9.7 Scalar Triple Products 9.10 Vector Triple products 9.11 Straight Line 9.11 Plane 9.14 Sphere 9.16 Vector Function of a Scalar Variable 9.21 Differentiation of Vector Functions 9.21 Scalar and Vector Point Function 9.22 Gradient of a Scalar Point Function 9.23 Level Surface 9.24 Directional Derivative of a Scalar Point Function 9.24 Tangent Plane and Normal to a Level Surface 9.26 Divergence and Curl of a Vector Point Function 9.28 Green’s Theorem in a Plane 9.46 Gauss’ Divergence Theorem 9.49 Stoke’s Theorem 9.52 Worked-Out Examples 9.17, 9.33, 9.56 Short and Long Answer Type Questions 9.69 Multiple-Choice Questions 9.72

Solution of 2010 WBUT Paper Solution of 2011 WBUT Paper Solution of 2012 WBUT Paper Solution of 2013 WBUT Paper Solution of 2014 WBUT Paper

SQP1.1–SQP1.13 SQP2.1–SQP2.14 SQP3.1–SQP3.10 SQP4.1–SQP4.7 SQP5.1–SQP5.8

Preface The highest form of pure thought is in Mathematics. Plato (428 BC–348 BC) Since mathematics forms the basis of any branch of engineering and technology, the West Bengal University of Technology has made a commendable endeavour by introducing different syllabi for mathematics at different semesters in the B.Tech. level. The current book has been written as per the latest WBUT syllabus for the firstyear, first-semester B.Tech. students. Our main objective of writing this book is to help students build upon the fundamental concepts which are also required for subjects studied in the higher semesters. Each and every topic of the book is lucidly explained and illustrated with different kinds of examples. Also, stepwise clarifications of different methods of solving problems are given.

Salient Features  

  



Full coverage of the WBUT syllabus (2010 Regulation) Lucid explanation of topics like Matrix, Infinite Series, Vector Algebra, Vector Calculus, Calculus of Functions of Several Variables Stepwise solutions to examples Solved WBUT questions from 2001-2009 incorporated within each chapter Solutions of WBUT examination papers from 2010-2014 are placed at the end of the book Rich pedagogy:  400 Solved Examples  315 Short and Long Answer Type Questions  220 Multiple Choice Questions

Chapter Organisation The contents of the book are divided into nine chapters. In Chapter 1, we first represent the fundamentals of matrices along with the notations and algebraic operations applicable on them. Here, we also discuss the determinant of a square matrix, singular and non-singular matrices, and the method of computing the inverse of a matrix along with its properties, orthogonal matrix and trace of a matrix. Chapter 2 deals with the concept of the rank of a matrix, matrix inversion

xiv

Preface

method, Cramer’s rule, consistency and inconsistency of a system of homogeneous and nonhomogeneous linear simultaneous equations, Eigen values and Eigen vectors, and the Cayley–Hamilton theorem and its applications. Chapter 3 discusses successive differentiation and Leibnitz’s theorem along with its applications. In Chapter 4, we present the very well-known three mean-value theorems, namely, Rolle’s, Lagrange’s and Cauchy’s mean-value theorems along with their wide range of applications in various fields. The series expansion theorems and formulas, namely, Taylor’s and Maclaurin’s series expansion, are also discussed in this chapter. Chapter 5 explains the concept of reduction formulas for integration and its applications. In Chapter 6, we introduce the concept of functions of several variables. Also, we describe the methods of differentiations and their applications towards optimisations of the functions. Chapter 7 deals with line integrals, double integrals and triple integrals. Chapter 8 basically covers preliminary ideas of real sequences and illustrative ideas of infinite series. Chapter 9 has been divided into three parts. In the first part of this chapter, we discuss vector algebra. The second part of the chapter deals with vector differentiations, gradient, divergence and curl. In the third part of the chapter, we give theorems on vector integrations (Green’s theorem, Divergence theorem, Stokes’ theorem) and their applications to physical problems. At the end of the each chapter, various kinds of solved examples covering all the topics, including 2001–2009 solved WBUT questions, are given. Numerous short and long-answer-type question and multiple-choice questions are given in the exercises of every chapter. Solutions of 2010 to 2014 WBUT examination papers are provided at the end of the book.

Acknowledgements We are, first and foremost, immensely grateful to our departmental colleagues of Siliguri Institute of Technology who offered noteworthy suggestions which helped us while correcting and improving the manuscript. We are also very thankful to the authors of various books which we consulted during the preparation of this book. We would like to thank the following reviewers who assessed various chapters of the script and contributed with their kind suggestions for improvement. Indranil Bhaumik

Academy of Technology, Hooghly

Asit Paria Sandip Paul Sarama Malik Das Nilanjan Dey

Budge Budge Institute of Technology, Kolkata Adamas Institute of Technology, 24 Parganas (N) Calcutta Institute of Technology, Howrah Calcutta Institute of Engineering and Management, Kolkata

Preface

xv

We owe the largest debt of gratitude to all those well wishers, including our family members, without whom this work would never have materialized—we acknowledge their encouragement, love, patience and support. Last, but not the least, we are very thankful to the entire team at McGraw Hill Education (India) for publishing this book within a short span of time. Sourav Kar Subrata Karmakar Publisher’s Note: Remember to write to us. We look forward to receiving your feedback, comments and ideas to enhance the quality of this book. You can reach us at [email protected]. Kindly mention the title and the authors’ names as the subject. In case you spot piracy of this book, please do let us know.

ROADMAP TO THE SYLLABUS

Engineering Mathematics-I This text is suitable for the Subject Code M101.

Module I Matrix Determinant of a square matrix; Minors and Cofactors; Laplace’s method of expansion of a determinant; Product of two determinants; Adjoint of a determinant; Jacobi’s theorem on adjoint of a determinant; Singular and nonsingular matrices; Adjoint of a matrix; Inverse of a nonsingular matrix and its properties; Orthogonal matrix and its properties, Trace of a matrix Rank of a matrix and its determination using elementary row and column operations; Solution of simultaneous linear equations by matrix inversion method; Consistency and inconsistency of a system of homogeneous and inhomogeneous linear simultaneous equations; Eigen values and Eigen vectors of a square matrix (of order 2 or 3); Eigen values of APTP; kA; AP-1P; Cayley–Hamilton theorem and its applications

GO TO

CHAPTER 1 CHAPTER 2

MATRIX I MATRIX II

Module II Successive Differentiation Higher-order derivatives of a function of single variable; Leibnitz’s theorem (statement only and its application; problems of the type of recurrence relations in derivatives of different orders and also to find ((yn)0) Mean-Value Theorems and Expansion of Functions Rolle’s theorem and its application; Mean-value theorems—Lagrange’s and Cauchy’s theorems and their application; Taylor’s theorem with Lagrange’s and Cauchy’s form of remainders and its application; Expansions of functions by Taylor’s and Maclaurin’s theorems; Maclaurin’s infinite series expansion of the functions sin x; cos x; ex; log(1 + x); (a + x )n, n being an integer or a fraction (assuming that the remainder Rn Æ 0 as n Æ • in each case)

xviii

ROADMAP TO THE SYLLABUS

Reduction Formula Reduction formulae both for indefinite and definite integrals of types Ú sinnx; Ú cosnx; Ú sinmx cosnx; Ú cosmx sin nx; Ú dx/(x2 + a2)n, m, n are positive integers

GO TO

CHAPTER 3 CHAPTER 4

SUCCESSIVE DIFFERENTIATION

CHAPTER 5

REDUCTION FORMULA

MEAN-VALUE THEOREMS AND EXPANSION OF FUNCTIONS

Module III Calculus of Functions of Several Variables Introduction to functions of several variables with examples; Knowledge of limit and continuity; Partial derivatives and related problems; Homogeneous functions and Euler’s theorem and related problems up to three variables; Chain rules; Differentiation of implicit functions; Total differentials and their related problems; Jacobians up to three variables and related problems; Maxima, minima and saddle points of functions and related problems; Concept of line integrals; Double and triple integrals

GO TO

CHAPTER 6

CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

CHAPTER 7

LINE INTEGRAL, DOUBLE INTEGRAL AND TRIPLE INTEGRAL

Module IV Infinite Series Preliminary ideas of sequence; Infinite series and their convergence/divergence; Infinite series of positive terms; Tests for convergence: Comparison test; Cauchy’s Root test; D’ Alembert’s Ratio test and Raabe’s test (statements and related problems on these tests); Alternating series; Leibnitz’s Test (statement; definition) illustrated by simple example; Absolute convergence and Conditional convergence

GO TO

CHAPTER 8

INFINITE SERIES

ROADMAP TO THE SYLLABUS

xix

Module V Vector Algebra and Vector Calculus Scalar and vector fields—definition and terminologies; Dot and cross products; Scalar and vector triple products and related problems; Equation of straight line; Plane and sphere; Vector function of a scalar variable; Differentiation of a vector function; Scalar and vector point functions; Gradient of a scalar point function; divergence and curl of a vector point function; Directional derivative (related problems on these topics) Green’s theorem; Gauss Divergence Theorem and Stokes’ theorem (statements and applications)

GO TO

CHAPTER 9

VECTOR ANALYSIS

CHAPTER

1 Matrix I 1.1

INTRODUCTION

Matrix algebra is a very essential part of mathematics. It has a wide range of applications in various branches of science and technology. Besides direct applications, we also borrow the concept of matrix notations for representing various systems in a compact manner. In this chapter, we first represent the fundamentals of matrices along with the notations and algebraic operations applicable on them. Next we discuss symmetric and skew-symmetric matrices with the help of the transpose property. Here, we shall also discuss a very important characteristic of matrices, namely, ‘determinant’, which is very useful for dealing with physical problems in science and technology. Here, we give different methods for computing determinants along with the various algebraic operations. Next, we describe the concept of singular and nonsingular matrices and the method of computing the inverse of a matrix along with its properties. In the last part, orthogonal matrix and trace of a matrix have been illustrated lucidly. Definition: A rectangular array of mn elements aij into m rows and n colunms where the elements aij belong to a field F enclosed by a pair of brackets, is said to be a matrix of order m × n over the field F. The m × n matrix is of the form ⎛ a11 ⎜ a21 ⎜ .. ⎜ ⎝ am

a12 a22 .. am 2

... a1n ⎞ ⎡ a11 ⎢a ... a2 n ⎟ or ⎢ 21 ⎟ .. .. ⎢ .. ⎟ ... amn ⎠ ⎢⎣ am

a12 a22 .. am 2

... a1n ⎤ ... a2 n ⎥ ⎥ .. .. ⎥ ... amn ⎥⎦

F is said to be the field of scalars. In particular, F is the field of real or complex numbers. The matrix is denoted by aij .

( )m × n

1.2

Engineering Mathematics-I

For example, ⎛ ⎞ (i) ⎜ 5 6⎟ is a matrix with 3 rows and 2 columns over a real field ⎜⎝ 8 0⎟⎠ 3 2 (ii)

⎛ 5 1 9⎞ ⎝ −4 6 1⎠ 2

⎛2 (iii) ⎜ 4 ⎜⎝ 8

1.2

1 9⎞ 6 5⎟ 7 2⎟⎠ 3

is a matrix with 2 rows and 3 columns over a real field 3

is a matrix with 3 rows and 3 columns over a real field 3

DIFFERENT TYPE OF MATRICES

1) Zero Matrix or Null Matrix: A matrix is called zero matrix if every element of it is 0. A null matrix of order m × n is denoted by Om × n . ⎛ 0 0⎞ For example, ⎜ 0 0⎟ is a zero matrix of order 3 2 and it is denoted by O3 2 . ⎜⎝ 0 0⎟⎠ ⎛ 0 0 0⎞ Also, ⎜ 0 0 0⎟ is a zero matrix of order 3 3 and it is denoted by O3 3 . ⎜⎝ 0 0 0⎟⎠ 2) Square Matrix: A matrix with equal number of rows and columns is called a square matrix. 5⎞ ⎛4 is a square matrix of order 2 2. For example, ⎝2 9⎠ 3) Diagonal Matrix: A square matrix is said to be a diagonal matrix if the elements other than the diagonal elements are all zero. ⎛ 1 0 0⎞ For example, ⎜ 0 4 0⎟ is a diagonal matrix of order 3 3. ⎜⎝ 0 0 3⎟⎠ 4) Scalar Matrix: A diagonal matrix is said to be a scalar matrix if all the diagonal elements are the same scalar. ⎛ 3 0 0⎞ For example, ⎜ 0 3 0⎟ is a scalar matrix of order 3 3. ⎜⎝ 0 0 3⎟⎠ 5) Identity Matrix: A scalar matrix with diagonal elements equal to 1 is called an identity matrix. ⎛ 1 0 0⎞ For example, ⎜ 0 1 0⎟ is an identity matrix of order 3 3. It is denoted by I 3. ⎜⎝ 0 0 1⎟⎠

1.3

Matrix I

6) Triangular Matrix: a) A square matrix ( ij ) is said to be an upper triangular matrix if all the elements below the diagonal are 0 . That is, aij = 0, i j . ⎛ 4 6 −8⎞ For example, ⎜ 0 1 3⎟ is an upper triangular matrix of order 3 3. ⎜⎝ 0 0 1⎟⎠ b) A square matrix ( ij ) is said to be a lower triangular matrix if all the elements above the diagonal are 0. That is, aij = 0, i j . ⎛ 4 0 0⎞ For example, ⎜ 7 1 0⎟ is a lower triangular matrix of order 3 3. ⎜⎝ −9 8 1⎟⎠ 7) Row Matrix and Column Matrix:

( )

Any matrix A aij is called a row matrix if m = 1, i.e., the matrix has only m×n one row. So its form is A ( a11 a12 a13 ... a1n )1× n .

( )m × n is called a column matrix if

Any matrix A aij only one column.

n = 1, i.e., the matrix has

⎛ a11 ⎞ ⎜ a21 ⎟ So its form is A = ⎜ a31 ⎟ ⎜ ⎟ ⎜ ... ⎟ ⎝ am1 ⎠ m ×1. For example, ( 2

3 0 5 8) is a row matrix.

⎛ 3⎞ ⎜ 0⎟ For example, A = ⎜ −6⎟ is a column matrix. ⎜ ⎟ ⎜ −3⎟ ⎝ 1⎠

1.3

ALGEBRAIC OPERATIONS ON MATRICES

1.3.1. Equality of Two Matrices Two matrices A and B are said to be equal if A and B have the same order and their corresponding elements are equal. Thus if A = aij and B (bij )m × n then A B if and only if aij bij for

( )m × n

i = 1, 2,

m; j = 1, 2,

n.

1.4

Engineering Mathematics-I

For example, A = d = 4, e

⎛a b ⎝d e

c⎞ ⎛ 1 2 3⎞ and B = are equal if a = 1, b = 2, c f⎠ ⎝ 4 5 6⎠

,

f = 6.

1.3.2. Multiplication by a Scalar The product of an m × n matrix A scalars is a matrix B bij = caij , i = 1, 2,

(aij )m × n by a scalar c where c

(bij )m × n defined by m; j

⎛ 2 For example, let A = ⎜ −3 ⎜⎝ 5

n and is written as B

12

1 4⎞ ⎛ 4 1 0⎟ then 2 = ⎜ −6 ⎜⎝ 10 0 1⎟⎠

F, the field of

cA c .

2 8⎞ 2 0⎟ . 0 2⎟⎠

1.3.3. Addition of Matrices If A C

(aij )m × n and B (bij )m × n then their sum (or difference) (cij )m × n where cij = aij ± bij , i = 1, 2, m; j = 1, 2, n.

A B is the matrix

Two matrices of the same order are said to be conformable for addition. For example, ⎛ 5 (i) If A = ⎜ 4 ⎜⎝ −7

1 8⎞ ⎛ −7 5 4⎞ 2 9⎟ and B = ⎜ −9 2 5⎟ ⎜⎝ 3 8 5⎟⎠ 5 1⎟⎠

⎛ 5 − 7 −1 + 5 8 + 4⎞ ⎛ −2 4 12⎞ then A B = ⎜ 4 − 9 2 + 2 9 + 5⎟ = ⎜ −5 4 14⎟ ⎜⎝ −7 + 3 5 + 8 1 + 5⎟⎠ ⎜⎝ −4 13 6⎟⎠ (ii) If A =

⎛ 2 i ⎝ 3 5i

then A B =

4 5i⎞ 8 7i 3i⎞ ⎛ −8 and B = 2 8i⎠ ⎝ −4 4 6i 7 i⎠ ⎛10 − 6i ⎝ 7 − 11i

4 2i⎞ where i = 9 7i ⎠

1.

Properties: a) Matrix addition is commutative, i.e., A B = B b) Matrix addition is associative, i.e., A ( B + C )

A. ( A + B) C .

c) Scalar multiplication is distributive over matrix addition, i.e., c( A B ) cA c ccB B. d) For any matrix Am × n , 0 Am e) For any matrix Am × n , Am

n

n

= Om × n . Om × n = Am × n .

1.5

Matrix I

1.3.4. Multiplication of matrices

( )m × n is of order m × n and B

If A = aij

(bij ) n × p is a matrix of order n × p then the

product AB is a matrix of order m × p and AB = C cij

∑ k =1aik bkj , n

Let A =

⎛ a11 ⎝ a21

a12 a22

i = 1, 2, a13 ⎞ a23 ⎠ 2

m; j = 1, 2,

3

⎛ b11 and B = ⎜ b21 ⎜⎝ b 31

(cij )m × p where

p. b12 ⎞ b22 ⎟ b32 ⎟⎠ 3

2

Here number of columns of A = 3 = number of rows of B. So multiplication is possible and the product AB is a 2 2 matrix, given by AB = =

⎛ a11 ⎝ a21

a12 a22

⎛b a13 ⎞ 11 . ⎜ b21 a23 ⎠ ⎜ ⎝ b31

⎛ a11b111 + a12 b221 + a13b331 ⎝ a21b111 + a22 b221 + a23b31

b12 ⎞ b22 ⎟ b32 ⎟⎠ a11b112 + a12 b222 + a13b332 ⎞ a21b12 a22 b22 a23b32 ⎠

Observation: a) The ij-th element of the product AB is obtained by multiplying the corresponding elements of the i-th row of A and the j-th column of B and adding such products. b) If the number of columns of A is not equal to the number of rows of B then AB is not defined. ⎛ 1 0⎞ ⎛ 1 0 −1⎞ and B = ⎜ 3 2⎟ ⎝ 0 2 3⎠ 2 3 ⎜⎝ 0 5⎟⎠ 3 2 Since number of columns of A = 3 = number of rows of B , the product AB is defined and is given by a 2 2 matrix ⎛ 1 0⎞ ⎛ 1 0 −1⎞ ⎜ AB = 3 2⎟ ⎝ 0 2 3⎠ ⎜ 5⎟⎠ ⎝0 Example 1

Let us consider two matrices

=

Ê 1 1 + 0 3 + ( 1) 0 1 0 + 0 (-2) + ( 1) 5ˆ Ê 1 -5ˆ =Á = 0 0 2 ¥ ( 2) 3 ¥ 5 ¯ ÁË 6 11¯ 2 ¥ 2 Ë 0 1 2¥3+3¥0

Again, since number of columns of B = 2 = number of rows of A, the product BA is defined and is given by a 3 3 matrix ⎛1 BA = ⎜ 3 ⎜⎝ 0

0⎞ ⎛ 1 0 −1⎞ 2⎟ ⎝ 0 2 3⎠ 5⎟⎠

1¥ 0 0 2 1 (-1) + 0 3 ˆ Ê 1 0 -1ˆ Ê 1 1+ 0 0 Á = 3 1 + ( 2) 0 3 0 + ( 2) 2 3 (-1) + ( 2) 3 = Á 3 4 -9 . Á Á 0 0 5¥2 0 ¥ ( 1) 5 ¥ 3 ¯ Ë 0 10 15¯ 3¥33 Ë 0 1 5¥0

1.6

Engineering Mathematics-I

Properties: (1) In general, matrix multiplication is not commutative, i.e., AB ≠ BA. From the previous example, it is obvious that AB ≠ BA. Also, in this case the orders of AB and BA are different. But in the next example we will see the fact that AB ≠ BA even when the orders of AB and BA are same. Let A =

Example 2 Then AB =

⎛1 ⎝2

⎛1 ⎝2

2⎞ ⎛ −5 12⎞ = 5⎠ ⎝ 18 −11⎠ 2

2⎞ ⎛ 3 . 3⎠ ⎝ 4 2⎞ ⎛ 1 . 5⎠ ⎝ 2

⎛3 and BA = ⎝4

2⎞ ⎛3 and B = 3⎠ 2 2 ⎝4

2⎞ ⎛ 7 = 3⎠ ⎝ −6 6

0⎞ 23⎠ 2

2⎞ 5⎠ 2

. 2

2

2

Here though the orders of AB and BA are same, AB ≠ BA. But in some special cases, AB may be same as BA, i.e., AB = BA, which follows from the next example: Let A =

Example 3 Then AB = and BA =

⎛1 ⎝0

⎛3 ⎝0

⎛1 ⎝0

2⎞ ⎛ 3 . 3⎠ ⎝ 0 2⎞ ⎛ 1 5⎠ ⎝ 0

2⎞ 3⎠ 2

and B = 2

2⎞ ⎛ 3 = 5⎠ ⎝ 0 2⎞ ⎛ 3 = 3⎠ ⎝ 0

12⎞ 15⎠ 2 12⎞ 15⎠ 2

⎛3 ⎝0

2⎞ 5⎠ 2

. 2

2

2

Here, AB = BA, which proves the above stated fact. (2) Matrix multiplication is associative, i.e., A( BC )

( AB )C .

(3) Matrix multiplication is distributive over matrix addition, i.e., A( B C )

AB

AC , provided both sides are defined.

(4) Matrix addition is distributive over matrix multiplication, i.e., ( )A = B BA A CA, provided both sides are defined. (5) For any square matrix A of order n × n , A I n = I n A = A. (6) For any square matrix A of order n × n, A On, n = On, n A = On, n . (7) The product of two non null matrices may result to a null matrix. This will be evident from the following example. Example 4

Ê 6 - 4ˆ 1 2ˆ , then Let A = Ê and B = ÁË - 3 2 ˜¯ ÁË 2 4˜¯ Ê 0 0ˆ 1 2ˆ Ê 6 - 4ˆ AB = Ê = Á ÁË 2 4˜¯ ÁË - 3 2 ˜¯ Ë 0 0˜¯

Therefore, we see that the product of two non null matrices results to a null matrix.

1.7

Matrix I

Example 5

Let A and B are two square matrices of the same order, examine

if the following holds (A + B)2 = A2 + 2AB + B2 Since, A and B are square matrices, A2 and B2 are both defined. Since, A and B are square matrices of same order, A + B and AB are both defined. Now, (A + B)2 = (A + B) (A + B) = A(A + B) + B(A + B) = A2 + AB + BA + B2 Since, in general matrix multiplication is not commutative, i.e., AB π BA, we have (A + B)2 π A2 + 2AB + B2 But, if the product of A and B are commutative i.e., AB = BA, then (A + B)2 = A2 + 2AB + B2 holds good Sol.

1.4

TRANSPOSE OF A MATRIX

( )

( )n × m obtained by convert-

The transpose of a matrix A aij is a matrix AT = a ji m×n ing rows into corresponding columns and vice-versa. ⎛ 1 5 −1⎞ For example, let us consider A = ⎝ −2 0 3⎠ 2 3 T

1 5 −1⎞ Then AT = ⎛ ⎝ −2 0 3⎠ 2

3

⎛ 1 =⎜ 5 ⎜⎝ −1

2⎞ 0⎟ 3⎟⎠ 3

2

Observation: The number of rows of AT = number of columns of A. and the number of columns of AT = number of rows of A. Properties: (a) ( T )T = A (b) ( )T = AT BT (c) ( )T = cAT , where c is a scalar (d) ( )T = cAT + dBT , where c and d are scalars (e) ( )T = BT AT Verification of the Above Properties ⎛ −1 2 1⎞ ⎛ 2 5 7⎞ Let us consider the two matrices, A = ⎜ 1 3 2⎟ and B = ⎜ 1 2 −1⎟ ⎜⎝ 5 4 1⎟⎠ ⎜⎝ 0 0 4⎟⎠ T

⎛ −1 2 1⎞ ⎛ −1 1 5⎞ (a) Here, AT = ⎜ 1 3 2⎟ = ⎜ 2 3 4⎟ ⎜⎝ 5 4 1⎟⎠ ⎜⎝ 1 2 1⎟⎠

1.8

Engineering Mathematics-I

⎛ −1 1 Now A =⎜ 2 3 ⎜⎝ 1 2 So the property (a) is verified. ⎛ −1 2 (b) Here, A B = ⎜ 1 3 ⎜⎝ 5 4

( )

T T

T

5⎞ ⎛ −1 2 1⎞ 4⎟ = ⎜ 1 3 2⎟ = A. ⎜⎝ 5 4 1⎟⎠ 1⎟⎠ 1⎞ ⎛ 2 5 7⎞ ⎛ −1 7 8⎞ 2⎟ + ⎜ 1 2 −1⎟ = ⎜ 2 5 1⎟ . 1⎟⎠ ⎜⎝ 0 0 4⎟⎠ ⎜⎝ 5 4 5⎟⎠ T

⎛ −1 7 8⎞ ⎛ −1 2 5⎞ ⎜ ⎟ ) = 2 5 1 = ⎜ 7 5 4⎟ . ⎜⎝ 5 4 5⎟⎠ ⎜⎝ 8 1 5⎟⎠ T

So, (

T

⎛ −1 2 1⎞ ⎛2 Again, A B = ⎜ 1 3 2⎟ + ⎜ 1 ⎜⎝ 5 4 1⎟⎠ ⎜⎝ 0 ⎛ −1 1 5⎞ ⎛ 2 = ⎜ 2 3 4⎟ + ⎜ 5 ⎜⎝ 1 2 1⎟⎠ ⎜⎝ 7 Hence, ( )T = AT BT . Similarly, we can show ( )T = AT Therefore, the property (b) is verified. T

T

5 2 0 1 2 1

T

7⎞ −1⎟ 4⎟⎠ 0⎞ ⎛ −1 2 5⎞ 0⎟ = ⎜ 7 5 4⎟ . 4⎟⎠ ⎜⎝ 8 1 5⎟⎠ BT .

⎛ − c 2c c⎞ (c) Here, cA = ⎜ c 3c 2c⎟ . ⎜⎝ 5c 4c c⎟⎠ T

⎛ − c 2c c⎞ ⎛ −c c 5c⎞ So (cA) = ⎜ c 3c 2c⎟ = ⎜ 2c 3c 4c⎟ ⎜⎝ 5c 4c c⎟⎠ ⎜⎝ c 2c c⎟⎠ ⎛ −1 1 5⎞ = c ⎜ 2 3 4⎟ = c ⋅ AT . ⎜⎝ 1 2 1⎟⎠ Hence, the property (c) is verified. T

⎛ −c 2c c⎞ ⎛ 2 d 5d 7d ⎞ (d) Here, cA + dB = ⎜ c 3c 2c⎟ + ⎜ d d −d⎟ ⎜⎝ 5c 4c c⎟⎠ ⎜⎝ 0 0 4d ⎟⎠ 7d ⎞ ⎛ − c + 2 d 2c + 5 =⎜ c+d 3c 2 d 2c − d ⎟ . ⎜⎝ 5 4c c + 4 d ⎟⎠ T

So,(cA + dB )

T

5c ⎞ ⎛ −cc d 2c 5d c 7d ⎞ ⎛ −cc d c d =⎜ c d 3c 2 d 2c d ⎟ = ⎜ 2c 5d 3c 2 d 4c ⎟ ⎜⎝ 5c ⎜⎝ c d 4c c d ⎟⎠ 2c d c d ⎟⎠

1.9

Matrix I

⎛ −c c 5c⎞ ⎛ 2 d = ⎜ 2 3c 4c⎟ + ⎜ 5 ⎜⎝ c 2c c⎟⎠ ⎜⎝ 7d

d 0⎞ ⎛ − c c 5c⎞ ⎛ 2 d 2d 0⎟ = ⎜ 2 3c 4c⎟ + ⎜ 5 d 4 d ⎟⎠ ⎜⎝ c 2c c⎟⎠ ⎜⎝ 7d

⎛ −1 1 5⎞ ⎛2 ⎜ ⎟ = c 2 3 4 + d ⎜5 ⎜⎝ 1 2 1⎟⎠ ⎜⎝ 7

1 0⎞ 2 0⎟ = cAT + dBT . 1 4⎟⎠

d 2d d

0⎞ 0⎟ 4 d ⎟⎠

Hence, the property (d) is verified. (e)

( AB)

T

⎡⎛ −1 2 1⎞ ⎛ 2 5 7⎞ ⎤ = ⎢⎜ 1 3 2⎟ . ⎜ 1 2 −1⎟ ⎥ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎣⎝ 5 4 1⎠ ⎝ 0 0 4⎠ ⎦

T

T

1 −5⎞ ⎛ 0 ⎛ 0 5 14⎞ ⎜ ⎟ = 5 11 8 = ⎜ −1 11 33⎟ . ⎜⎝ 14 33 35⎟⎠ ⎜⎝ −5 8 35⎟⎠ ⎛ 2 1 0⎞ ⎛ −1 1 5⎞ Again, BT AT = ⎜ 5 2 0⎟ . ⎜ 2 3 4⎟ ⎜⎝ 7 −1 4⎟⎠ ⎜⎝ 1 2 1⎟⎠ ⎛ 0 5 14⎞ = ⎜ −1 11 33⎟ . ⎜⎝ −5 8 35⎟⎠ So, (

1.5

)T = BT AT and the property (e) is verified.

SYMMETRIC AND SKEW-SYMMETRIC MATRICES

A square matrix A is said to be symmetric if AT A , i.e., A = (aij ) n × n is symmetric if aij a ji . Examples of symmetric matrices are 7i ⎞ ⎛ 1 i 2−3 4+5 ⎛ 2 5 7⎞ ⎛ a h g ⎞ ⎜ 2 3i 2 6i 3 − 5i ⎟ ⎜ 5 4 6⎟ , ⎜ h b f ⎟ , ⎜ −7 + i⎟ 6 4i ⎜⎝ 7 6 9⎟⎠ ⎜⎝ g f c ⎟⎠ ⎜ 4 + 5i −6 ⎟ 3 − 5 −7 + i 2 − i ⎠ ⎝ 7 T A square matrix A is said to be skew-symmetric if A A, i.e., A = (aij )n × n is skew-symmetric if aij a ji . Examples of skew-symmetric matrices are ⎛ 0 ⎜ −5 ⎜⎝ 0

5 0⎞ 0 6⎟ , 6 0⎟⎠

⎛ 0 ⎜ −9 ⎜⎝ 3

9 −3⎞ 0 6⎟ . 6 0⎟⎠

Properties: (i) If A and B are two symmetric matrices of the same order then their addition A B is also symmetric.

1.10

Engineering Mathematics-I

(ii) Suppose A and B are two symmetric matrices of the same order, then their product AB is also symmetic, provided AB = BA. (iii) Diagonal elements of any skew-symmetric matrix are all zero. Theorem 1.1: For any square matrix A, A AT is symmetric and A AT is skewsymmetric.

(

)

T

( ) ( )

T

Proof: Since A AT = ( A) AT = AT A A AT , we can say that A AT is symmetric. T T T AT = AT A = −( A AT ), therefore A AT is Again A AT = ( A) skew-symmetric. Hence, the theorem is proved.

(

)

T

Theorem 1.2: Any square matrix can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix. 1 1 Proof: For any square matrix A, we can write A (A AT ) (A AT ) . 2 2 1 T ) is also From the last theorem, we know A AT is symmetric, so ( 2 symmetric. Also, we have from the last theorem that A AT is skew-symmetric and so 1 T ( ) is also skew-symmetric. 2 Therefore, A can be expressed as the sum of a symmetric and a skew-symmetric matrix. Hence, the theorem is proved. ⎛ 1 3 5⎞ For example, let us express ⎜ 2 4 9⎟ as the sum of symmetric and skew-symmetric ⎜⎝ 5 7 8⎟⎠ matrices. T

⎛ 1 3 5⎞ ⎛ 1 3 5⎞ ⎛ 1 2 5⎞ T ⎜ ⎟ ⎜ ⎟ Here, A = 2 4 9 and so A = 2 4 9 = ⎜ 3 4 7⎟ ⎜⎝ 5 7 8⎟⎠ ⎜⎝ 5 7 8⎟⎠ ⎜⎝ 5 9 8⎟⎠ ⎛ 1 3 5⎞ ⎛ 1 2 5⎞ ⎛ 2 5 10⎞ Now A AT = ⎜ 2 4 9⎟ + ⎜ 3 4 7⎟ = ⎜ 5 8 16⎟ ⎜⎝ 5 7 8⎟⎠ ⎜⎝ 5 9 8⎟⎠ ⎜⎝ 10 16 16⎟⎠ 1 0⎞ ⎛ 1 3 5⎞ ⎛ 1 2 5⎞ ⎛ 0 ⎜ ⎟ ⎜ ⎟ ⎜ and A A = 2 4 9 − 3 4 7 = −1 0 2⎟ . ⎜⎝ 5 7 8⎟⎠ ⎜⎝ 5 9 8⎟⎠ ⎜⎝ 0 −2 0⎟⎠ T

1 1 T ) is symmetric and ( From the last theorem, we have ( 2 2 skew-symmetric. 1 1 Since A (A AT ) (A AT ), we can write 2 2

T

) is

Matrix I

A=

⎛ 2 5 10⎞ ⎛ 0 1 1 ⎜ 5 8 16⎟ + ⎜ −1 2 ⎜ 10 16 16⎟ 2 ⎜ 0 ⎝ ⎠ ⎝

⎛ 2 5 1 ⎜ 5 8 2 ⎜ 10 16 ⎝ Hence, A can be matrix. ⇒ A=

1.11

1 0⎞ 0 2⎟ 2 0⎟⎠

10⎞ 1 0⎞ ⎛ 0 1 16⎟ + ⎜ −1 0 2⎟ 16⎟⎠ 2 ⎜⎝ 0 −2 0⎟⎠ expressed as the sum of a symmetric and a skew-symmetric

Theorem 1.3: The product of a matrix with its transpose results in a symmetric matrix. Proof: Beyond the scope of the book.

Ê 2 3ˆ Verify that the product of the matrix Á 4 5˜ and its transpose results Example 6 Á ˜ Ë 6 7¯ to a symmetric matrix. Sol. Here we have

Ê 2 3ˆ A = Á 4 5˜ Á ˜ Ë 6 7¯

and so

Ê 2 4 6ˆ AT = Á Ë 3 5 7 ˜¯

Now,

A.AT

Ê 2 3ˆ Ê 13 23 33ˆ Ê 2 4 6ˆ Á ˜ = 4 5 Á = Á 23 41 59˜ Á ˜ Ë 3 5 7 ˜¯ Á ˜ Ë 6 7¯ Ë 33 59 85¯

which is a symmetric matrix.

1.6

SOME SPECIAL TYPES OF MATRICES

(1) Idempotent Matrix: Any A2 = A. ⎛ −1 For example, let A = ⎜ 3 ⎜⎝ 5

Ê -1 Then A = Á 3 Á Ë 5 2

square matrix A is called an idempotent matrix if 1 −1⎞ 3 3⎟ 5 5⎟⎠

1 -1ˆ Ê -1 3 3 .Á 3 Á 5 5¯ Ë 5

Ê -1 1 -1ˆ = Á 3 3 3 = A. Á Ë 5 5 5¯ So A is idempotent.

1 -1ˆ 3 3 5 5¯

1.12

Engineering Mathematics-I

2) Nilpotent Matrix: Any square matrix A is called a nilpotent matrix of index p if A p = O, where p is the least positive integer. 3 2⎞ ⎛ 5 For example, let A = ⎜ 15 −9 6⎟ ⎜⎝ 10 −6 4⎟⎠ 3 2⎞ ⎛ 5 3 2⎞ ⎛ 5 Then A2 = ⎜ 15 −9 6⎟ . ⎜ 15 −9 6⎟ ⎜⎝ 10 −6 4⎟⎠ ⎜⎝ 10 −6 4⎟⎠ ⎛ 0 0 0⎞ = ⎜ 0 0 0⎟ = O ⎜⎝ 0 0 0⎟⎠ So A is a nilpotent matrix of index 2. 3) Involutary Matrix: A square matrix A is said to be involutary if 5 8 ⎛ −5 ⎜ For example, let us show that A = 3 5 ⎜⎝ 1 2 Here, 5 8 0⎞ ⎛ −5 5 8 0⎞ ⎛ −5 A2 = ⎜ 3 5 0⎟ ⎜ 3 5 0⎟ ⎜⎝ 1 2 −1⎟⎠ ⎜⎝ 1 2 −1⎟⎠ ⎛ 25 − 24 + 0 = ⎜ −15 + 15 + 0 ⎜⎝ −5 5 6 −1 1

A2 = I . 0⎞ 0⎟ is involutary. −1⎟⎠

40 40 0 0 0 0⎞ 24 25 0 0 0 0⎟ 8 + 10 − 2 0 + 0 + 1⎟⎠

⎛ 1 0 0⎞ = ⎜ 0 1 0⎟ ⎜⎝ 0 0 1⎟⎠ Therefore, the matrix A is involutary. Example 7

If A and B are two matrices such that AB = A and BA = B, then

prove that AT and BT are idempotent. Sol.

Since AB = A, we have

( AB)T

= AT ⇒ BT AT

AT ...(1)

Again BA = B, so from (1)

( BA)T AT

⇒ AT AT = AT (by(1)) ⇒

( )

2

(

AT ⇒ AT BT AT AT

)

AT

1.13

Matrix I

So, it is proved that AT is idempotent. Similarly, since BA = B, we have

( BA)T

= BT ⇒ AT BT

BT ...(2)

Again, AB = A, so from (1)

( AB)T BT

(

BT ⇒ BT AT BT

)

BT

⇒ BT BT = BT (by(1)) ⇒

( )

2

BT

So, it is also proved that BT is idempotent.

1.7

DETERMINANT OF A SQUARE MATRIX

Definition Let M be the set of all square matrices of order n × n with real or complex entries then the determinant of any matrix of M is a function from the set M to any scalar field of real or complex numbers, i.e., determinant of a square matrix is a function which assigns to each matrix a scalar value. Determinant of a square matrix A = (aij )n × n is denoted by det A or A and we say the order of the determinant is n. ⎛ a11 ⎜a Let A = ⎜ 21 ... ⎜ ⎝ an

a12 a22 ... an 2

... a1n ⎞ ... a2 n ⎟ , then det A = ... ... ⎟ ⎟ ... ann ⎠

a11 a21 ... an

a12 a22 ... an 2

... a1n ... a2 n . ... ... ... ann

1 2 3 ⎛ 1 2 3⎞ ⎜ ⎟ For example, let A = 4 5 6 , Then det A = 4 5 6 is a determinant of order 3. ⎜⎝ 7 8 9⎟⎠ 7 8 9

1.8

PROPERTIES OF THE DETERMINANTS

Property 1: When any two rows or columns of a determinant are identical then the value of the determinant is zero. 1 2 3 For example, 4 5 6 = 0, since 1st and 3rd rows are identical. 1 2 3 1 1 4 2 2 5 = 0, since 1st and 2nd columns are identical. 3 3 6

1.14

Engineering Mathematics-I

Property 2: If all the entries in any one of the rows or columns are zero then the value of the determinant is zero. 1 2 3 For example, 4 5 6 = 0, since all the elements in the 3rd row are zero. 0 0 0 1 0 4 2 0 5 = 0, since all the elements in the 2nd column are zero. 3 0 6 Property 3: The determinant of any square matrix and its transpose have the same value, i.e., there will be no effect in the value of the determinant if we change the rows into columns and columns into rows. 1 2 4 1 6 9 For example, 6 5 7 = −2 5 8 since rows and columns are interchanged. 9 8 5 4 7 5 Property 4: The value of a determinant alters its sign when any two adjacent rows or columns are interchanged. 1 2 4 6 5 7 5 7 = ( −1) 1) 1 2 4 since 1st and 2nd rows are For example, 6 9 8 5 9 8 5 interchanged. 1 6 9

2 4 1 4 −2 5 7 = ( −1) 6 7 5 since 2nd and 3rd columns are interchanged. 8 5 9 5 8

Property 5: If any row or column is multiplied by any scalar then the value of the determinant is multiplied by the same scalar. 1 2 −2 4 1 5 7 = 2× 6 For example, 6 2 9 2 8 5 9 column are multiplied by 2. 1 2 4 1 6 3 5 3 7 3 = 3× 6 9 8 5 9

2 4 5 7 since the all the elements of the 1st 8 5

2 4 5 7 since the all the elements of the 2nd row are 8 5

multiplied by 3. Property 6: When the elements of any row (or any column) of a determinant are expressed as a sum of two quantities, then the determinant is the sum of two different determinants containing the terms of the sum as a row (or column) respectively. 1 2 −2 4 1 5 7=6 For example, 6 3 9 4 8 5 9

2 4 2 5 7+ 3 8 5 4

2 4 5 7 since the all the elements 8 5

of the 1st column are expressed as a sum of two quantities.

1.15

Matrix I

1 2 4 1 6 2 5 4 7 5=6 9 8 5 9

2 4 1 5 7+2 8 5 9

2 4 4 5 since the all the elements of the 8 5

2nd row are expressed as a sum of two quantities. Property 7: If any row (or column) in the determinant is replaced by summation of two or more rows (or columns) then it does not effect the value of the determinant. 2 For example, 3 4

2 4 2 3 −2 2 5 4 7 5 7= 3 5 7 , 1st row is replaced by the sum 8 5 4 8 5

of 1st and 2nd rows. [ R1′ → R1 + R2 ] 1 6 9

2 4 1 5 7=6 8 5 9

2+4 4 5 + 7 7 , 2nd column is replaced by the sum of 2nd and 3rd 8+5 5

columns. [C2′ → C2 + C3 ]. Property 8: If any row (or column) in the determinant is a scalar multiple of any other row (or column) then the value of the determinant is zero. 2 For example, 3 4 1 3 9

1.9

2 4 6 12 = 0, 2nd row is three multiples of the 1st row. 8 5

MINORS AND COFACTORS OF A DETERMINANT

Let A = (aij )n × n

1.9.1

2 4 5 6 = 0, since 3rd column is two multiples of the 1st column. 8 8

a11 a21 ... and its determinant is given by det A = ai1 ... an1

a12 a22 ... ai 2 ... an 2

... a1 j ... a2 j ... ... ... aij ... ... ... anj

... ... ... ...

a1n a2 n ... . ain ... ann

Minor

The minor of any entry aij of det A is defined to be that determinant obtained by deleting the corresponding row and column intersecting at the entry aij . The minor of aij is denoted by Mij and is given by deleting i -th row and j -th column from det A as follows:

1.16

Engineering Mathematics-I

a11 a21 ... Mij = a( a(

1)1 1)1

... an1

a12 a22 ... a( a(

1)2 1)2

... an 2

... a1( j −1) ... a2( j −1) ... ... ... a(i 1)( j −1) ... a(i 1)( j −1) ... ... ... an( j 1)

a1( j +1) a2( j +1) ...

... ... ...

a1n a2 n ...

a(i −1) a(i −1) n 1 ( j 1) a(i 1)( j +1) ... a( 1) n ... ... ... an( j +1) ... ann n

Observation If det A is of order n then minor of any element aij is of order (

1.9.2

1)

Cofactor

The cofactor of any element aij = ( −1)i + j × (minor of any element aij ). The cofactor of any element aij is denoted by Aij. So, the cofactor corresponding to aij is given by Aij = ( −1)i j × Mij . Example 8 Let det A =

1 2 3 4

Minor of 1 (a11 term) is obtained by deleting 1st row and 1st column from det A as M11 = 4. Cofactor of 1 (a11 term) is given by A11 = ( −1) 1)1 1 × M11 = ( −1) 1)1 1 × 4 = 4. Minor of 2 (a12 term) is obtained by deleting 1st row and 2nd column from det A as M12 = 3. Cofactor of 2 (a12 term) is given by A12 = ( −1) 1)1 2 × M12 = ( −1) 1)1 2 × 3 = −3. Minor of 3 (a21 term) is obtained by deleting 2nd row and 1st column from det A as M21 = 2. Cofactor of 3 (a21 term) is given by A21 = ( −1) 1)2 1 × M21 = ( −1) 1)2 1 × 2 = −2. Minor of 4 (a22 term) is obtained by deleting 2nd row and 2nd column from det A as M22 = 1. Cofactor of 4 (a22 term) is given by A22 = ( −1) 1)2 2 × M22 = ( −1) 1)2 2 × 1 = 1. Example 9 1 2 3 Let det A = 4 5 6 . 7 8 9 Minor of 1 (a11 term) is obtained by deleting 1st row and 1st column from det A as M11 =

5 6 . 8 9

Cofactor of 1 (a11 term) is given by A11 = ( −1)1 1 × M11 = ( −1)1 1 ×

5 6 . 8 9

1.17

Matrix I

Minor of 2 (a12 term) is obtained by deleting 1st row and 2nd column from det A as M12 =

4 6 . 7 9

4 6 . 7 9 Minor of 4 (a21 term) is obtained by deleting 2nd row and 1st column from the Cofactor of 2 (a12 term) is given by A12 = ( −1)1 2 3 . 8 9 Cofactor of 4 (a21 term) is given by A21 = ( −1)2

2

× M12 = ( −1)1

2

×

1

× M21 = ( −1)2

1

×

det A as M21 =

1.10 1.10.1

2 3 . 8 9

EXPANSION OF A DETERMINANT Case 1

Let A = (aij )1

1

= (a11 ); then det A

a11 = a11 .

For example, let A = (2), then det A = 2 = 2.

1.10.2

Case 2

Let A = (aij )2 2 ; then det A =

a11 a21

a12 . a22

(i) Row-wise Expansion: Expanding about the 1st row, we get the value of the determinant as det A = a11 (cofactor of a11 ) a12 (cofactor of a12 ) = a11 × A11 a12 × A12 Similarly, expanding about the 2nd row we get the value of the determinant as det A = a21 ( cofactor of a21 ) a22 ( cofactor of a22 ) = a21 × A21 a22 × A22 . (ii) Column-wise Expansion: Expanding about the 1st column, we get the value of the determinant as det A = a11 ( cofactor of a11 ) a21 ( cofactor of a21 ) = a11 × A11 a21 × A21 . Similarly, expanding about the 2nd column, we get the value of the determinant as det A = a12 ( cofactor of a12 ) a22 ( cofactor of a22 ) = a12 × A12 a22 × A22 . Note: a Any 2nd order determinant can be evaluated directly as det A = 11 a21 (a11 a22 ) − (a12 a21 ). Example 10

Let det A =

1 2 . 3 4

Here, a11 = 1 and the corresponding cofactor A11 = 4. a12 = 2 and the corresponding cofactor A12 = 3

a12 = a22

1.18

Engineering Mathematics-I

a21 = 3 and the corresponding cofactor A21 = 2 a22 = 4 and the corresponding cofactor A22 = 1 (i) Row-wise expansion: Expanding about the 1st row, we get det A = a11 ( cofactor of a11 ) a12 ( cofactor of a12 ) = a11 × A11 a12 × A12 = 1 × 4 2 × ( 3) = −2. Similarly, expanding about the 2nd row, we get det A = a21 ( cofactor of a21 ) a22 ( cofactor of a22 ) = a21 × A21 a22 × A22 = = 3 × ( 2) 4 × 1 = −2. (ii) Column-wise expansion: Expanding about the 1st column, we get det A = a11 ( cofactor of a11 ) a21 ( cofactor of a21 ) = a11 × A11 a21 × A21 = 1 × 4 3 × ( 2) = −2. Similarly, expanding about the 2nd column, we get det A = a12 ( cofactor of a12 ) a22 ( cofactor of a22 ) = a12 × A12 a22 × A22 = = 2 × ( 3) 4 × 1 = −2.

1.10.3 Case 3 a11 Let A = (aij ) 3 3; then det A = a21 a31

a12 a22 a32

a13 a23 . a33

(i) Row-wise Expansion: Expanding about the 1st row, we get the value of the determinant as det A = a11 ( cofactor of a11 ) a12 (cofactor of a12 ) a13 ( cofactor of a13 ) = a11 × A11 a12 × A12 a13 × A13 . Similarly, expansions can be done about the 2nd row and 3rd row. (ii) Column-wise Expansion: Expanding about the 1st column, we get the value of the determinant as det A = a11 (cofactor of a11 ) a21 ( cofactor of a21 ) a31 ( cofactor of a31 ) = a11 × A11 a21 × A21 a31 × A31 . Similarly, expansion can be done about the 2nd column and 3rd column. 1 0 2 Let us consider det A = 2 3 0 0 4 5 (i) Row-wise expansion: Expanding about the 1st row, we get det A = a11 (cofactor of a11 ) a12 ( cofactor of a12 ) a13 ( cofactor of a13 ) = a11 A11 + a12 A12 + a13 A13 Example 11

1.19

Matrix I

3 0 2 0 2 3 + 0 ( −1)1 2 + 2 ( −1)1 3 4 5 0 5 0 4 = 15 0 + 16 = 31. Similarly, expanding about the 2nd row and 3rd row we can get det A = 31. = 1 × ( 1)1

1

(ii) Column-wise expansion: Expanding about the 1st column, we get det A = a11 ( cofactor of a11 ) a21 (cofactor of a21 ) a31 (cofactor of a31 ) = a11 A11 + a21 A21 + a31 A31 3 0 0 2 0 2 + 2 ( −1)2 1 + 0 ( −1)3 1 4 5 4 5 3 0 = 15 16 0 = 31 Similarly, expanding about the 2nd column and 3rd column we can get det A = 31. = 1 × ( 1)1

1.10.4 A = (aij )n

1

Generalisation of the above Cases n

⎛ a11 ⎜a = ⎜ 21 ... ⎜ ⎝ an

a12 a22 ... an 2

... a1n ⎞ ... a2 n ⎟ then det A = ... ... ⎟ ⎟ ... ann ⎠

a11 a21 ... an

a12 a22 ... an 2

... a1n ... a2 n ... ... ... ann

can be evaluated by expanding the determinant about any one of the n rows or n columns. So we can conclude the facts by stating the following theorems. Theorem 1.4: The determinant of any square matrix can be evaluated by adding the products of the elements of any row or column and their corresponding cofactors. Theorem 1.5: If we consider the products of the elements of any row (or column) and the cofactors corresponding to the other row (or column), then summation of such products are always zero. Proof: Beyond of scope of the book. 1 0 2 Let us consider det A = 2 3 0 . 0 4 5

Example 12

Summation of the products of the elements of the 1st row and the cofactors corresponding to the 2nd row = 11 (cofactor of a21 ) a12 ( cofactor of a22 ) a13 ( cofactor of a23 ) = a11 A21 + a12 A22 + a13 A23 0 2 + 0 ( −1)2 4 5 = 8 + 0 8 = 0. = 1 × ( 1)2

1

2

1 2 + 2 ( −1)2 0 5

3

1 0 0 4

1.20

Engineering Mathematics-I

1.11 1.11.1

LAPLACE METHOD OF EXPANSION OF A DETERMINANT Complementary Minor

Consider A = (aij )n × n . Then D det A = aij is a determinant of order n. If we n×n delete r number of rows and r number of columns from D, the remaining determenant is of order ( ) and is said to be a minor of order ( ) of D. When all the rows and columns of a minor M are deleted from the determinant D then the remaining determinant is called a complimentary minor of M. a11 a21 Consider D = a31 a41

a12 a22 a32 a42

a13 a23 a33 a43

a14 a24 . a34 a44

Now if we delete the 2 d , 3rd rows and 3 d , 4th column from D, we have the minor as a a12 M23, 34 = 11 . a41 a42 Also, the complementary minor of M23, 34 is obtained by deleting all the rows and columns corresponding to M23, 34 in D. a a24 So, complementary minor of M23, 34 = 233 , (obtained by deleting the 1 , a33 a34 4th rows and 1 , 2nd columns from D). a a34 a11 a12 and 33 are also the examples of complementary minors. a43 a44 a21 a22 a31 a41

1.11.2

a a32 and 13 a23 a42

a14 are also the examples of complementary minors. a24

Algebraic Complement of a Minor

Consider M to be a minor of order r obtained by i1 i2 , ..., ir rows and j1 j2 , ..., jr columns and M be its complementory minor. Then the algebraic complement of M is ( 1) a11 a12 a13 a14 a21 a22 a23 a24 Consider D = . Then a31 a32 a33 a34 a41 a42 a43 a44

(i +i + 1 2

+ i ) ( j +,, j + r 1 2

+j ) r

(i) Algebraic complement of

a11 a41

a a12 is ( 1)(1+ 4)+ (1+ 2) 23 a33 a42

a24 a34

(ii) Algebraic complement of

a11 a21

a a12 is ( 1)(1+ 2)+ (1+ 2) 33 a43 a22

a34 a44

× M.

1.21

Matrix I

(iii) Algebraic complement of

1.11.3

a31 a41

a a32 is ( 1)(3+ 4)+ (1+ 2) 13 a23 a42

a14 a24

Laplace Expansion

The value of a determinant D

aij

n×n

can be expressed as the sum of the products of

all minors of order r and their respective algebraic complements. a11 a12 a13 a14 a21 a22 a23 a24 by the minors of order 2 selecting Let us try to expand D = a31 a32 a33 a34 a41 a42 a43 a44 from the first two rows as a a34 a11 a13 a a34 a a12 D = 11 × ( −1)(1+ 2)+ (1+ 2) 33 + × ( −1) 1 (1+ 2)+ (1+ 3) 32 a43 a44 a21 a23 a42 a44 a21 a22 a a33 a12 a13 a a14 a a34 + 11 × ( −1)(1+ 2)+ (1+ 4) 32 + × ( −1))(1+ 2) + (2 + 3) 31 a42 a43 a22 a23 a21 a24 a41 a44 +

a12 a22

a a14 × ( −1)(1+ 2)+ (2 + 4) 31 a41 a24

a33 a13 + a43 a23

a14 a × ( −1))(1+ 2) + (3+ 4) 31 a24 a41

a32 . a42

Note: The above can be expanded considering minors of any order and forming from any set of rows. But it is always easy to expand like above. Laplace Expansion can be done in terms of any sets of columns also. Example 13 2 6 D= 1 −5

Let us calculate the determinant 3 2 9 4

1 7 8 2

2 6 Expanding D = 1 −5 D=

2 6

5 2 using Laplace expansion. 3 7 3 2 9 4

1 7 8 2

5 2 as above, we have 3 7

3 8 3 2 1 9 3 × ( 1)(1+ 2)+ (1+ 2) + × ( 1)(1+ 2)+ (1+ 3) 2 2 7 6 7 4 7

+

2 5 9 8 −3 1 1 3 × ( −1)(1+ 2)+ (1+ 4) + × ( −1)(1+ 2)+ (2 + 3) 6 −2 4 2 2 7 −5 7

+

5 1 8 1 5 1 9 −3 × ( −1)(1+ 2)+ (2 + 4) + × ( −1)(1+ 2)+ (3+ 4) 2 −2 −5 2 7 −2 −5 4

= 22 50 8 51 15 14 23 22 4 × 42 − 37 × 49 = 1249.

1.22

Engineering Mathematics-I

1.12

PRODUCT OF DETERMINANTS

Let D

aij

n n

and D

bij

n n

be two determinants of order n Then the product

is a determinant of order n

1.12.1

Four Processes of Multiplication

1) Row–Columnwise Multiplication: a

=

a a

n

b

n

... a1n b a2 n b × ... ... ... ann

a12 a22 2

a b11 + a b11 +

+ .. + ..

b12 b22 ... 2

... b1n b2 n = ... bnn

12 + a b + a b12 + + .. ... .a bn 2

... + ..a bn1

... a b ... a b ... ... a b

nn n

.. n

2) Row-Row-wise Multiplication: a

=

a a

n

b

a12 a22

... a1n ... a2 n

2

... ann

+ + 11

n

×

b b

+ +

11

b12 b22

... b1n ... b2 n

2

... bnn

=

+a b + + + 21 ...

... a b ... a b ...

21

... + .a b1

.a b2

nn n

... n

3) Column-Row wise multiplication: D1 ¥ D2 = | aij | n ¥ n ¥ | bij |n ¥ n . a11 a21 = ... an1

a12 a22 ... an 2

... a1n b11 ... a2 n b21 × ... ... ... bn1 ... ann

a11b11 + a21b12 + ..an1b1n a11b21 + a21b22 + ..an1b2 n ... a11bn1 + a22 bn 2 + ..an1bnn

b12 b22 ... bn 2

... b1n ... b2 n = ... ... ... bnn

a12 b11 + a22 b12 + ..an 2 b1n a12 b21 + a22 b22 + ..an 2 b2 n ... a12 b1n + a22 bn + ..an 2 bnn

... a1n b11 + a2 n b12 + ..ann b1n ... a1n b21 + a2 n b22 + ..ann b2 n . ... ... ... a1n bn1 + a2 n bn 2 + ..ann bnn

1.23

Matrix I

4) Column-column wise multiplication: D1 ¥ D2 = | aij |n ¥ n ¥ | bij |n ¥ n . a11 a21 = ... an1

a12 a22 ... an 2

b11 ... a1n b21 ... a2 n ¥ ... ... ... bn1 ... ann

a11b11 + a21b21 + ..an1bn1 a11b12 + a21b22 + ..an1bn 2 ... a11b1n + a21b2 n + ..an1bnn

... b1n ... b2 n = ... ... ... bnn

a12 b11 + a22 b21 + ..an 2 bn1 a12 b12 + a22 b22 + ..an 2 bn 2 ... a12 b1n + a22 b2 n + ..an 2 bnn

1 Let us consider D1 = −5 0

Example 14

Then D1

b12 b22 ... bn 2

1 D2 = −5 0

0 2 2 3 0×4 2 1 3

... a1n b11 + a21bn 2 + ..ann bn1 ... a1n b12 + a21b22 + ..ann bn 2 ... ... ... a1n b1n + a21b2 n + ..ann bnn

0 2 2 and 3 0 D2 = 4 2 1 3

1 0 3 1 0 7

1 0 3 1 0 7

(i) By row–columnwise multiplication, we have D1

1× 2 0 4 + 2 3 1 1 + 0 ( −3) + 2 0 1 0 + 0 1+ 2 7 D2 = ( 5) 2 + 3 4 + 0 3 ( 5) 1 + 3 × ( −3) + 0 × 0 ( −5) × 0 + 3 × 1 + 0 × 7 0 × 2 + ( −2) × 4 + 1 × 3 0 × 1 + ( −2) × ( −3) + 1 × 0 0 × 0 + ( −2) × 1 + 1 × 7 10 = 2 −5

1 14 14 3 = −1541. 6 5

(ii) By row-row-wise multiplication, we have 1 2 + 0 1+ 2 0 1 4 + 0 ( −3) + 2 1 1 3+0 0+2 7 D1 D2 = ( 5) 2 + 3 1 + 0 0 ( 5) 4 + 3 × ( −3) + 0 × 1 ( −5) × 3 + 3 × 0 + 0 × 7 0 × 2 + ( −2) × 1 + 1 × 0 0 × 4 + ( −2) × ( −3) + 1 × 1 0 × 3 + ( −2) × 0 + 1 × 7 2 = −7 7 −2

6 29 7

17 15 = −1541. 7

So we have the same result in each of the cases. Similarly, if we apply the other two methods of multiplication, we will get the same result. Theorem 1.6: Let A and B be two square matrices of order n × n. Then (i) det( (ii) det(

)=

n

d ( A) det

) = det( A) det( B )

1.24

Engineering Mathematics-I

1.13

ADJOINT OF A DETERMINANT

Let us consider D

aij

n×n

a11 a21 = ... an

a12 a22 ... an 2

... a1n ... a2 n and Aij be the cofactor of aij in D. ... ... ... ann

Then adjoint of D is defined as A11 A12 ... A1n A21 A22 ... A2 n D Aij = . n×n ... ... ... ... An An 2 ... Ann 1 0 2 For example, let D = 2 3 0 . Then the adjoint of D is 0 4 5 3 4 0 D= − 4 0 3

0 5 2 5 2 0

2 0 1 0 1 − 2 −

Statement 1.3.1

0 5 2 5 2 0

2 0 1 − 0 1 2

3 4 15 −10 8 0 = 8 5 −4 . 4 −6 4 3 0 3

Jacobi’s Theorem on Adjoint of a Determinant

Let D ≠ 0 be a determinant of order n and D be its adjoint. Then D = D n −1. Proof: Beyond the scope of this book. Corollary: For n = 3, the Jacobis theorem becomes D = D 2 . A11 A12 A13 a11 a12 a13 Proof: Here, D = a21 a22 a23 and adjoint of D is D = A21 A22 A23 , where A31 A32 A33 a31 a32 a33 Aij is the cofactor of aij in D. a11 a12 a13 A11 A12 A13 Now D D = a21 a22 a23 × A21 A22 A23 a31 a32 a33 A31 A32 A333 a11 A11 + a12 A12 + a13 A13 a11 A21 + a12 A22 + a13 A23 a11 A31 + a12 A32 + a13 A33 = a21 A11 + a22 A12 + a23 A13 a21 A21 + a22 A22 + a23 A23 a21 A31 + a22 A32 + a23 A33 a31 A11 + a32 A12 + a33 A13 a31 A21 + a32 A22 + a33 A23 a31 A31 + a32 A32 3 + a33 A33 D = 0 0 = D3 .

0 0

0 0 D

1.25

Matrix I

So, D D = D3 ⇒ D = D 2 (for D ≠ 0).

1.14

SINGULAR AND NON-SINGULAR MATRICES

Any square matrix A = (aij ) n × n is said to be nonsingular iff det( A) = aij Otherwise, it is singular. For example,

n×n

≠ 0.

⎛ 1 0 2⎞ (i) ⎜ 2 3 0⎟ is a nonsingular matrix since its determinant is nonzero ⎜⎝ 0 4 5⎟⎠ ⎛ 1 0 2⎞ (ii) ⎜ 2 3 0⎟ is a singular matrix since its determinant is zero ⎜⎝ 0 0 0⎟⎠

1.15

ADJOINT OF A MATRIX

Let us consider any square matrix

A = (aij )n

n

⎛ a11 ⎜a = ⎜ 21 ... ⎜ ⎝ an

a12 a22 ... an 2

... a1n ⎞ ... a2 n ⎟ ... ... ⎟ ⎟ ... ann ⎠

and Aij be the cofactor of aij in det A. Then adjoint of the matrix A is denoted by adj( A) and is defined as the transpose of the matrix ( ij )n × n .

So, adj ( A) = ( Aij )Tn

n

⎛ A11 ⎜A = ⎜ 21 ... ⎜ ⎝ An

A12 A22 ... An 2

T

... A1n ⎞ ... A2 n ⎟ . ... ... ⎟ ⎟ ... Ann ⎠

⎛ 1 0 2⎞ For example, let A = ⎜ 2 3 0⎟ . Then the adjoint of the matrix A is ⎜⎝ 0 4 5⎟⎠ ⎛ 3 ⎜ 4 ⎜ 0 adj( A) = ⎜⎜ − 4 ⎜ ⎜ 0 ⎜ 3 ⎝

0 5 2 5 2 0

2 0 1 0 1 − 2

−

0 5 2 5 2 0

2 0 1 − 0 1 2

T

3⎞ 4⎟ T ⎟ 8⎞ 8 −6 ⎞ ⎛ 15 ⎛ 15 −10 0⎟ ⎜ 8 ⎟ = ⎜ −10 − = 5 4 5 4 ⎟. 4⎟ ⎜ ⎟ ⎜ ⎟ 4 3⎠ 3⎠ ⎟ ⎝ 8 −4 ⎝ −6 0⎟ 3 ⎟⎠

1.26

1.15.1

Engineering Mathematics-I

Properties

(1) For any square matrix A, adj ( AT ) = [adj ( A)]T . (2) For any square matrix A of order n × n, adj (cA) = c n 1 [adj ( A)], where c is any scalar. Theorem 1.7: For any square matrix A of order n × n, always A i [ adj ( A)] = [ adj ( A)] A = det A I n Proof: Beyond of scope of the book. Example 15

From the previous example, we have

⎛ 15 ⎛ 1 0 2⎞ A = ⎜ 2 3 0⎟ and adj( A) = ⎜ −10 ⎜ ⎜⎝ 0 4 5⎟⎠ ⎝ 8

8 5 4

6⎞ 4 ⎟. ⎟ 3⎠

Also, it is easy to check that det A = 31. ⎛ 31 0 0 ⎞ ⎛1 0 0⎞ Now A ⋅ [ adj ( A)] = ⎜ 0 31 0 ⎟ = 31 × ⎜ 0 1 0 ⎟ = det A ⋅ I 3 . ⎜ ⎟ ⎜ ⎟ ⎝ 0 0 31 ⎠ ⎝0 0 1⎠ ⎛ 31 0 0 ⎞ ⎛1 0 0⎞ Again [ adj ( A)] A = ⎜ 0 31 0 ⎟ = 31 × ⎜ 0 1 0 ⎟ = det A ⋅ I 3 . ⎜ ⎟ ⎜ ⎟ ⎝ 0 0 31 ⎠ ⎝0 0 1⎠ Hence, the above theorem is verified.

1.16

INVERSE OF A NONSINGULAR MATRIX

If for a nonsingular square matrix A of order n there exists a non-singular square matrix B of order n such that AB = BA = I n , then B is said to be the inverse of A. If inverse exists for A then we say the matrix A is inverible and inverse of A is denoted by A−1 . Now from the last theorem we have A.[ adj ( A)] = [ adj ( A)]. A = det A.I n . So, A ⋅

1 1 [ adj ( A)] = [ adj ( A)] A = I n det A det A

1 [ adj ( A)] is the inverse of the Therefore, from the definition we can say det A matrix A. Hence A−1 =

1 [ adj ( A)] and it satisfies A A−1 = A det A

1

A = In .

Note: So it is obvious from the above that inverse exists for a matrix A iff det A ≠ 0 , i.e., iff A is non-singular.

1.27

Matrix I

⎛ 1 0 2⎞ Let us consider A = ⎜ 2 3 0⎟ . ⎜⎝ 0 4 5⎟⎠

Example 16

⎛ 15 From the previous example, we have det A = 31 and adj( A) = ⎜ −10 ⎜ ⎝ 8

8 5 4

6⎞ 4 ⎟. ⎟ 3⎠

Since det A = 31 ≠ 0 , the given matrix is nonsingular and so A−1 exists. ⎛ 15 8 −6⎞ 1 1 Now A−1 = [ adj ( A)] = ⎜ −10 5 4⎟ . det A 31 ⎜ 8 −4 3⎟⎠ ⎝ Theorem 1.8: Inverse of a nonsingular square matrix is always unique. Proof: Beyond the scope of the syllabus.

1.16.1

Properties

( )

(1) If A−1 exists for A then A

−1

= A.

(2) If A and B are two invertible matrices then AB is also invertible and

( AB )−1 = B

1

A−1 .

( )

(3) If A is invertible then AT is also invertible and AT Let us verify the above properties: Example 17

Let us consider two matrices A =

⎛2 ⎝0

T

3⎞ ⎛ 2 0⎞ = ⎝ −3 1⎠ 1⎠ T

and adj ( B) =

⎛3 ⎛ 3 0⎞ = ⎝0 ⎝ −2 1⎠

2⎞ . 1⎠

1 1 ⎛ 2 0⎞ [ adj ( A)] = det A 2 ⎝ −3 1⎠ 1 1 ⎛ 3 −2⎞ = [ adj ( B)] = . 1⎠ det A 3 ⎝0

So A−1 = and B −1

Verification of Property (1) We have from above, A =

1 ⎛ 2 0⎞ ⎛ 1 0⎞ and A−1 = . ⎝ 3 2⎠ 2 ⎝ −3 1⎠

Now det( A−1 ) = 1 and adj ( A 1 ) =

T

( ).

= A

T

⎛ 1 2⎞ ⎛ 1 0⎞ and B = . ⎝ 0 3⎠ ⎝ 3 2⎠

Here, det A = 2 ≠ 0 and det B = 3 ≠ 0. Also adj ( A) =

−1

⎛ 1 3⎞ ⎛ 1 0⎞ = . ⎝ 0 2⎠ ⎝ 3 2⎠

1.28

Engineering Mathematics-I

( )

⎛ 1 0⎞ [ adj ( A −1 )] = = A. ⎝ 3 2⎠ det ( A −11 ) Hence, the property is verified. So A−

−1

1

=

Verification of Property (2) ⎛ 1 2⎞ ⎛ 1 0⎞ We have from above A = and B = . ⎝ 0 3⎠ ⎝ 3 2⎠ Also, A−1 =

1 ⎛ 2 0⎞ 1 ⎛ 3 −2⎞ and B −1 = . 1⎠ 2 ⎝ −3 1⎠ 3 ⎝0

⎛ 1 0⎞ ⎛ 1 2⎞ ⎛ 1 2 ⎞ = ⎝ 3 2⎠ ⎝ 0 3⎠ ⎝ 3 12⎠ Since det( AB) = 6 0, AB is invertible. Now AB =

T

⎛ 12 −3⎞ ⎛ 12 −2⎞ = . ⎝ −2 1⎠ ⎝ −3 1⎠ 1 1 ⎛ 12 −2⎞ So ( )−1 = [ adj ( AB)] = . 1⎠ det ( ) 6 ⎝ −3 1 ⎛ 3 −2⎞ 1 ⎛ 2 0⎞ 1 ⎛ 12 −2⎞ × Again B 1 A−1 = = 1⎠ 2 ⎝ −3 1⎠ 6 ⎝ −3 1⎠ 3 ⎝0 adj ( AB) =

Hence ( AB )−1 = B 1 A−1 which verifies property (2). Verification of Property (3) 1 ⎛ 2 0⎞ ⎛ 1 0⎞ ⎛ 2 0⎞ det A = 2 , adj( A) = , and A−1 = . ⎝ 3 2⎠ ⎝ −3 1⎠ 2 ⎝ −3 1⎠ 1 ⎛ 2 −3⎞ = . 1⎠ 2 ⎝0

We have A =

( )

T

So, A−

⎛ 1 3⎞ , det( AT ) = 2 ⎝ 0 2⎠

Now AT =

0, so AT is also invertible T

and adj ( AT ) = (adj ( A))T = So (

T −1

)

=

1 det (

T

)

[ adj ( AT )] =

( ) (A )

Hence, AT

1.17

−1

⎛2 ⎛ 2 0⎞ = ⎝0 ⎝ −3 1⎠

T

1 ⎛2 2 ⎝0

3⎞ 1⎠

3⎞ 1⎠

and so property (3) is verified.

ORTHOGONAL MATRIX

Any square matrix A = (aij )n × n is called orthogonal if it satisfies AAT = I n . Theorem 1.9: For an orthogonal matrix An×n , always AT A = I n . Proof: Beyond the scope of the book.

1.29

Matrix I

From the above, we can say for any orthogonal matrix of order n × n, AAT = AT A = I n . ⎛1 1 For example, let us consider the matrix A = ⎜ 2 3 ⎜2 ⎝ ⎛ 1 2 2⎞ 1 So AT = ⎜ −2 2 1 2⎟ . 3⎜ 2 2 1⎟⎠ ⎝ ⎛1 1 Now AAT = ⎜ 2 3 ⎜2 ⎝ =

2 2⎞ ⎛ 1 1 1 −2⎟ × ⎜ −2 2 1⎟⎠ 3 ⎜⎝ 2

2 2⎞ 1 −2⎟ 2 1⎟⎠

2 2⎞ 1 2⎟ 2 1⎟⎠

⎛ 9 0 0⎞ ⎛ 1 0 0⎞ 1 ⎜ 0 9 0⎟ = ⎜ 0 1 0⎟ = I 3. 9 ⎜ 0 0 9⎟ ⎜ 0 0 1⎟ ⎝ ⎠ ⎝ ⎠

So AAT = I 3 . Therefore, A is an orthogonal matrix. ⎛ 1 1 Again, A A = ⎜ −2 2 3⎜ 2 ⎝ T

=

2 2⎞ ⎛1 1 ⎟ 1 2 × ⎜2 2 1⎟⎠ 3 ⎜⎝ 2

2 2⎞ 1 −2⎟ 2 1⎟⎠

⎛ 9 0 0⎞ ⎛ 1 0 0⎞ 1 ⎜ 0 9 0⎟ = ⎜ 0 1 0⎟ = I 3. 9 ⎜ 0 0 9⎟ ⎜ 0 0 1⎟ ⎝ ⎠ ⎝ ⎠

Therefore, AAT = AT A = I 3 . Theorem 1.10: Any orthogonal matrix A is nonsingular and its determinant is given by det( A) = ±1. [WBUT 2003] Proof: For any orthogonal matrix A, we have AAT = I n . T

So, det (

) = det ( I n ) = 1.

⇒ d t ⋅ det AT = 1 ⇒ (d

)2 = 1

⇒ d A = ±1 ≠ 0. So the orthogonal matrix A is nonsingular and its determinant is given by det( A) = ±1. Hence, the theorem is proved. Example 18 ⎛1 1 A = ⎜2 3 ⎜2 ⎝

We have from the previous example that the matrix 2 2⎞ 1 −2⎟ is orthogonal. 2 1⎟⎠

Here, det A = 1.

1.30

Engineering Mathematics-I

1.17.1 Inverse of an Orthogonal Matrix From the above, it is clear that for any orthogonal matrix A , A−1 exists since it is nonsingular. Also, we have for any orthogonal matrix A of order n × n, AAT = AT A = I n . So, by the definition of inverse, we can say A 1 = AT . Example 19 and det A = 1.

⎛1 1 From the previous two examples, A = ⎜ 2 3 ⎜2 ⎝

So, A−1 exists and is given by A

2 2⎞ 1 −2⎟ is orthogonal 2 1⎟⎠

⎛ 1 2 2⎞ 1 AT = ⎜ −2 −1 2⎟ . 3 ⎜ 2 −2 1⎟ ⎝ ⎠

1

Theorem 1.11: Transpose and inverse of an orthogonal matrix is again orthogonal. Proof: Since for any orthogonal matrix A of order n × n, AAT = AT A = I n . Then AAT = AT A = I n ⇒ AT A = AAT = I n ⇒ AT ( AT )T = ( AT )T AT = I n . Hence by the definition, AT is orthogonal. Since for any orthogonal matrix A, A 1 AT holds, A−1 is also orthogonal. Note: The product of two orthogonal matrices is again orthogonal.

1.18

TRACE OF A MAT A RIX

Let A be any square matrix of order n × n , i.e., A = (aij )n n . Then trace of A, denoted by tr A, is the sum of the principal diagonal elements of A. Consider A = (aij )n

then tr A

n

⎛ a11 ⎜a = ⎜ 21 ... ⎜ ⎝ an

a11 + a22 +

Properties: (1) tr A tr B tr ( (2) tr AT tr A (3) tr ( ) = tr ( ).

a12 a22 ... an 2

... a1n ⎞ ... a2 n ⎟ , ... ... ⎟ ⎟ ... ann ⎠

+ ann . ).

Proof: Beyond the scope of this book. Example 20

Let A =

⎛ 1 0⎞ ⎛ 1 3⎞ and B = ⎝ 4 3⎠ ⎝ 0 4⎠

then tr A = 1 + 3 = 4 and tr B = 1 + 4 = 5

1.31

Matrix I

So, tr A tr B = 4 + 5 = 9. ⎛ 2 3⎞ Now, A B = ⎝ 4 7⎠ )= 2+7= 9

So, tr (

Hence tr A tr B tr ( ), which verifies the property (1). ⎛ 1 4⎞ T Again A = and tr AT = 1 + 3 = 4. ⎝ 0 3⎠ So, tr AT

tr A, which verifies the property (2). ⎛ 1 0⎞ ⎛ 1 3⎞ ⎛ 1 3⎞ Here AB = = ⎝ 4 3⎠ ⎝ 0 4⎠ ⎝ 4 24⎠

) = 1 + 24 = 25 ⎛ 1 3⎞ ⎛ 1 0⎞ ⎛ 13 9⎞ Again BA = = ⎝ 0 4⎠ ⎝ 4 3⎠ ⎝ 16 12⎠ So, tr (

So, tr (

) = 13 12 = 25.

Hence tr (

) = tr (

), which verifies the property (3).

WORKED-OUT EXAMPLES Example 1.1

Find if it is possible to form AB and BA , stating with reasons where

the operations do not hold when, A=

Sol.

⎛4 ⎝3

⎛ 2 3⎞ 2 −1⎞ and B = ⎜ −3 0⎟ 7 1⎠ ⎜⎝ −1 5⎟⎠

[WBUT-2004]

The order of matrices A and B are respectively (2 × 3) and (3 × 2). Since the number of columns of the matrix A and the number of rows of the matrix B are same, therefore AB is possible and is given by the following 2 2 matrix: AB =

=

⎛4 ⎝3

⎛ 2 3⎞ 2 −1⎞ ⎜ −3 0⎟ 7 1⎠ ⎜ ⎝ −1 5⎟⎠

⎛ 8 6 + 1 12 5⎞ ⎛ 3 7 ⎞ = . ⎝ 6 21 1 9 5 ⎠ ⎝ 26 14⎠

Again, since the number of columns of the matrix B and number of rows of the matrix A are same, therefore BA is also possible and is given by following 3 3 matrix ⎛ 2 3⎞ ⎛ 4 2 −1⎞ BA = ⎜ −3 0⎟ 7 1⎠ ⎜⎝ −1 5⎟⎠ ⎝ 3

1.32

Engineering Mathematics-I

⎛ 8 9 = ⎜ −12 ⎜⎝ −4 4 15

4 21 −6 2 − 35

⎛ 17 −17 = ⎜ −12 −6 ⎜⎝ 11 −37

2 + 3⎞ 3⎟ 1 5⎟⎠

1⎞ 3⎟ 6⎟⎠

Example 1.2 Prove that P t AP is a symmetric or a skew-symmetric matrix according to whether A is symmetric or skew-symmetric. [WBUT-2009] Sol.

Let A be a symmetric matrix, i.e., At Now, Bt

( P t AP )t = P t At ( P t )t

A and B = P t AP.

P t AP = B

Therefore, P t AP is a symmetric matrix. Again, let A be a skew-symmetric matrix, i.e , At Now, Bt

( P t AP )t = P t At ( P t )t = P t ( − A)P = − P t AP

A and B = P t AP. B

t

Therefore, P AP is a skew-symmetric matrix. Example 1.3

Find the matrices A and B such that

A 2B = Sol.

⎛ 1 3⎞ ⎛ 1 2⎞ and 2 A B = ⎝ −1 2⎠ ⎝ 4 −1⎠

Here, A 2B =

⎛ 1 3⎞ ⎝ −1 2⎠

...(1)

⎛1 ⎝4

...(2)

and 2

=

2⎞ 1⎠

Multiplying (2) by 2 and adding to (1), we have (

2 B) 2(2 A B) =

or , 5 A =

⎛ 3 7⎞ ⎝ 7 0⎠

⎛3 ⎜ or , A = ⎜ 5 7 ⎜ ⎝5

7⎞ 5⎟ ⎟ 0⎟ ⎠

⎛ 1 3⎞ ⎛1 +2 ⎝ −1 2⎠ ⎝4

2⎞ 1⎠

Matrix I

1.33

From (2) we have, ⎛3 ⎜ B = 2⎜ 5 7 ⎜ ⎝5 ⎛ 1 ⎜ =⎜ 5 −6 ⎜ ⎝ 5

7⎞ 5⎟ − ⎛ 1 ⎟ ⎝4 0⎟ ⎠ 4⎞ 5⎟ ⎟ 1⎟ ⎠

Example 1.4

Find the matrices A and B such that AT

d 3A − B

A 3B = 2II 3 Sol.

2⎞ 1⎠

Here, A 3 B = 2 I 3 and

...(1)

= 4AT

3

...(2)

Multiplying (2) by 3 and adding to (1), we have (

3 B) 3(3 A B) = 2

or, 10 = 2

3

12 AT

12 AT

3

or, 5 A I 3 = 6 AT

...(3)

Therefore, transposing both sides, 6

( )

T T

= (5 A − I 3 )

T

or, 6 A = 5 AT − I 3 or, 6 A I 3 = 5 AT

...(4)

Multiplying (3) by 5 and (1) by 6 and subtracting, we have 5 (5 A I 3 ) − 6 (6A 6 A I 3 ) = 30 or, 11 So, A

11

3

T

30 30A AT

=0 ...(5)

I3

From (1) and (5), we obtain −I 3 + 3 B = 2 I 3 or, B = I 3 . Hence, A

⎛ −1 I3 = ⎜ 0 ⎜⎝ 0

0 0⎞ 1 0⎟ and B 0 −1⎟⎠

⎛ 1 0 0⎞ I 3 = ⎜ 0 1 0⎟ ⎜⎝ 0 0 1⎟⎠

1.34

Engineering Mathematics-I

Example 1.5

Sol.

⎛ −1 If A = ⎜ 3 ⎜⎝ 5

A matrix A Here, ⎛ −1 2 A =⎜ 3 ⎜⎝ 5

1 −1⎞ 3 3⎟ then show that A is an idempotent matrix. 5 5⎟⎠ [WBUT-2003]

is said to be an idempotent matrix if A2 = A. 1 −1⎞ ⎛ −1 3 3⎟ ⎜ 3 5 5⎟⎠ ⎜⎝ 5

1 −1⎞ 3 3⎟ 5 5⎟⎠

1− 3 5 1 3−5 ⎞ ⎛ 1 3−5 ⎜ = −3 3 9 +15 15 3 9 − 15 −3 9 + 15 ⎟ ⎜⎝ 5 15 25 5 + 15 − 25 −5 5 15 25⎟⎠ ⎛ −1 =⎜ 3 ⎜⎝ 5

1 −1⎞ 3 3⎟ = A 5 5⎟⎠

Therefore, A is an idempotent matrix.

Example 1.6

Sol.

⎛ 1 Show that the matrix ⎜ −3 ⎜⎝ −4

⎛ 1 Let, A = ⎜ −3 ⎜⎝ −4 Now, ⎛ 1 1 A2 = ⎜ −3 3 ⎜⎝ −4 4

1 1⎞ 3 −3⎟ is a nilpotent matrix. 4 −4⎟⎠ [WBUT-2005]

1 1⎞ 3 −3⎟ 4 −4⎟⎠ 1⎞ ⎛ 1 −3⎟ ⎜ −3 −4⎟⎠ ⎜⎝ −4

1 1⎞ 3 −3⎟ 4 −4⎟⎠

1− 3 4 1 3−4 ⎞ ⎛ 1 3−4 = ⎜ −3 3 9 + 12 3 9 − 12 −3 9 + 12 ⎟ ⎜⎝ −4 4 12 16 4 + 12 − 16 −4 12 + 1 16⎟⎠ ⎛ 0 0 0⎞ = ⎜ 0 0 0⎟ = O ⎜⎝ 0 0 0⎟⎠ Therefore, A is a nilpotent matrix. Example 1.7 If A is an idempotent matrix then show that B idempotent. Hence, show that AB = BA = O. Sol.

Since A is an idempotent matrix, we have A2 = A. Here, we are to show B2 = B.

I − A is also

1.35

Matrix I

(I

2 Now B

A)

2

( I − A) ( I − A)

A ⋅ A = I − A A A2

=I

I ⋅ A A⋅ I

=

2 A + A (sinceA2 = A)

=I A=B So, B I − A is also idempotent. Also, AB = A.( I − A) A ⋅ I A ⋅ A A − A2

A A O

and BA = ( I

A − A O.

Example 1.8

A) ⋅ A

I ⋅ A A⋅ A

A A2

Show that

1 a 1 1 1 1 1+ b 1 1 ⎛ 1 1 1 1⎞ = abcd 1 + + + + ⎟ 1 1 1 c 1 ⎝ a b c d⎠ 1 1 1 1+ d Sol.

[WBUT-2002]

Here, 1 a 1 1 1 1 1+ b 1 1 1 1 1 c 1 1 1 1 1+ d 1+

= abcd

1+

= abcd

1 b 1 c 1 d

1 a

1 a 1+ 1 c 1 d

1 b

1 a 1 b 1+ 1 d

1 c

1 a 1 b 1 c 1+

[Dividing first, second, third, fourth rows by a, b, c, d respectively] 1 d

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + + + 1+ + + + 1+ + + + 1+ + + + a b c d a b c d a b c d a b c d 1 1 1 1 1+ b b b b 1 1 1 1 1+ c c c c 1 1 1 1 1+ d d d d

[ R1′ → R1 + R2 + R3 + R4 ]

1.36

Engineering Mathematics-I

1 1 1 1 1 1 1 1 1+ b b b b ⎛ 1 1 1 1⎞ = abcd 1 + + + + ⎟ 1 1 1 1 ⎝ a b c d⎠ 1+ c c c c 1 1 1 1 1+ d d d d 1 1 ⎛ 1 1 1 1⎞ b = abcd 1 + + + + ⎟ 1 ⎝ a b c d⎠ c 1 d 1 0 0 ⎛ 1 1 1 1⎞ = abcd 1 + + + + ⎟ 0 1 0 ⎝ a b c d⎠ 0 0 1

0 0 0 1 0 0 0 1 0 0 0 1

[C2′ → C2 − C1 , C3′ → C3 − C1 , C4′ → C4 − C1 ]

[Expanding the determinant about its first row ]

⎛ 1 1 1 1⎞ = abcd 1 + + + + ⎟ ⎝ a b c d⎠

Example 1.9

Sol.

0 −a Evaluate −b b −cc

a b 0 d d 0 e −f

c e by Laplace expansion method. f 0 [WBUT-2003, 2007]

Expanding the determinant by Laplace method in terms of minors of second order, we have, 0 −a −b b −cc = ( −1)1

a b 0 d d 0 e −f 2+1 2

c e f 0

0 a 0 −a 0 − f

f + ( 1)1 0

2+11 3

0 b −d − a d −e

+ ( −1)1+ 2+1+ 4

0 −d 0 a b −b + ( −1)1+ 2+ 2+3 − a e −e − f 0 d −

+ ( −1)1+ 2+ 2+ 4

a c −b 0 b + ( −1)1+ 2+3+ 4 0 e −c − f d

f 0 f 0

c −b − d e −c −e

1.37

Matrix I

= a 2 f 2 − abef + adcf + dacf − abef e = a2 f 2

22abef abef + 2 d f

(be cd )2

(be cd )2

= (af − be + cd )2 If Ai , Bi and Ci be the cofactors of ai , bi , ci ( = 1, 2, 3) in

Example 1.10 a1 Δ = a2 a3

b1 b2 b3

c1 B1 C1 c2 then show that B2 C2 c3 B3 C3

C1 C2 C3

A1 A2 A3

A1 A2 A3

B1 B2 = 2 Δ 2 B3 [WBUT-2003]

Sol.

By Jacobi’s theorem for 3rd order determinant, we have A1 B1 C1 A2 B2 C2 = Δ 2 A3 B3 C3 Now, B1 C1 B2 C2 B3 C3

C1 C2 C3

A1 A2 A3

A1 A2 A3

B1 B2 B3

B1 C1 + A1 = B2 C2 + A2 B3 C3 + A3

A1 + B1 C1 C1 + A1 A2 + B2 + C2 C2 + A2 A3 + B3 C3 C3 + A3

A1 + B1 A2 + B2 A3 + B3

B1 C1 + A1 = B2 C2 + A2 B3 C3 + A3

A1 A2 [C3′ A3

A1 A2 A3

B1 = B2 B3

C1 C2 C3

B1 = ( −1) B2 B3

A1 A2 [C2′ A3

C1 C3 − C1 ] + C2 C3

C1 C2 − C3 ] + C2 C3

A1 C1 A1 C1 A2 C2 + ( 1) A2 C2 A3 C3 A3 C3

A1 A2 A3

A1 B1 A2 B2 [C2′ A3 B3

B1 B2 [C3′ B3

C2 − C1 ]

C3 − C2 ]

B1 B2 B3

[Interchanging 2nd and 3rd column in the first and interchanging 1st and 2nd column in the second determinant] A1 = ( −1)2 A2 A3

B1 B2 B3

C1 A1 C2 + ( 1)2 A2 C3 A3

B1 B2 B3

C1 C2 C3

1.38

Engineering Mathematics-I

[Interchanging 1st and 2nd column in the first and interchanging 2nd and 3rd column in the second determinant ] A1 = 2 A2 A3

B1 B2 B3

C1 C2 = 2 Δ 2 . C3

Example 1.11 (i)

Define symmetric and skew-symmetric determinants.

(ii) Show that every skew-symmetric determinant of odd order is zero. [WBUT-2004] Sol.

(i) If A be a symmetric matrix then det A is called a symmetric determinant. If A be a skew-symmetric matrix then det A is called a skew-symmetric determinant. (ii) Let us consider a skew-symmetric matrix A = (aij )n × n of odd order n. Then the skew-symmetric determinant of odd order n is given by a11 a21 det A = ... an

a12 a22 ... an 2

− a11 −a = 12 ... − a1n

− a21 − a22 ... − a2 n

a11 a12 = (−1) ... a1n n

... a1n ... a2 n ... ... ... ann ... − an1 ... − an 2 s e aij sinc ... ... ... − ann

a21 a22 ... a2 n

a ji

... an1 ... an 2 [Taking i common ... ... ... ann

) from each row]

( )

i.e., det A = ( −1) n ⋅ det AT Since n is odd, we have det

det A

i.e., 2 det A = 0 i.e., det A = 0 Hence any skew-symmetric determinant of odd order is zero.

1.39

Matrix I

Example 1.12

Prove that

a +1 a a a a a+2 a a ⎛ a a a a⎞ = 24 ⎜ 1 + + + + ⎟ a a a+3 a ⎝ 1 2 3 4⎠ a a a a+4

[WBUT-2004]

a +1 a a a a a+2 a a Here, a a a+3 a a a a a+4

Sol.

1+

= (1.2.3.4)

1+

= 24

a 2 a 3 a 4

a 1

a 1 1+

a 2

a 3 a 4

a 1 a 2 1+

a 3

a 4

a 1 a 2 a 3 1+

[Dividing first, second, third, fourth rows by 1, 2, 3, 4 respectively] a 4

a a a a a a a a a a a a a a a a + + + 1+ + + + 1+ + + + 1+ + + + 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 a a a a 1+ 2 2 2 2 a a a a 1+ 3 3 3 3 a a a a 1+ 4 4 4 4

[ R1′ → R1 + R2 + R3 + R4 ]

⎛ a a a a⎞ = 24 1 + + + + ⎟ ⎝ 1 2 3 4⎠

1 1 1 1 a a a a 1+ 2 2 2 2 a a a a 1+ 3 3 3 3 a a a a 1+ 4 4 4 4

1.40

Engineering Mathematics-I

1 a ⎛ a a a a⎞ 2 = 24 1 + + + + ⎟ a ⎝ 1 2 3 4⎠ 3 a 4

0 0 0 1 0 0 0 1 0

[C2′ → C2 − C1 , C3′ → C3 − C1 , C4′ → C4 − C1 ]

0 0 1

1 0 0 ⎛ a a a a⎞ = 24 1 + + + + ⎟ 0 1 0 ⎝ 1 2 3 4⎠ 0 0 1

[Expanding the determinant aabout its first row]

⎛ a a a a⎞ = 24 1 + + + + ⎟ . ⎝ 1 2 3 4⎠ )2

(b Example 1.13

2

Prove that

b c2

a2 ( c a )2 c2

a2 b2 = 2abc ( 2 ( b)

b

)3

[WBUT-2004, 2008, 2009] )2

(b b2 c2

Sol.

)2

(b 2

=

b c2

)2

(b =

a2 ( c a )2 c2

2

b c2

a 2 (b ( c a )2 0 ( (c

b

2

)2 b2

)( b)( 0 )2

(b = (a b + c)

a2 b2 ( b )2

2

b c2

a 2 − (b 0 (a + b) b )2

C2′

) ( b ) b) 0 0 (a + b c) b

2bc 2c 2c 2b 2 = (a b + c) b (c b) 0 R1′ 2 c 0 (a + b c) 2

C2 − C1 , C3′ → C3 − C1

c2

) ( b )( b ) 0 b) (a + b c)(a b − c)

b

( (c

)2

R1 − (

2

3)

1.41

Matrix I

=

=

(

(

1 b )2 2bc b2 bc c2

= 2(

1 b c) b2 c2 )2

= 2(

b

−1 b)b 0 0 (a + b c)c 0 b2

b) c2 )

c2

(

C2′

C2 + C1 , C3′ → C3 C1

b )

b2 [Expanding the determinant (ac bc) about itts first row].

)2 (abc 2 + b2 c 2 + a 2 bc + ab2 c − b2 c 2 )

= 2 abc(

b

)2 (

= 2 abc (

b

)3

Example 1.14

(c

( b

(

bC2 , C3′ → cC3

1

0

2

= 2(

Sol.

cb 2cb 2 −2bc b)b C2′ 0 0 (a + b c)c

2bc b )2 2 b (c bc 2 c

b

)

1 a a 2 − bc Prove without expanding 1 b b2 − ac = 0 1 c c 2 − ab

[WBUT-2005].

1 a a 2 − bc Here 1 b b2 − ac 1 c c 2 − ab 1 a = 0 b a b2 0 c a c2 1 a = 0 b 0

a 2 − bc ac a 2 + bc R2′ ab a 2 + bc

a 2 − bc ((bb a )(a b + c) ((cc a )(a b + c)

1 a a 2 − bc = (b a )(c a ) 0 1 ( b ) 0 1 ( b )

R2 − R1 , R3′ → R3

R1

1.42

Engineering Mathematics-I

= (b a )(c a )

1 (a b c) [Expanding the determinant about firs r t column] 1 (a b c)

= (b a )(c a )(a b + c)

1 1 1 1

= (b a )(c a )(a b + c) 0 =0

Example 1.15

bc − a 2 Prove that ca − b2 ab − c 2

ca b2 abb c 2 bc a 2 b

ab − c 2 bc − a 2 = (a 3 ca − b2

b3 + c 3

3abc b )2

[WBUT-2006]

Sol.

a b c Let D = b c a c a b Then expanding we have D = (a 3 + b3

c3 − 3abc b )

Now adjoint of D is given by c a a b D= −

−

b c a b b c c a

bc − a 2 = ca − b2 ab − c 2

b a c b

b c c a

a c c b −

−

a c b a

ca b2 ab c 2 bc a 2

a b c a a b b c

ab − c 2 bc − a 2 ca − b2

Now from Jacobi’s theorem for 3rd order determinant, we have D bc − a 2 i.e., ca − b 2 ab − c 2 = (a3

ca b 2 ab b c2 bbc a 2

b3 + c 3 − 3

ab − c 2 2 bc − a 2 = ⎡⎣ − (a 3 b3 + c3 − 3abc) ⎤⎦ ca − b 2 )2 .

Hence the result is proved.

D2

1.43

Matrix I

b2 Example 1.16

c2

a2 c2 + a2 c2

b2 c2

Prove that

a2 b2 a

2

= 4a 2 b2 c 2 b

2

[WBUT-2006] b2 Sol.

Here

c2 2

a2 c + a2 c2

c2 = −c 2 a 2 b2

a2 c2 c2

b2

b2 c + a2 b2

a

b2 c2 + a2 b2

c2 c 2 C1′ a 2 + b2

2

a a2

a2 b2

2

b c2

b2 =

c2 2

2

[Transposing the determinant] b

0 = −2c 2 −2b2

b2 c2 + a2 b2

a

0 = 2 c2 b2

b2 c + a2 b2

c2 c2 a 2 + b2

0 = 2 c2 b2

b2 a2 0

2

⎡ c2 = 2 ⎢ −b 2 2 b ⎢⎣

c2 c2 2

2

C1′ b

C1 − C2

C1 − C3

2

c2 c 2 C2′ = C2 − C1 a 2 + b2 2 c2 2 c + c a 2 + b2 b2

a2 ⎤ ⎥ [Expanding the determinantt 0 ⎥⎦ a about its first row.].

= 2 ⎡⎣ −b2 { 2 (a 2 + b2 ) b2 2 } − c 2 a 2 b2 ⎤⎦ = 2{−c 2 a 2 b2 − c 2 a 2 b2 } = 4a 2 b2 c 2

1.44

Engineering Mathematics-I

Example 1.17

Prove that

( ( (

)2 )2 )2

( ( (

2

( ( (

)2 )2 )2

( ( (

2

)2 )2 = 2( x )2

( ( (

y y z z x

b b c c a).

[WBUT-2008] Sol.

) )2 )2

( = (

( (

)2 )2 (

) )2 )2

( ( (

)2 )2

( ( )2

2

) )2 )2

)2 ( ) − ( )2

( (

2a x 2a y − ( )2

+z (

( = ( x z )( y z ) (

)2 2

2 ( )( )2

) (

x+ y+

2a 2a

2 2

2

( (

( (

(

) 1

1 c) )

2

− C1 ,

)

(z

2

)2

0 2(a − c) 2 −(

)

a (

− R2

0

= ( x z )( y z )(

= ( x z )( y z )( x

) 2c ) z c)2

1 + (

2

− R3

( (

1 = ( x z )( y z )(

)2 − (

(

)2

− C1 y)

2(a b) −(

2( )

− R3 ,

2c ) 2c )2

(

( ) y+ 2 (z b

)2 )2

2c x z ) c)( y z ) ( )2

2

) 2a )2

= ( x z )( y z ) (

)2 ( )2 − ( ( )2

( (

2

2

[Expanding the determinant about its first row.]

) −(

)2

)2

1.45

Matrix I

= ( x z )( y z )( x

2(a b) 2( ) (a b)(2 z a − b) (a − c)(2 z − a c) )

1 1 (2 z a − b) (2 z

= 2(

)( y

)(

y)(

b)(

= 2(

)( y

)(

y)(

b)(a c)[(2 z a − c)) (2

= 2(

)( y

)(

y)(

b)(

)(b

)

= 2(

)( y

)(

y)(

b)(b

)(

)

Example 1.18

Sol.

y)

x+ p q Solve the equation q x+r r p

x+ p q q x+r r p

) b)]

r p =0 x+q

r p =0 x+q

x+ p+q+r or , q r

x+ p+q+r x+r p

1 or , ( x + p + q + r ) q r

1 x+r p

1 or , ( x + p + q + rr)) q r

0 x+r q p−r

or , ( x + p + q + r )

R1 + R2

0 p q = 0[C2′ x+q r

C2 − C1, C3′ → C3 C1 ]

p−q =0 x+q −r q2

r 2 + pq + qr + rp) = 0

Therefore, x

( p + q + r) r)

Example 1.19

R3 ]

1 p =0 x+q

x+r −q p−r

or , ( x + p + q + r )( x 2 − p2

x+ p+q+r p = 0 [ R1′ x+q

x = ± p2 + q 2 + r 2 − pq − qr − rp

Prove that

2 2 bc −ca ab c + a bc ca − ab = a 2 −bc ca ab c2

a2 a 2 + b2 b2

c2 b2 b2

⎡a 0 c⎤ =⎢a b 0⎥ ⎢ ⎥ c2 ⎣ 0 b c ⎦

2

1.46

Sol.

Engineering Mathematics-I

a 0 c Let us consider D = a b 0 0 b c 2

a 0 c a 0 c a 0 c Then D 2 = a b 0 = a b 0 a b 0 0 b c 0 b c 0 b c c2 + a2 or , D = a 2 c2 2

a2 a + b2 b2

c2 b2

2

b

2

[Multiplying y row−wise] c

...(1)

2

Again adjoint of D is given by D=

bc −ca ab bc ca − ab −bc ca ab

Since by Jacobi’s theorem for 3rd order determinant, D D2 we obtain bc −ca ab ⎡ a 0 c ⎤ bc ca − ab = ⎢ a b 0 ⎥ ⎢ ⎥ −bc ca ab ⎣ 0 b c ⎦

2

...(2)

Combining (1) and (2), we have 2 2 bc −ca ab c + a bc ca − ab = a 2 −bc ca ab c2

Example 1.20

Sol.

1 1 Here, 1 1

1 1 Prove that 1 1 bcd b + c d cda c + d a dab d a + b abc a b + c

a2 a 2 + b2 b2

c2 b2 b2

bcd b + c d cda c + d a dab d a + b abc a b + c a2 b2 c2 d2

⎡a 0 c⎤ =⎢a b 0⎥ ⎢ ⎥ c2 ⎣ 0 b c ⎦

a2 1 a a2 b2 1 b b2 = c2 1 c c2 d2 1 d d2

2

a3 b3 c3 d3

1.47

Matrix I

1 1 = 1 1

bcd cda dab abc

a + b c + d a a2 b + c d + a b b2 c d + a b − c c2 d a + b c − d d2

1 1 = 1 1

bcd cda dab abc

a+b c+d b+c d+a c d+a b d a+b c

1 1 = (a b + c d ) 1 1

bcd cda dab abc

a 2 1 bcd b2 1 cda − c 2 1 dab d 2 1 abc 1 1 1 1

a a2 b b2 c c2 d d2

a2 a abcd a 2 1 b bcda b2 b2 − 2 abcd c cdab c 2 c d dabc d 2 d2

a3 b3 c3 d3

[Multiplying first, second, third, fourth rows by a, b, c, d respectively] a abcd b =0− abcd c d

1 1 1 1

a2 b2 c2 d2

a3 b3 c3 d3

1 1 = 1 1

a a2 b b2 c c2 d d2

a3 b3 [Interchanging 1st and 2nd d columns] c3 d3

1 1 = 1 1

a a2 b b2 c c2 d d2

a3 b3 c3 d3

Example 1.21

Prove that

1 + a 2 − b2 2ab −2b 2 2 2ab 1− a + b 2a = (1 2 2 2b 2a 1− a − b Sol.

1 + a 2 − b2 2ab −2b 2 2 2ab 1− a + b 2a 2b 2a 1 − a 2 − b2

2

2 3

)

1.48

Engineering Mathematics-I

1+ =

0

2b 2a

1+

0 (1 +

−

) 1 )2 0

= (1

−

[

0 1

2b 2a 1

1 ) 0

= (1

Sol.

−

= (1

)

= (1

)3

Example 1.22

a b 0 0 a b b a

0 1

2

2b 2a − a2

− bR1 ]

[ 2

+

ab b2 a 2ab b a

2 a a b 0 Show that 0 a b = b b a 2 ab 2

a b 0 a b 0 0 a b 0 a b [By row wise multiplication b a b a

2 ab b2 a 2 ab b a 2 ab b a

Example 1.23

Sol.

1 n 1 (n 1 (n 1 (n

1 n 1 (n Show that 1 (n 1 (n n2

1 0 n 1 (n n 1 (n 1 (n + − n

) ) )

n2 ) ) )

n3 (n + 1)3 = 12 (n + 2)3 (n + 3 3

n3 (n + 1)3 (n + 2)3 (n + 3 3 0

0 (n (n (n +

) +

− −

n + 1)2 + 2)2 + 3)2

1.49

Matrix I

[C2′ → C2 − nC1 , C3′ → C3 − nC2 , C4′ → C4 − nC3 ] 1)2 2)2 3)2

1 ( n 1 = 2 2( 2) 2( 3 3( 3) 3(

1)2 2)2 [ R2′ → R2 − R1 ; R3′ → R3 − R1 ] 3)2

1 (n +1) 1) ( = 2 ⋅ 3 ⋅ 1 (n + 2) ( 1 (n + 3) (

1 (n +1) 1) ( 1)2 = 2 ⋅3 0 1 2 3 0 2 4 8 = 2 ⋅3 3 [4 Example 1.24

Sol.

8 − 2(2

3)] = 12

0 Without expanding, prove that − a −b b

0 Here, D = − a −b b

a b 0 c =0 c 0

[WBUT-2007]

a b 0 c c 0

0 a b = − −a 0 c [Transposing] −b −c 0 = −D or , 2 D = 0 or , D = 0. Example 1.25

Sol.

Ê 0 1 - 2ˆ If the matrix Á - 1 0 3 is singular then find the value of l . Á Ë l -3 0 ¯ [WBUT-2004]

0 1 Since the matrix is singular, we have, −1 0 l −3 or , 1( 3l ) 2(6) = 0 or , 3l = 12 or , l = 4.

2 3 =0 0

1.50

Engineering Mathematics-I

⎛1 Find the inverse of the matrix ⎜ 3 ⎜⎝ 0 ⎛ 1 0 −1⎞ Let A = ⎜ 3 4 5⎟ . ⎜⎝ 0 6 7⎟⎠

0 4 6

Example 1.26

Sol.

1 Then det A = 3 0

−1⎞ 5⎟ if it exists. 7⎟⎠

−1 5 = 20 ≠ 0. 7

0 4 6

Since det A ≠ 0, A−1 exists. Now the adjoint of the matrix A is given by ⎛ ⎜ ⎜ ⎜ ⎜ adj( A) = ⎜ ⎜− ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

T

4 −6 6

5 7

0 −6 6

1 7

0 4

1 5

⎛2 = ⎜6 ⎜ ⎝4 Hence A−1 =

A−1

21 7 8

−

−

3 0

5 7

1 0

1 7

1 3

1 5

T

18 ⎞ ⎛ 2 6 ⎟ 6 = ⎜ 21 −7 7 ⎟ ⎜ 4⎠ ⎝ −18 6

4⎞ 8⎟. ⎟ 4⎠

1 adj ( A) det A

⎛ ⎜ 1 ⎜ ⎜ 10 ⎛ 2 6 4⎞ ⎜ 1 ⎜ = ⋅ 21 −7 −8 ⎟ = ⎜ 21 20 ⎜ −18 6 4 ⎟ ⎜ ⎝ ⎠ ⎜ 20 ⎜ ⎜ −9 ⎜ ⎝ 10

Example 1.27

⎞ 3 4⎟ ⎟ 0 6⎟ ⎟ 1 0 ⎟⎟ − 0 6⎟ ⎟ ⎟ 1 0⎟ 3 4 ⎟⎠

⎛2 For the matrix A = ⎜ 1 ⎜⎝ 0

A3 6 A2 12 A 10 I = O. Hence, find A−1 .

3 10 −7 20 3 10

⎞ 1 ⎟ ⎟ 5 ⎟ ⎟ −2 ⎟ ⎟ 5 ⎟ ⎟ 1 ⎟ ⎟ 5 ⎠

3 4⎞ 0 1⎟ , prove that 1 4⎟⎠

1.51

Matrix I

Sol.

⎛2 Here A = ⎜ 1 ⎜⎝ 0

3 4⎞ 0 1⎟ . 1 4⎟⎠

Therefore, ⎛2 A2 = ⎜ 1 ⎜⎝ 0

3 4⎞ ⎛ 2 0 1⎟ ⎜ 1 1 4⎟⎠ ⎜⎝ 0

3 4⎞ ⎛ 1 0 1⎟ = ⎜ 2 1 4⎟⎠ ⎜⎝ −1 1

10 21⎞ 4 8⎟ 4 15⎟⎠

and ⎛ 1 A2 ⋅ A = ⎜ 2 ⎜ 1 ⎝ −1

A3

10 21 ⎞ ⎛ 2 4 8⎟⎜ 1 ⎟⎜ 4 15 ⎠ ⎝ 0

3 4 ⎞ ⎛ −88 0 1⎟ = ⎜ 0 ⎟ ⎜ 1 4 ⎠ ⎝ −66

24 78 ⎞ 14 36 ⎟ ⎟ 12 52 ⎠

Now A3

8 ⎛ −8 6 A2 12 A 10 I = ⎜ 0 ⎜⎝ −6 6

24 78⎞ ⎛ 1 14 36⎟ − 6 ⎜ 2 ⎜⎝ −11 12 52⎟⎠

10 21⎞ 4 8⎟ 4 15⎟⎠

⎛ 2 −3 4⎞ ⎛ 1 0 0⎞ ⎜ ⎟ +12 1 0 1 − 10 ⎜ 0 1 0⎟ ⎜⎝ 0 −1 4⎟⎠ ⎜⎝ 0 0 1⎟⎠ ⎛ 0 0 0⎞ = ⎜ 0 0 0⎟ = O ⎜⎝ 0 0 0⎟⎠ Hence, we have A3

6 A2 12 A 10 I = O.

Now we are to find A−1 . A3

6 A2 12 A 10 I = O

or , A( A2 − 6A 6 A 12 I ) = 10 I ⎡1 2 ⎤ ( A 6 A 12 I ) = I 10 ⎣ ⎦ Hence by the definition of inverse, we have or , A.

⎡1 2 ⎤ ( A − 6 A 12 I )⎥ 10 ⎣ ⎦ ⎡⎛ 1 −10 21⎞ ⎛ 2 −3 4⎞ ⎛ 1 0 0⎞ ⎤ 1 A−1 = ⎢⎜ 2 −4 8⎟ − 6 ⎜ 1 0 1⎟ + 12 ⎜ 0 1 0⎟ ⎥ 10 ⎢⎜ −1 −4 15⎟ ⎜⎝ 0 −1 4⎟⎠ ⎜⎝ 0 0 1⎟⎠ ⎥ ⎠ ⎣⎝ ⎦ ⎛ 1 8 −3 ⎞ 1 i.e., A−1 = ⎜ −4 8 2 ⎟ ⎟ 10 ⎜ −1 2 3⎠ ⎝ A

1

1.52

Engineering Mathematics-I

⎛0 Determine the values of a, b, c for which the matrix ⎜ a ⎜⎝ a

Example 1.28 is orthogonal. Sol.

2b c⎞ b − c⎟ b c⎟⎠

Let ⎛0 A = ⎜a ⎜⎝ a

2b c⎞ b − c⎟ b c⎟⎠

If A is an orthogonal matrix then AAT = I which implies ⎛0 ⎜a ⎜⎝ a ⎛ 4b 2 c 2 or , ⎜⎜ 2b2 c 2 ⎜⎝ −2b2 c 2

2b c⎞ ⎛ 0 b − c⎟ ⎜ 2b b c⎟⎠ ⎜⎝ c

2b 2 c 2 a 2 b2 + c 2 a 2 b2 − c 2 2

a a ⎞ ⎛ 1 0 0⎞ b − b⎟ = ⎜ 0 1 0⎟ c c⎟⎠ ⎜⎝ 0 0 1⎟⎠

−2b2 c 2 ⎞ ⎛ 1 0 0⎞ a 2 − b2 c 2 ⎟⎟ = ⎜ 0 1 0⎟ ⎜ ⎟ a b + c ⎟⎠ ⎝ 0 0 1⎠ 2

2

2

Equating the corresponding entries of 1st row, we have 4b 2

c2 = 1

...(1)

2b 2

c2 = 0

...(2)

2

...(3)

a

2

2

b +c =1

Adding (1) and (2), we get 6

2

Putting b = ± c2

1 1 ⇒b=± 6 6

= 1 ⇒ b2 =

b2 =

1 6

in (2), we get

1 1 ⇒c=± 3 3

Putting the value of b and c in (3), we obtain a2

b2 − c 2 = 1 −

1 1 1 1 − = ⇒a=± 6 3 2 2

Therefore, a

1 2

;b=±

1 6

;c=±

1 3

1.53

Matrix I

Example 1.29 find its inverse. Sol.

⎛ cos q Show that the matrix ⎜ 0 ⎜⎝ − sin q

⎛ cos q Let A = ⎜ 0 ⎜⎝ − sin q

sin i q⎞ 1 0 ⎟ is orthogonal and hence cos q ⎟⎠

0 sin i q⎞ 1 0 ⎟ 0 cos q ⎟⎠

Now, ⎛ cos q AA = ⎜ 0 ⎜⎝ − sin q

0 sin i q ⎞ ⎛ cos q 1 0 ⎟⎜ 0 0 cos q ⎟⎠ ⎜⎝ − sin q

T

⎛ cos q =⎜ 0 ⎜⎝ − sin q

0 sin i q ⎞ ⎛ cos q 1 0 ⎟⎜ 0 0 cos q ⎟⎠ ⎜⎝ sin q

⎛ cos 2q sin 2q =⎜ 0 ⎜ 0 ⎝

0 sin i q⎞ 1 0 ⎟ 0 cos q ⎟⎠

T

sin i q⎞ 0 ⎟ cos q ⎟⎠

0 1 0

⎞ 0 0 ⎟ 1 0 2 2 ⎟ 0 cos q sin q ⎠

⎛ 1 0 0⎞ = ⎜ 0 1 0⎟ = I ⎜⎝ 0 0 1⎟⎠ Since AAT = I , A is an orthogonal matrix. Again for any othogonal matrix A, we know A A

−1

T

⎛ cos q =⎜ 0 ⎜⎝ − sin q

Example 1.30

1

= AT . Hence,

0 − sin i q⎞ 1 0 ⎟ 0 cos q ⎟⎠

0 sin i q⎞ ⎛ cos q ⎟ 1 0 =⎜ 0 ⎟ ⎜⎝ sin q 0 cos q ⎠

If A be a skew-symmetric and (

) be a nonsingular matrix

−1

then show that B = ( I − A)( I + A) is orthogonal. [WBUT-2009] Sol.

A square matrix is orthogonal if A AT = I n Now, )−1}{( I

A) 1}T

B BT = {(

)(

= (I

A)( I

A) 1{(

)−1}T (I (I

= (I

A)( I

A) 1{(

)T } 1 (I (I T

= (I

A)( I

A) 1 ( I T

AT ) 1 ( I T

= (I

A)( I

A) 1 ( I

A)( I

AT ) 1 ( I

A)T , since ( At ), since (

)T = BT AT T

AT ) AT ), since I T = I

)

1

= ( A 1 )T

1.54

Engineering Mathematics-I

= (I

A)( I

A) 1 ( I

A) 1 ( I

A), since AT = A

= (I

A){(

)(

)} 1 ((II

A), since (

)−1 = B

11.

A −1

Again, (

) = I + A A A2 = ( I A) ( I + A)) = (

)(

)(

)

Hence, from above B. BT = ( I

A){( I

A)( I 1

= ( I A)( I =I I=I

A) ( I

A)} A )}−1 (I (I 1

A) ( I

A) A)

Therefore, B = ( I − A)( I + A)−1 is orthogonal.

EXERCISES

Short and Long Answer Type Questions 1. Find the matrices A ⎛ 1 2 A 2 B = ⎜ 6 −3 ⎜⎝ −5 3

and B if 0⎞ ⎛ 2 −1 5⎞ 3⎟ and 2 A B = ⎜ 2 −1 6⎟ ⎜⎝ 0 1⎟⎠ 1 2⎟⎠ ⎡ ⎛ 1 ⎢ Ans : A = ⎜ 2 ⎢ ⎜ ⎝ −1 ⎣

2. For the matrices A = a) (

)T = AT

b) (

)T = BT AT

1 −1 ⎞ ⎤ 1 0 ⎟⎥ ⎟⎥ 1 0 ⎠⎦

0 2⎞ ⎛ 0 1 3 ⎟ and B = ⎜ 2 ⎟ ⎜ 1 1⎠ ⎝ −2

⎛ 3 2⎞ ⎛ 0 1⎞ and B = , verify the following: ⎝ −1 2⎠ ⎝ −1 2⎠

BT

3. Express the following matrices as the sum of symmetric and skew-symmetric matrices. a)

⎛ 2 4⎞ ⎝ 3 2⎠ ⎡ ⎛ ⎢ ⎜2 ⎢ Ans : Symmertric matrix: ⎜ ⎢ ⎜⎜ 7 ⎢⎣ ⎝2

7⎞ 2 ⎟⎟ , Skew-symmertric matrix: 2 ⎟⎟ ⎠

⎛ ⎜ 0 ⎜ ⎜⎜ −1 ⎝ 2

1 ⎞⎤ 2 ⎟⎟ ⎥⎥ 0 ⎟⎟ ⎥ ⎠ ⎥⎦

1.55

Matrix I

⎛ 1 2 3⎞ b) ⎜ 4 5 6⎟ ⎜⎝ 7 8 9⎟⎠ ⎡ ⎛1 3 5⎞ ⎢ Ans : Symmertric matrix: ⎜ 3 5 7 ⎟ , Skew-symmertric matrix: ⎢ ⎜ ⎟ ⎝5 7 9⎠ ⎣ ⎛ 3 c) ⎜ −1 ⎜⎝ 5

⎛0 ⎜1 ⎜ ⎝2

1 −2 ⎞ ⎤ 0 −1 ⎟ ⎥ ⎟⎥ 1 0 ⎠⎦

1 5⎞ 2 4⎟ 4 1⎟⎠ 1 5⎞ ⎛ 0 0 0 ⎞⎤ ⎟ 2 4 , Skew-symmertric matrix: ⎜ 0 0 0 ⎟ ⎥ ⎜ ⎟⎥ ⎟ 4 1⎠ ⎝ 0 0 0 ⎠⎦

⎡ ⎛ 3 ⎢ Ans : Symmertric matrix: ⎜ −1 ⎢ ⎜ ⎝ 5 ⎣

4. If A is a skew-symmetric matrix then show that A2 is symmetric. Also, verify this with the matrix 1 −2⎞ ⎛ 0 A = ⎜ −1 0 3⎟ ⎜⎝ 2 3 0⎟⎠ 5. If A( ) =

q ⎛ cosq q ⎝ sinq

i q⎞ , prove that q ⎠

A(q ) A(j ) = A(j ) A(q ) = A(

j)

[WBUT-2006]

⎛ 0 6. If A = ⎜ 1 ⎜⎝ −1

4 3⎞ 3 −3⎟ then show that A2 = I , i.e., A is involutory. 4 4⎟⎠

⎛ −1 7. If A = ⎜ 3 ⎜⎝ 5

1 −1⎞ 3 3⎟ then show that A2 = A, i.e., A is idempotent. 5 5⎟⎠

⎛1 8. If A = ⎜ 3 ⎜⎝ 4 9. Show that

2 3⎞ 2 1⎟ then prove that A2 2 1⎟⎠

⎛a (u v w ) × ⎜ h ⎜⎝ g

h b f

23 3 A − 40 I

O.

g⎞ ⎛ u ⎞ f ⎟ × ⎜ v ⎟ = au2 + bv 2 + cw2 + 2huv 2huv + 2 fvw 2 gwu c ⎟⎠ ⎜⎝ w⎟⎠

10. Prove without expanding the following determinants: a b b c c a a) b c c a a b = 0. c a a b b c

1.56

Engineering Mathematics-I

bc a 2 a 2 bc ab ca b) b2 ca b2 = ab ca bc . c 2 c 2 ab ca bc ab 1 a a 2 − bc c) 1 b b2 − ac = 0. 1 c c 2 − ab

d)

1 p 1 1 ⎛ 1 1 1⎞ 1 1+ q 1 = pqr 1 + + + ⎟ . ⎝ p q r⎠ 1 1 1+ r

a b c e) a 2 b2 c 2 = (b cc)( )(c a )(a b)( b)(bc + ca + ab). bc ca ab

11. Prove that (l + p + q + r ) is a factor of

l+p q q l+r r p

r p . l +q

12. Solve the following for x : x +1 2 3 (i) 1 x +1 3 =0 3 −6 x + 1 x3 a3 (ii) b3 a 3 c3 a3

13. Prove that

x2 b2 c2

⎡ Ans : x ⎣

x b = 0. c

a x c b c b−x a = 0 if ( b a c x

⎡ ⎢ Ans : x ⎣

b

) = 0.

−1 cos C cos B 14. If A B + C = p , then prove that cos C −1 1 cos A = 0 cos B cos A −1

15. Show that (

1, 1 ± i 7 ⎤ ⎦

3) is a factor of the determinant

x+3 4 5 5 x+3 5 5 −4 x + 3

b, c,

a3 ⎤ ⎥ bc ⎦

1.57

Matrix I

⎛ ⎜ ⎜ 16. Show that the matrix ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 17. For A = ⎜ −1 ⎜⎝ −2 2 18. Verify (i) (

1

1

3 1

6 −2

3 1

6 1

3

6

−1 ⎞ 2⎟ ⎟ 0 ⎟ is an orthogonal matrix. ⎟ 1 ⎟ ⎟ 2⎠

1 2⎞ 0 3⎟ , show that ( 3 0⎟⎠ ) −11 = B

1

A−1 (ii) (

)−1 is an orthogonal matrix.

)( T

1

)

( )

= A

1 T

for the matrices

⎛ 2 1 2⎞ ⎛ 4 1 2⎞ ⎜ ⎟ A = −1 0 4 and B = ⎜ −1 0 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −2 0 0 ⎠ ⎝ −2 4 0 ⎠ ⎛ 2 1 2⎞ 19. If A = ⎜ 3 0 4⎟ , find the matrix B3 ⎜⎝ −2 0 8⎟⎠

3

for which AB = I 3 .

20. Examine whether the following matrices A and B are conformable for addition and multiplication. If so, find A B, AB, BA. (i) A =

⎛ 3 1⎞ ⎛ 1 2 2⎞ , B = ⎜ 0 0⎟ ⎝ 0 6 0⎠ ⎜⎝ 2 5⎟⎠ ⎡ ⎛ 3 12 6 ⎞ ⎤ ⎢ Ans : AB = ⎛ 7 11 ⎞ , BA = ⎜ 0 0 0 ⎟ ⎥ ⎜0 0⎟ ⎢ ⎜ ⎟⎥ ⎝ ⎠ ⎝ 2 34 4 ⎠ ⎦ ⎣

⎛ a 0 b⎞ ⎛ − a b⎞ (ii) A = ⎜ b c 0⎟ , B = ⎜ 0 0⎟ ⎜⎝ 0 c a⎟⎠ ⎜⎝ −c a⎟⎠ ⎡ ⎛ −aa 2 bc ⎢ ⎜ ⎢ Ans : AB = ⎜ −ab ⎢ ⎜ −ac ⎝ ⎣ 21. If A I

⎛ 1 3 4⎞ ⎜ −1 1 3⎟ then evaluate ( ⎜⎝ −2 −3 1⎟⎠

matrix of order 3 .

)(

ab ⎞ ⎤ ⎟⎥ b2 ⎟ ⎥ a 2 ⎟⎠ ⎥ ⎦

) where I is the identity ⎡ 9 ⎞⎤ ⎛ −12 −12 ⎢ Ans : ⎜ −6 −13 −4 ⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎝ 3 −6 −18 ⎠ ⎦ ⎣

1.58

Engineering Mathematics-I

22. Find the matrices A and B if 2

3B =

⎛ 3 1⎞ ⎛ 8 3⎞ , A BT = ⎝ 3 3⎠ ⎝ 7 6⎠ ⎡ ⎛1 0⎞ ⎛ 2 1 ⎞⎤ ⎢ Ans : A = ⎜ 2 3 ⎟ and B = ⎜ 1 0 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣

23. If A B = 2 BT and 3

2 B = I 3 , find the matrices A and B . ⎡ ⎛1 ⎢ ⎜5 ⎢ ⎜ ⎢ Ans : A = ⎜ 0 ⎢ ⎜ ⎢ ⎜ ⎢ ⎜0 ⎝ ⎣

0 1 5 0

⎞ ⎛1 0⎟ ⎜5 ⎟ ⎜ 0⎟ ,B =⎜0 ⎟ ⎜ ⎜ 1⎟ ⎟ ⎜0 5⎠ ⎝

0 1 5 0

⎞⎤ 0 ⎟⎥ ⎟⎥ 0 ⎟⎥ ⎟⎥ 1 ⎟⎥ ⎟⎥ 5 ⎠⎦

24. Find the matrices A and B such that ⎛ 1 1 1⎞ T 3 = 2 3 and 2 A + B = ⎜ 0 1 0⎟ ⎜⎝ 1 0 1⎟⎠ ⎡ ⎛3 ⎢ ⎜5 ⎢ ⎜ ⎢ Ans : A = ⎜ 3 ⎢ ⎜5 ⎢ ⎜1 ⎢ ⎜ ⎝5 ⎣ ⎛ 0 1 2⎞ 25. If A = ⎜ 2 0 1⎟ show that A3 ⎜⎝ 1 2 0⎟⎠

1⎞ ⎛ −1 ⎜ 5 5 ⎟⎟ ⎜ −6 0 ⎟ and B = ⎜ ⎜ 5 ⎟ ⎜ 3 3⎟ ⎜ ⎟ 5⎠ ⎝ 5

−2 5 3 5 0

6 A 9I3

9 5 −1 5 0

⎞⎤ ⎟⎥ ⎟⎥ 0 ⎟⎥ ⎟⎥ −1 ⎟ ⎥ ⎟⎥ 5 ⎠⎦ 3 5

O. Hence obtain a matrix B such

that BA = I 3 26. Find all real matrices A =

⎛ a b⎞ such that A2 ⎝ c d⎠

O

⎡ ⎛a ⎢ Ans : ⎜ c ⎝ ⎣

b⎞ , where a 2 a ⎟⎠

⎤ bc = 0.⎥ ⎦

27. Express the following matrices as the sum of symmetric and skew-symmetric matrices. ⎛ 1 a 1⎞ i) ⎜ b 1 1⎟ ⎜⎝ 1 1 c⎟⎠

⎛a ii) ⎜ b ⎜⎝ g

a b g

b⎞ g⎟ a ⎟⎠

1.59

Matrix I

⎛ 2 28. Show that ⎜ −1 ⎜⎝ 1

3 5⎞ 4 5⎟ is idempotent. 3 −4⎟⎠

⎛ 1 29. Show that ⎜ 5 ⎜⎝ −2 2

1 3⎞ 2 6⎟ is nilpotent of index 3. 1 −3⎟⎠

30. If A = An =

⎛3 ⎝1

4⎞ , prove by the method of induction that 1⎠

⎛ 1 2 n −4 n ⎞ ⎝ n 1 2 n⎠

where n is any positive integer. 31. Prove without expanding the following: a3 2 a) a a 1 ( b)

3a 2 3a a + 2a 2a + 1 2a 1 a + 2 3 3 2

b )2 ca bc

p3 ( p 1)3 c) ( p 2)3 ( p 3)3 b2 d)

ca (b

)

bc ab

2

ab p2 ( p + 1)3 ( p + 1)3 ( p + 1)3

1 1 = (a − 1)6 1 1

(

= 2 abc(

b

)3

)2

p ( p 1)3 ( p 1)3 ( p + 1)3

1 1 = 12 1 1

c2 + 1 c2 b2 + 1 b c 2 2 2 2 c +1 a + c +1 a +1 c + a = (ab + bc + ca )2 2 2 2 2 b a +1 a b +1 a b a+b 3 b c c+a

a b c y b q x y z e) x y z = x a p = p q r p q r z c r a b c y+z x−z f) y − z z + x z y z x g)

1 y+z y2 + z2

x y y x = 8 xyz x+y

1 z+ x z2 + x2

1 x + y = ( y z )( z x )( x x 2 + y2

y)

1.60

Engineering Mathematics-I

a a2 h) b b2 c c2 i)

a3 + 1 b3 + 1 = (a b)( b)(b c)(c − aa)( )(abc + 1) 3 c +1

a b 0 b c b+c 0 = (a b + c)(ab bc ca ) 0 a c+a

x2 j) y 2 z2

x 2 − (y ( y z )2 y 2 − (z ( z x )2 2 z − (x ( x y )2 1 z+ x z2 + x2

yz zx = ( x xy

y)( y z )( z − x )( x + y + z )( x 2 + y 2 + z 2 )

k)

1 y+z y2 + z2

1 x + y = (x x 2 + y2

l)

a b ax + by b c bx + cy = (b2 ax + by bx + cy 0

y)( y z )( z x )

ac ax 2

bxy + cyy 2 )

32. Using product of determinants, show that 1 (a b ) a) cos(a b ) 1 cos((g ) cos( b g ) a2 b) b2 c2

bc b c 2 2 ab b2 ca a 2 2bc c 2 abb b2 2ca a 2

( c) ( y (

)2 )2 )2

( b )2 ( y b )2 ( b )2

( (y (

(g a ) (b g ) = 0 1 ca ab = (a 3 + b3 + c3 − 3abc b )2 bc )2 )2 = 2( x − y)( y − z )( z − x )(a − b)(b − c)(c − a ) )2 [WBUT-2008]

33. Using Laplace method of expansion, prove the following: x y a) u −vv

y −u u x v v x u y

a −a b) −a a −a a

b c b c b c b −c

v u = ( x 2 + v2 − y2 y x d d = 8abcd d d

u 2 )2

1.61

Matrix I

1 2 c) 3 4 a b d) c d

2 3 4 5

3 4 5 6

4 5 =0 6 7

b −a b a b −a = 4(a 2 d c d c d c

b2 )(c (c 2

d2 )

−1 0 0 a 0 −1 0 b = 1 − ax − by − cz e) 0 0 −1 c x y z −1 1 34. Prove that the determinant x 2 x

x 1 x2

x2 x is a perfect square. 1

35. Verify a) adj ( AT ) = (adjA j )T and b) A ⋅ adj ( A) = adj ( A) A = det A I ⎛ 1 0 1⎞ for the matrix A = ⎜ 2 4 6⎟ . ⎜⎝ 1 2 1⎟⎠ 36. Find the adjoint of the following matrices: ⎛1 a) ⎜ 2 ⎜⎝ 3

1 1⎞ 1 3⎟ 2 −1⎟⎠

⎡ 3 4 ⎞⎤ ⎛ −5 ⎢ Ans : ⎜ 11 −4 −1 ⎟ ⎥ ⎢ ⎜ ⎟⎥ 1 −3 ⎠ ⎦ ⎝ 7 ⎣

⎛2 b) ⎜ 0 ⎜⎝ 2

1 3⎞ 2 0⎟ 1 1⎟⎠

⎡ ⎛ 2 ⎢ Ans : ⎜ 0 ⎢ ⎜ ⎝ −44 ⎣

4 4 4

6 ⎞⎤ 0 ⎟⎥ ⎟⎥ 4 ⎠⎦

1 1 1

1⎞⎤ 1⎟⎥ ⎟⎥ 3 ⎠⎦

⎛ 2 2 0⎞ 37. Find the matrix A such that det A = 2 and adj A = ⎜ 2 5 1⎟ . ⎜⎝ 0 1 1⎟⎠ ⎡ ⎛ 2 ⎢ Ans : A = ⎜ −11 ⎢ ⎜ ⎝ 1 ⎣

1.62

Engineering Mathematics-I

38. Find the inverse of the following matrices: ⎡ −5 1 ⎞ ⎤ ⎛ ⎢ ⎜3 2 2 ⎟⎟ ⎥⎥ ⎢ ⎜ ⎢ Ans : ⎜ −3 4 −1 ⎟ ⎥ −3 1 ⎟ ⎥ ⎢ ⎜ ⎜1 ⎟ ⎢ 2 2 ⎠ ⎥⎦ ⎝ ⎣ ⎡ ⎛ 7 2 −3 ⎞ ⎤ ⎢ Ans : 1 ⎜ −2 1 4 ⎟ ⎥ ⎢ 11 ⎜ 1 5 −2 ⎟ ⎥ ⎝ ⎠⎦ ⎣

⎛ 1 1 1⎞ a) ⎜ 1 2 3⎟ ⎜⎝ 1 4 9⎟⎠ ⎛ 2 1 −1⎞ b) ⎜ 0 1 2⎟ ⎜⎝ 1 3 −1⎟⎠

39. A and B are real orthogonal matrices of the same order and det Show that A B is a singular matrix. ⎛0 40. Determine the value of a, b, c so that ⎜ a ⎜⎝ a

det B = 0.

2b c⎞ b −c⎟ is orthogonal. b c⎟⎠

1 1 1 ⎤ ⎡ , b=± , c=± ⎥ ⎢ Ans : a = ± 6 2 3⎦ ⎣

Multiple-Choice Questions ⎛ 1 2 3⎞ 1. The matrix ⎜ 2 4 6⎟ is a ⎜⎝ 3 6 5⎟⎠ a) symmetric matrix c) diagonal matrix

b) skew-symmetric matrix d) none of these

2. If A is a non-null square matrix then A AT is a a) symmetric matrix b) skew-symmetric matrix c) null matrix d) none of these 3. If A is a non-null square matrix then A AT is a b) skew-symmetric matrix a) symmetric matrix c) null matrix d) none of these 4. If A = a)

⎛ 1 1⎞ then ( ⎝ 1 1⎠

⎛0 ⎝1

2 T

1⎞ 1⎠

c) 2A

) = b) I 2 d) none of these

T

5. (2 A 3 ) is equal to 3BT 3

a) 2 c) 4

T

9B 9 T

b) 2

T

3BT 3

d) none of these

1.63

Matrix I

)T is equal to

6. (

a) AT BT c) BT AT 7. If A =

b) AT BT d) none of these

⎛ −1 0⎞ then A At = ⎝ 0 1⎠

a) I 2 c)

b) A

⎛ 0 1⎞ ⎝ 1 0⎠

8. If A =

⎛2 ⎝1

a) O 9. If

d) none of these 1⎞ then A2 3⎠

7I =

b) 2A

c) 3A

d) 5A

⎛ 2 k ⎞ ⎛ 0 1⎞ ⎛ 1 2⎞ = then the value of k is ⎝ 1 3⎠ ⎝ −1 0⎠ ⎝ −3 1⎠

a) −5

b) 0

d) −1

c) 5

10. If A is a symmetric as well as skew-symmetric then A is a/an a) diagonal matrix b) null matrix d) none of these c) Identity matrix 11. If A is an idempotent matrix then I A is a/an a) nilpotent matrix b) idempotent matrix c) involutory matrix d) none of these 12. If A and B are two square matrices of same order such that 2 ( )2 = 2 2 AB then b) A2 B d) none of these

a) A BT c) AB = BA

1 1 1 0 is 3 2

x 13. The cofactor of x in the determinant 2 1 a) −2

b) 4

14. The adjoint of the determinant a)

1 2 6 3

b)

6 3 1 2

c) 2 2 1 is 3 6 c)

−6 3 1 −2

d) 0

d)

6 −1

1 2 3 15. The value of the determinant 2 3 4 is 3 4 5 a) 1

b) −1

c) 2

d) 0

3 2

1.64

Engineering Mathematics-I

1 w w2 16. If w is the cube root of unity then the value of the determinant w w 2 1 is w2 w 1 a) w 2

b) w

c) 1+ w

d) 0

17. The value of a skew-symmetric determinant of odd order is always a) 0 b) 1 c) −1 d) none of these 18. The roots of the equation a) 1, 2, 3

x +1 0 0 x−2 0 0

b) −1, 2, 3

0 0 = 0 are x −3

c) 1, − 2, 3

d) −1, − 2, 3

a b ax + b 19. b c bx + c = 0 if ax + b bbx + c 0 a) a, b, c are in AP c)

b) a, b, c are in GP

1 1 1 , , are in AP a b c

d) none of these 0 a b 3 x 2 = 0 then b 3x 0 a = 1 −a a

20. If a b are the roots of the equation x 2 a) 6

b)

3 2

c) −6

d) 3

[WBUT-2007]

1 2 3 21. If 4 6 = 0 then the value of a is 7 8 9 a) 5 22. If

b) either −2 or 1

c) 1

d) not −2

b c c b c c+a a = kabc, then k = b a a b

a) 3

b) 1

c) 4

d) 2

c) 405

d) −1

100 101 102 23. The value of 105 106 107 is 110 111 112 a) 2

b) 0

24. If det( A 3 3 ) = 4 then det(2 a) 32

b) 16

3 3)

is equal to c) 8

d) 4

[WBUT-2005]

1.65

Matrix I

25. adj (2 A3 3 ) is equal to a) 32.adj ( )

b) 8.adj ( )

c) 4.adj ( )

d) 2.adj ( )

c) adj( A)

d) [ adj( A)]

26. adj ( AT ) is equal to a) 3.adj ( )

( )

b) adj A−

T

27. For a 3rd order determinant D, its adjoint determinant is equal to a) D 2

b) D

c) D3

d) D 4

c) 6

d) 4.

⎛ 3 0 0⎞ 28. The trace of A = ⎜ 0 1 0⎟ is ⎜⎝ 0 0 2⎟⎠ a) 7

b) 5

29. The trace of AT is same as a) trace of A c) [trace of A]T

b) trace of A−1 d) none of these

30. For an orthogonal matrix A, A−1 is same as a) A b) AT c) adj A 31. For any nonsingular matrix A, a)

(A ) −

T

(A )

b) AT

T −1

is same as

c) A

d) none of these

32. For any orthogonal matrix A, det A is equal to a) 0 b) 1 c) ±1 Answers: 1. (a) 2. (a) 10. (b) 11. (b) 19. (b) 20. (c) 28. (c) 29. (a)

3. (b) 12. (c) 21. (a) 30. (b)

4. (c) 13. (c) 22. (c) 31. (a)

d) none of these

5. (b) 14. (d) 23. (b) 32. (c)

6. (c) 15. (d) 24. (a)

d) −1

7. (a) 16. (d) 25. (c)

8. (d) 17. (a) 26. (d)

9. (d) 18. (b) 27. (a)

CHAPTER

2 Matrix II 2.1

INTRODUCTION

In this chapter, first we deal with the concept of the rank of a matrix and also the process of determination of rank. Next, we discuss the matrix inversion method, Cramer’s rule and also the consistency and inconsistency of a system of homogeneous and nonhomogeneous linear simultaneous equations. Then we represent the methods of determination for Eigen values and Eigen vectors of a square matrix and also the Cayley–Hamilton theorem and its applications. In the last part, we discuss the diagonalisation of a square matrix which is included as further-reading, material for interested students.

2.2

RANK OF A MATRIX

Let A be a nonzero matrix of order m × n. The rank of A is defined to be r if r is the greatest positive integer such that A has at least one nonzero minor of order r. Important Observations (i) The rank of a null matrix is zero. (ii) Rank of n-th order identity matrix is n. (iii) If the rank of A be r, every minor of order greater than r is zero. (iv) For a nonzero m × n matrix A, 0 < rank A {m , n}. (v) For an n-th order square matrix A, the rank of A is n if det( A) rank of A is less than n if det( A) = 0. (vi) Rank of A = Rank of AT .

0 and

2.2

Engineering Mathematics-I

Example 1 ⎛3 2⎞ Let A = ⎜ ⎟ . Here, det( A) ⎝1 4⎠ rank of A = 2.

0. Since highest-order nonzero minor is of order 2,

Example 2 ⎛0 2 3 4⎞ Let A = ⎜ 0 5 1 1 ⎟ . Here, the highest-order minors are 3rd order minors, and ⎜ ⎟ ⎝0 0 0 0⎠ they are ⎛0 2 3⎞ ⎛0 2 4⎞ ⎛0 3 4⎞ ⎛2 3 4⎞ ⎜ 0 5 1 ⎟, ⎜ 0 5 1 ⎟, ⎜ 0 1 1 ⎟, ⎜ 5 1 1 ⎟. ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0 0 0⎠ ⎝0 0 0⎠ ⎝0 0 0⎠ ⎝0 0 0⎠ They are all singular matrices. So, rank of A < 3. So we have to search for at least one 2nd order nonzero minor if it exists. 2 3 ⎛ 2 3⎞ Now, we have ⎜ ⎟ as a 2nd order nonzero minor since 5 1 ≠ 0. ⎝ 5 1⎠ Hence, rank of A = 2

2.3

ELEMENTARY ROW AND COLUMN OPERATIONS

Let A be a nonzero matrix of order m × n. Elementary row (or column) operations on A are of the following three kinds: (i) Interchanging of any two rows (or columns) of A. [Notation: Rij (or Cij ) stands for interchanging of the i th row and j th row (or of the i th column and jth column).] (ii) Multiplication of a row (or column) by a nonzero quantity. [Notation: d Ri (or d Ci ) stands for multiplication of the i th row by d (or of the i th column by d).] (iii) Addition of scalar multiple of a row (or column ) to another row (or column). [Notation: Ri d ⋅ R j (or Ci d ⋅ C j ) stands for addition of the d multiple of the j th row to the i th row (or d multiple of the j th column to the i th column).] Example 3 ⎛2 ⎜3 ⎜ ⎝1

2 4 2 5 1 3

3⎞ 0 ⎟ ⎯R12 → ⎟ 4⎠

⎛3 ⎜2 ⎜ ⎝1

2 5 2 4 1 3

0⎞ 3⎟, ⎟ 4⎠

2.3

Matrix II

⎛ 2 4 −2 ⎜3 5 2 ⎜ 1 ⎝1 3

⎛2 ⎜3 ⎜ ⎝1

2 4 2 5 1 3

3⎞ 0 ⎟ ⎯C23 → ⎟ 4⎠

⎛2 ⎜3 ⎜ ⎝1

2 4⎞ 2 5 ⎟ ⎯2 R2 → ⎟ 1 3⎠

⎛2 ⎜6 ⎜ ⎝1

2 4⎞ 4 10 ⎟ , ⎟ 1 3⎠

⎛2 ⎜3 ⎜ ⎝1

2 4⎞ 2 5 ⎟ ⎯3C1 → ⎟ 1 3⎠

⎛6 ⎜9 ⎜ ⎝3

2 4⎞ 2 5⎟ ⎟ 1 3⎠

3⎞ ⎛ −2 4 ⎜ 2 5 0⎟ R2 2 R3 → ⎜ ⎟ ⎯ ⎝ 1 3 −4 ⎠ 3⎞ ⎛ −2 4 ⎜ 2 5 0⎟ C1 ⎜ ⎟ ⎯ ⎝ 1 3 −4 ⎠

2.4

C3

3⎞ 0⎟ ⎟ 4⎠

3⎞ ⎛ −2 4 ⎜ 4 11 −8 ⎟ , ⎜ ⎟ ⎝ 1 3 −4 ⎠

3⎞ ⎛ 4 4 ⎜ 2 5 0⎟ → ⎜ ⎟ ⎝ −7 3 −4 ⎠

ROW EQUIVALENT AND COLUMN EQUIVALENT MATRICES

Suppose a matrix Bm×n is obtained by performing a finite number of elementary row (or column) operations on another matrix Am×n . Then Am×n and Bm×n are said to be row equivalent (or column equivalent). Example 4 ⎛2 A = ⎜3 ⎜ ⎝1

2 4 2 5 1 3

3⎞ 0 ⎟ ⎯R12 → ⎟ 4⎠

⎛3 ⎜2 ⎜ ⎝1

2 5 2 4 1 3

0⎞ 3⎟ ⎟ 4⎠

4 11 −8 ⎞ ⎛3 2 5 R2 +3R3 ⎜ ⎟ 2 4 3 ⎯ → 5 1 13 ⎟ ⎜ 1 3 4⎠ ⎝1 1 3 A and B are row equivalent.

⎛5 R1 2 R3 ⎯ → ⎜2 ⎜ ⎝1

0⎞ 9 ⎟ = B. ⎟ 4⎠

Example 5 ⎛2 A = ⎜3 ⎜ ⎝1 C ⎯ 2

2 4 2 5 1 3 C3

3⎞ 0 ⎟ ⎯C23 → ⎟ 4⎠

⎛ 2 0 −2 →⎜ 3 9 2 ⎜ 1 ⎝1 5

⎛ 2 4 −2 ⎜3 5 2 ⎜ 1 ⎝1 3

3⎞ 0⎟ ⎟ 4⎠

3⎞ 0 ⎟ ⎯C1 ( )C2 → ⎟ 4⎠

A and B are column equivalent.

3⎞ ⎛ −2 4 −2 ⎜ −2 5 2 0 ⎟ = B. ⎜ ⎟ 1 −4 ⎠ ⎝ −2 3

2.4

Engineering Mathematics-I

2.5

ROW-REDUCED ECHELON MATRIX

A matix A is called a row-reduced echelon matrix (or row-echelon matrix) if (i) all nonzero rows precede all zero rows of A (ii) in a nonzero row, the first nonzero element is 1 (called the leading 1) (iii) all the columns which contain the leading 1 of some row have all other elements zero (iv) for each nonzero row, if the leading element of row i occurs in column pi then p1 < p2 < p3 < … Example 6 The following matrices are row-reduced echelon matrices. ⎛1 0 0⎞ ⎛1 0 4 0⎞ ⎜ 0 1 0 ⎟, ⎜ 0 1 1 0 ⎟, ⎜ ⎟ ⎜ ⎟ ⎝0 0 1⎠ ⎝0 0 0 0⎠

⎛0 ⎜0 ⎜ ⎜⎜ 0 ⎝0

1 0 0 0

0 1 0 0

3 2 0 0

0⎞ 0⎟ ⎟. 1⎟ 0 ⎟⎠

Theorem 2.1: A matrix can be made row equivalent to a row-reduced echelon matrix by elementary row operations. Proof: Beyond the scope of this book. Theorem 2.2: Two row equivalent matrices have the same rank. Proof: Beyond the scope of this book. Theorem 2.3: If a row-reduced echelon matrix A has r nonzero rows then rank A = r. Proof: Beyond the scope of this book.

2.6

Steps: Step 1 Step 2 Step 3 Step 4

DETERMINATION OF RANK OF MATRIX BY ELIMENTARY OPERATIONS

Apply elementary row operations on the matrix. Convert the matrix to a row-reduced echelon matrix. Count the number of nonzero rows. The value obtained in Step 3 is the rank.

Example 7

⎛1 Let us find the rank of A = ⎜ 1 ⎜ ⎝3

1 1 1

1⎞ 1⎟ . ⎟ 1⎠

[WBUT 2006]

We apply elementary row operations on A to reduce it to a row-echelon matrix.

2.5

Matrix II

⎛1 A = ⎜1 ⎜ ⎝3

1⎞ R2 R1 , R3 3R1 1⎟ ⎯⎯⎯⎯⎯⎯ ⎯ ⎯→ ⎟ 1⎠

1 1 1

⎛1 0 1 R3 R2 , R1 R2 ⎜0 2 2 ⎯⎯ ⎯⎯ ⎯⎯⎯⎯⎯ → ⎜0 0 ⎝ The row-reduced echelon matrix So, Rank B = 2 and hence Rank

1 1⎞ 2 −2 ⎟ ⎟ 2 −2 ⎠

0⎞ ⎛ 1 ⎞ ⎛1 0 0⎞ 2 ⎟ ⎜⎝ − 2 ⎟⎠ R2 ⎜ 0 1 1 ⎟ = B ⎟ →⎜ 0 0 0⎟ 0⎠ ⎯ ⎝ ⎠ B has the two nonzero rows. A = 2.

⎛1 ⎜3 Find the rank of A = ⎜ ⎜⎜ 0 ⎝1

Example 8

⎛1 ⎜0 ⎜ ⎝0

2 1 0⎞ 6 12 9 ⎟ ⎟. 0 5 8⎟ 2 2 1 ⎟⎠

We apply elementary row operations on A to reduce it to a row-echelon matrix ⎛1 ⎜3 A=⎜ ⎜⎜ 0 ⎝1 1 R2 ⎯9 →

2 1 0⎞ 6 12 9 ⎟ R2 3R R1 , R4 R1 ⎟ 0 5 8 ⎟ ⎯⎯ ⎯⎯⎯⎯⎯→ 2 2 1 ⎟⎠ ⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

⎛1 1 ⎜ R2 ⎜ 0 3 ⎯ →⎜ 0 ⎜0 ⎝

2 0 0 0 2 0 0 0

1 1 5 1 0 1 0 0

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

2 0 0 0

1 9 5 1

0⎞ 9⎟ ⎟ 8⎟ 1 ⎟⎠

0⎞ 1⎟ R4 R2 , R3 R1 , R1 R2 ⎟ 8 ⎟ ⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯→ 1 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

1⎞ 1⎟ R2 R3 , R1 R3 ⎟ → 1⎟ ⎯⎯ ⎯⎯⎯⎯⎯ ⎟ 0⎠

0 1 0 0

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

2 0 0 0

2 0 0 1 0 0 0 0

1⎞ 1⎟ ⎟ 3⎟ 0 ⎟⎠

0⎞ 0⎟ ⎟ = B. 1⎟ 0 ⎟⎠

The row-reduced echelon matrix B has 3 non-zero rows. So, Rank B = 3 and hence Rank A = 3.

2.7

SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS BY MATRIX INVERSION METHOD

Let us consider the system of n linear equations involving n unknowns: a11 x1 + a12 x2 +

+ a1n xn = b1

a21 x1 + a22 x2 + + a2 n xn = b2 .............................. an x1 + an 2 x2 +

+ ann xn = bn

2.6

Engineering Mathematics-I

⎛ a11 ⎜a Let A = ⎜ 21 ⎜⎜ … ⎝ an ⎛ b1 ⎞ ⎜b ⎟ B=⎜ 2⎟. ⎜⎜ … ⎟⎟ ⎝ bn ⎠

a12 a22 … an 2

… a1n ⎞ ⎛ x1 ⎞ ⎜x ⎟ … a2 n ⎟ ⎟ , called the coefficient matrix, X = ⎜ 2 ⎟ and … …⎟ ⎜⎜ … ⎟⎟ … ann ⎟⎠ ⎝ xn ⎠

Then the above system of equations can be written in the form AX = B. Now if det( A) 0, then the system AX = B has the unique solution X = A−1 B. Example 9

Let us solve the following system of equations:

2 x 33yy + 4 = −4 x+z =0 − y + 4z = 2

[WBUT 2005]

⎛2 Here, the coefficient matrix A = ⎜ 1 ⎜ ⎝0

3 4⎞ ⎛ −4 ⎞ ⎛ x⎞ ⎟ ⎜ ⎟ 0 1 , X = y and B = ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ 1 4⎠ ⎝ 2⎠ ⎝z⎠

Then the given system of equations can be written as AX = B. 2 Now det( A) = 1 0

3 4 0 1 = 10 ≠ 0. 1 4

So the system has a unique solution X = A−1 B. ⎛ 1 8 −3 ⎞ Here, adj( A) = ⎜ −4 8 2 ⎟ . ⎜ ⎟ 3⎠ ⎝ −1 2 Now A−1 = Hence X

⎛ 1 8 −3 ⎞ 1 1 adj( A), or, A−1 = ⎜ − 4 8 2 ⎟ . det ( A) 10 ⎜ −1 2 3 ⎟ ⎝ ⎠ −1 A B

⎛ x⎞ 1 or, ⎜ y ⎟ = ⎜ ⎟ 10 ⎝z⎠

⎛ 1 8 −3 ⎞ ⎛ −4 ⎞ ⎜ −4 8 2 ⎟ ⎜ 0 ⎟ ⎜ ⎟⎜ ⎟ ⎝ −1 2 3 ⎠ ⎝ 2 ⎠

⎛ −10 ⎞ ⎛ −1 ⎞ 1 ⎜ 20 ⎟ = ⎜ 2 ⎟ . 10 ⎜ 10 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ So the solutions are x = 1, y = 2, z = 1. =

Matrix II

2.8

2.7

SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS BY CRAMER’S RULE

Let us consider the system of linear equations involving n unknowns: a11 x1 + a12 x2 + + a1n xn = b1 a21 x1 + a22 x2 + + a2 n xn = b2 .............................. an x1 + an 2 x2 +

+ ann xn = bn

⎛ a11 a12 … a1n ⎞ ⎛ x1 ⎞ ⎜a ⎟ ⎜x ⎟ a … a 22 2n Let A = ⎜ 21 ⎟ , called the coefficient matrix, X = ⎜ 2 ⎟ and … … … … ⎜⎜ ⎟⎟ ⎜⎜ … ⎟⎟ a a … a n2 nn ⎠ ⎝ n ⎝ xn ⎠ ⎛ b1 ⎞ ⎜b ⎟ B = ⎜ 2 ⎟. ⎜⎜ … ⎟⎟ ⎝ bn ⎠ Then the above system of equations can be written in the form AX = B. Now if det( A) 0, then there exists a unique solution of the system AX = B and is given by b1 a12 a1n det A1 1 b2 a22 a2 n x1 = = det A det A … … … … bn an 2 … ann a11 b1 a1n det A2 1 a21 b2 a2 n x2 = = det A det A … … … … an bn … ann ............................................... a11 det An 1 a21 xn = = det A det A … an

a12 a22 … an 2

… … … …

b1 b2 … bn

Observations: (1) When det( A) 0 then the system AX = B has a unique solution. (2) When det( A) = 0 and at least one of det A1 det A2 , ..., det An is nonzero, then the above system has no solution. (3) When det( A) = 0 and det A1 det A2 = = det An = 0 then the above system has an infinite number of solutions.

2.8

Engineering Mathematics-I

Example 10

Let us solve the following system by Cramer’s rule:

x + 2 y + 3z = 6 2x 4 y + z = 7 3 x 2 y + 9 = 14. If we write the above equations in the form AX = B then ⎛ 6⎞ ⎛1 2 3⎞ ⎛ x⎞ the coefficient matrix A = ⎜ 2 4 1 ⎟ , X = ⎜ y ⎟ and B = ⎜ 7 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 14 ⎠ ⎝ 3 2 9⎠ ⎝z⎠ 1 2 3 Here, det A = 2 4 1 = −20 ≠ 0. 3 2 9 So the system has a unique solution. 6 2 3 Now det A1 = 7 4 1 = −20. 14 2 9 1 6 3 det A2 = 2 7 1 = 20 3 14 9 1 2 6 det A3 = 2 4 7 = 20 3 2 14 det A1 −20 = = 1. det A −20 det A2 −20 y= = = 1. det A −20 det A2 −20 z= = = 1. det A −20

So, x =

Hence the solution is (1, 1, 1) , which is unique.

2.9

SYSTEM OF HOMOGENEOUS AND NONHOMOGENEOUS LINEAR EQUATIONS

Let us consider the following system of m linear equations with n unknowns a11 x1 + a12 x2 +

+ a1n xn = b1

a21 x1 + a22 x2 + + a2 n xn = b2 .............................. am x1 + am 2 x2 +

+ amn xn = bm

Matrix II

2.9

The above system is called a homogeneous system of linear equations if all the bi's are zero and the system is called a nonhomogeneous system of linear equations if at least one bi is nonzero.

2.9.1. Matrix Representation of a System of Homogeneous and Nonhomogeneous Linear Equations ⎛ a11 ⎜a Let A = ⎜ 21 ⎜⎜ … ⎝ am ⎛ b1 ⎞ ⎜b ⎟ B = ⎜ 2 ⎟. ⎜⎜ … ⎟⎟ ⎝ bm ⎠

a12 a22 … am 2

… a1n ⎞ ⎛ x1 ⎞ ⎜x ⎟ … a2 n ⎟ ⎟ , called the co-efficient matrix, X = ⎜ 2 ⎟ and … …⎟ ⎜⎜ … ⎟⎟ … amn ⎟⎠ ⎝ xn ⎠

Then the above system of equations can be written in the form AX = B. ⎛ a11 ⎜a Here, A = ⎜ 21 ⎜⎜ … ⎝ am

b1 ⎞ b2 ⎟ ⎟ is called the augmented matrix. …⎟ bm ⎟⎠ ⎛0⎞ ⎜0⎟ The system AX = B is homogeneous for B = ⎜ ⎟ , otherwise the system is ⎜⎜… ⎟⎟ ⎝0⎠ nonhomogeneous. So, the homogeneous system can be written in the form AX = O, where O is the null matrix or zero matrix. Example 11

a12 a22 … am 2

… a1n … a2 n … … … amn

The following is a homogeneous system of 3 linear equations with 4

unknowns: x 3y 3y + 4 = 0 x+z+w=0 − y + 4 z − 2w = 0 ⎛2 Here, the coefficient matrix A = ⎜ 1 ⎜ ⎝0 ⎛2 and augmented matrix A = ⎜ 1 ⎜ ⎝0

3 4 0 1 1 4

⎛x⎞ 0⎞ ⎛0⎞ ⎜ y⎟ ⎟ 1 , X = ⎜ ⎟, B = ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜⎜ z ⎟⎟ 2⎠ ⎝0⎠ w ⎝ ⎠ 0 0⎞ 1 0 ⎟. ⎟ 2 0⎠

3 4 0 1 1 4

2.10

Engineering Mathematics-I

Example 12

The following is a nonhomogeneous system of 3 linear equations

with 3 unknowns: 5x 4 y + 4 = 2 x+z =0 3x 2 y + 4 z = 0 ⎛5 Here the coefficient matrix A = ⎜ 1 ⎜ ⎝3 ⎛2 and augmented matrix A = ⎜ 1 ⎜ ⎝0

2.10

4 4⎞ ⎛2⎞ ⎛ x⎞ 0 1⎟ , X = ⎜ y ⎟ , B = ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ 2 4⎠ ⎝0⎠ ⎝z⎠

3 4 2⎞ 0 1 0 ⎟. ⎟ 1 4 0⎠

CONSISTENCY AND INCONSISTECY OF THE SYSTEM OF LINEAR EQUATIONS

We say the system of linear equations is consistent if it has a solution. On the other hand, the system is called inconsistent if it has no solution. For example, i) The following system 5x 4 y = 2 x y=0 has the solution (2, 2). So the system is consistent. ii) The following system 8x 4 y = 2 2x y = 1 has no solution. So the system is inconsistent.

2.11

EXISTENCE OF THE SOLUTION OF HOMOGENEOUS SYSTEM

Let us consider the following homogeneous system of m linear equations with n unknowns: a11 x1 + a12 x2 + + a1n xn = 0 a21 x1 + a22 x2 + + a2 n xn = 0 .............................. am x1 + am 2 x2 +

+ amn xn = 0

It is very important to keep in mind that the homogeneous system is always consistent, since it has always a solution of the form ( x1 x2 , xn ) = (0, 0, 0, … , 0) .

Matrix II

2.11

This solution is called a trivial solution. The solutions other than the trivial are known as nontrivial solutions. Example 13

The following homogeneous system of 2 linear equations with 3

unknowns 3x 2 y + z = 0 x + y − 3z = 0 has a trivial solution (0, 0, 0). Also it is very interesting to see that (1, 2, 1), (2, 4, 2), (3, 6, 3), etc., are also the solutions of the system. In fact the system has a solution of the form of k(1, 2, 1). Actually, these are nontrivial solutions. Theorem 2.4: In a homogeneous system with m equations and n unknowns, if the number of equations are less than the number of unknowns (i.e., m < n) then the system has a nontrivial (non-zero) solution. In fact, there exists infinitely many solutions. Proof: Beyond the scope of the book. Example 14

Let us solve the following homogeneous system:

x + 2y − z = 0 2x y 2z = 0 First of all, it is clear that it has a trivial (or, zero) solution (0, 0, 0). So the system is consistent. Here number of equations ( ) = 2 and number of unknowns ( ) = 3. So, m < n. Therefore, nontrivial solution also exists. ⎛ x⎞ ⎛ 1 2 −1 ⎞ ⎜ y ⎟ and Now if we write the system as AX = O, then A = ⎜ , X = ⎟ ⎜ ⎟ ⎝ 2 1 −2 ⎠ ⎝z⎠ ⎛0⎞ O = ⎜ ⎟ , null matrix. ⎝0⎠ Here, we apply elementary row operations to convert A to a row-echelon matrix. ⎛1 2 A=⎜ ⎝2 1 ⎛1⎞ ⎜ ⎟ R2 ⎝ 3⎠ ⎯ →

−1 ⎞ R2 2 R1 ⎛ 1 2 ⎯ → ⎜ −2 ⎟⎠ 3 ⎝0 ⎛ 1 2 −1 ⎞ R 2R2 ⎜0 1 0⎟ ⎯ 1 → ⎝ ⎠

−1 ⎞ 0 ⎟⎠

⎛ 1 0 −1 ⎞ ⎜ 0 1 0 ⎟. ⎝ ⎠ ⎛ x⎞ ⎛ 1 0 −1 ⎞ ⎜ ⎟ ⎛ 0 ⎞ y =⎜ ⎟ So the given system is equivalent to ⎜ ⎟ ⎝0 1 0⎠ ⎜ z ⎟ ⎝0⎠ ⎝ ⎠ i.e., x z=0 y=0

2.12

Engineering Mathematics-I

Let we choose z = k , then x = k , where k is any arbitrary constant. Therefore, the solution is ( x, y, z ) = (k , 0, k ) = k (1, 0, 1), which is nontrivial. Hence, the system has infinitely many solutions. Theorem 2.5: In a homogeneous system with n equations and n unknowns, if the rank of the coefficient matrix is less than n then the system has a non-trivial (nonzero) solution. In fact, there exists infinitely many solutions. Proof: Beyond the scope of the book. Example 15

Solve the following homogeneous system:

x + y + 3z = 0 2x

y

=0

3x 2 y + 4 = 0 First of all, it is clear that it has a trivial (or, zero) solution (0, 0, 0). So the system is consistent. Here number of equations ( ) = 3 and number of unknowns ( ) = 3. ⎛1 1 3⎞ Now if we write the system as AX = O then A = ⎜ 2 1 1 ⎟ , ⎜ ⎟ ⎝3 2 4⎠ ⎛ x⎞ ⎛0⎞ X = ⎜ y ⎟ and O = ⎜ 0 ⎟ , null matrix. ⎜ ⎟ ⎜ ⎟ ⎝z⎠ ⎝0⎠ Here, we apply elementary row operations to convert A to a row-echelon matrix. ⎛1 1 3⎞ R 2R 2 R1, R3 3R1 A = ⎜ 2 1 1 ⎟ ⎯⎯2 ⎯⎯⎯⎯⎯ ⎯ ⎯→ ⎜ ⎟ 3 2 4 ⎝ ⎠ ⎛1 ⎜ R3 R2 ⎯ → ⎜0 ⎝0

⎛1 ⎜0 ⎜ ⎝0

1 3⎞ 1 −5 ⎟ ⎯( ) R2 → ⎟ 0 0⎠

1 3⎞ 1 −5 ⎟ ⎟ 1 −5 ⎠ ⎛1 1 3⎞ ⎛ 1 0 −2 ⎞ ⎜ 0 1 5 ⎟ ⎯R1 − 2 R2 → ⎜ 0 1 5 ⎟ . ⎜ ⎟ ⎜ ⎟ ⎝0 0 0⎠ ⎝0 0 0⎠

So, the rank of A = 2 (< 3, number of unknowns ). Therefore, nontrivial solution exists. ⎛ 1 0 −2 ⎞ ⎛ x ⎞ ⎛ 0 ⎞ So the given system is equivalent to ⎜ 0 1 5 ⎟ ⎜ y ⎟ = ⎜ 0 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝0 0 0⎠ ⎝ z ⎠ ⎝0⎠

2.13

Matrix II

i.e., x 2z = 0 y + 5z = 0 Let us choose z = k , then x k and y k where k is any arbitrary constant. Therefore, the solution is ( x, y, z ) = (2 k , − 5k , 2 k ) = (2, − 5, 2), which is nontrivial. Hence, the system has infinitely many solutions.

2.12

EXISTENCE OF THE SOLUTION OF A NON-HOMOGENEOUS SYSTEM

Let us consider the following non-homogeneous system of m linear equations with n unknowns a11 x1 + a12 x2 +

+ a1n xn = b1

a21 x1 + a22 x2 + + a2 n xn = b2 .............................. am x1 + am 2 x2 +

+ amn xn = bm

If we write the above system of equations in the form of AX = B ⎛ a11 ⎜a then coefficient matrix A = ⎜ 21 ⎜⎜ … ⎝ am ⎛ a11 ⎜a augmented matrix A = ⎜ 21 ⎜⎜ … ⎝ am

a12 a22 … am 2

a12 a22 … am 2

… a1n ⎞ ⎛ x1 ⎞ ⎛ b1 ⎞ ⎟ ⎜ ⎟ ⎜b ⎟ … a2 n x ⎟ , X = ⎜ 2 ⎟ , B = ⎜ 2 ⎟ and … …⎟ ⎜⎜ … ⎟⎟ ⎜⎜ … ⎟⎟ … amn ⎟⎠ ⎝ xn ⎠ ⎝ bm ⎠

… a1n … a2 n … … … amn

b1 ⎞ b2 ⎟ ⎟. …⎟ bm ⎟⎠

Theorem 2.6: A nonhomogeneous system AX = B is consistent iff rank A = rank A. In other words, solution exists for a nonhomogeneous system AX = B iff rank A = rank A, otherwise the system has no solution. Proof: Beyond the scope of the book. Example 16 + 2y = 5 2 x 5 y = 11 3 x 7 y = 17

Let us consider the system

2.14

Engineering Mathematics-I

⎛1 2⎞ Here, the coefficient matrix is A = ⎜ 2 5 ⎟ and the augmented matrix ⎜ ⎟ ⎝ 3 7⎠ ⎛ 1 2 5⎞ A = ⎜ 2 5 11 ⎟ . ⎜ ⎟ ⎝ 3 7 17 ⎠

is

Now we apply elementary row operations to the augmented matrix A ⎛ 1 2 5⎞ R2 2R 2 R1, R3 3R1 A = ⎜ 2 5 11⎟ ⎯⎯⎯⎯⎯⎯⎯ ⎯ ⎯→ ⎜ ⎟ 3 7 17 ⎝ ⎠ ⎛ 1 0 3⎞ ⎜ ⎟ ææææææÆ ⎜ 0 1 1 ⎟ ⎝0 0 1⎠ R1 2R R 2, R3

R2

⎛1 2 5⎞ ⎜0 1 1⎟ ⎜ ⎟ ⎝0 1 2⎠

R1 3R R 3 , R2

æ

R3

Æ

⎛1 0 0⎞ ⎜0 1 0⎟ ⎜ ⎟ ⎝0 0 1⎠

⎛1 0⎞ So from the above, A is row equivalent to ⎜ 0 1 ⎟ and A is row equivalent to ⎜ ⎟ ⎝0 0⎠ ⎛1 0 0⎞ ⎜0 1 0⎟ ⎜ ⎟ ⎝0 0 1⎠ and therefore rank of A = 2 and rank of A = 3. Hence rank A ≠ rank A and correspondingly, the system is not consistent, i.e., the system has no solution. Theorem 2.7: Consider a nonhomogeneous system AX = B with m linear equations and n unknowns which is consistent (i.e., rank A = rank A). Then the following cases hold. i) The sytem has a unique solution (i.e., only one solution) if rank A = rank A n when m n or m > n. ii) The sytem has infinitely many solutions if a) rank A = rank A m n or m > n and b) rank A = rank A m when m < n. Proof: Beyond the scope of the book. Example 17

Let us consider the system

+ y+ z =1 2x

y 2z = 1

x + 2 y + 3z = 0

n when

2.15

Matrix II

⎛1 1 1⎞ Here, the coefficient matrix is A = ⎜ 2 1 2 ⎟ and ⎜ ⎟ ⎝1 2 3⎠ ⎛1 1 1 1⎞ the augmented matrix is A = ⎜ 2 1 2 1 ⎟ . ⎜ ⎟ ⎝1 2 3 0⎠ Now we apply elementary row operations to the augmented matrix A. ⎛1 1 1 1⎞ R 2R 2 R1, R3 3R1 A = ⎜ 2 1 2 1 ⎟ ⎯⎯2 ⎯⎯⎯⎯⎯⎯ → ⎜ ⎟ 1 2 3 0 ⎝ ⎠

⎛1 ⎜0 ⎜ ⎝0

1 1 1 0 1 2

⎛1 1 1 ( − ) R2 ⎯ → ⎜0 1 0 ⎜ ⎝0 1 2

1⎞ R2 , R1 3R2 1 ⎟ ⎯R ⎯3 ⎯⎯⎯⎯⎯ ⎯⎯ → ⎟ 1⎠

⎛1⎞ ⎛1 0 1 ⎜ ⎟ R3 ⎝2⎠ ⎯ → ⎜⎜ 0 1 0 ⎝0 0 1

0⎞ R −R 1⎟ ⎯ 1 3 → ⎟ 1⎠

1⎞ 1⎟ ⎟ 1⎠

⎛1 0 1 ⎜0 1 0 ⎜ ⎝0 0 2

⎛1 0 0 ⎜0 1 0 ⎜ ⎝0 0 1

0⎞ 1⎟ ⎟ 2⎠

1⎞ 1⎟ ⎟ 1⎠

⎛1 0 0⎞ So from the above, A is row equivalent to ⎜ 0 1 0 ⎟ ⎜ ⎟ ⎝0 0 1⎠ ⎛1 0 0 and A is row equivalent to ⎜ 0 1 0 ⎜ ⎝0 0 1

1⎞ 1⎟ . ⎟ 1⎠

and therefore, rank of A = 3 and rank of A = 3. Hence, rank A = rank A = 3 (i.e., rank A = rank A = number of unknowns) and according to case (i) of the above theorem, the system is consistent and the system has a unique solution. ⎛ 1 0 0 ⎞ ⎛ x ⎞ ⎛ 1⎞ Now the above system is equivalent to ⎜ 0 1 0 ⎟ ⎜ y ⎟ = ⎜ 1 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ 0 0 1 ⎠ ⎝ z ⎠ ⎝ −1 ⎠ i.e., x =1 y =1 z= 1

2.16

Engineering Mathematics-I

Hence we have the unique solution (1, 1, −1) for the given system. Example 18

Let us consider the system

x+ y+z =6 x + 2 y + 3 z = 14 x + 4 y + 7 z = 30 ⎛1 1 1 ⎞ Here, the coefficient matrix is A = ⎜ 1 2 3 ⎟ and ⎜ ⎟ ⎝1 4 7 ⎠ ⎛1 1 1 6 ⎞ the augmented matrix is A = ⎜ 1 2 3 14 ⎟ . ⎜ ⎟ ⎝ 1 4 7 30 ⎠ Now we apply elementary row operations to the augmented matrix A ⎛1 1 1 6 ⎞ R2 R1, R3 R1 A = ⎜1 2 3 14 ⎟ ⎯⎯⎯⎯⎯⎯⎯ → ⎜ ⎟ 1 4 7 30 ⎝ ⎠ 1 ⎛ 1 0 −1 ⎜0 1 2 R3 3R R2 , R1 R2 ⎯⎯ ⎯⎯⎯⎯⎯ ⎯⎯ → ⎜ ⎝0 0 0

⎛ 1 1 1 6⎞ ⎜ 0 1 2 8⎟ ⎜ ⎟ ⎝ 0 3 6 24 ⎠

2⎞ 8⎟. ⎟ 0⎠

⎛ 1 0 −1 ⎞ So from the above, A is row equivalent to ⎜ 0 1 2 ⎟ ⎜ ⎟ ⎝0 0 0⎠ 1 ⎛ 1 0 −1 and A is row equivalent to ⎜ 0 1 2 ⎜ ⎝0 0 0

2⎞ 8⎟. ⎟ 0⎠

and therefore, rank of A = 2 and rank of A = 2. Hence, rank A = rank A = 2 < 3 (i.e., rank A = rank A < number of unknowns) and according to case (ii) (a) of the above theorem, the system is consistent and the system has infinitely many solutions. ⎛ 1 0 −1 ⎞ ⎛ x ⎞ ⎛ −2 ⎞ Now the above system is equivalent to ⎜ 0 1 2 ⎟ ⎜ y ⎟ = ⎜ 8 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝0 0 0⎠ ⎝ z ⎠ ⎝ 0⎠ i.e., x z = −2 y+z =8

2.17

Matrix II

Let z = k , then x

k 2 and y

k 8, where k is any arbitary constant.

So ( x, y, z ) = (k − 2 2, − k 8, k ) = ( 2, 8, 0)

(1, − 1, 1).

Hence we have infinitely many solutions for the given system. Example 19

Let us consider the system

x + 2y + z = 2 2 x 5y 5y + 3 = 5 ⎛1 2 1⎞ Here, the coefficient matrix is A = ⎜ ⎟ and ⎝ 2 5 3⎠ ⎛1 2 1 2⎞ the augmented matrix is A = ⎜ ⎟. ⎝2 5 3 5⎠ Now we apply elementary row operations to the augmented matrix A ⎛ 1 0 −1 0 ⎞ ⎛ 1 2 1 2 ⎞ R2 2 R1 ⎛ 1 2 1 2 ⎞ R1 − 2 R2 → ⎜ → ⎜ A=⎜ ⎟. ⎟⎯ ⎟⎯ ⎝ 0 1 1 1⎠ ⎝2 5 3 5⎠ ⎝0 1 1 1⎠ ⎛ 1 0 −1 ⎞ So from the above, A is row equivalent to ⎜ ⎟ ⎝ 0 1 1⎠ ⎛ 1 0 −1 0 ⎞ and A is row equivalent to ⎜ ⎟. ⎝ 0 1 1 1⎠ and therefore, rank of A = 2 and rank of A = 2. Hence, rank A = rank A = 2 (i.e., rank A = rank A ≤ number of equations) and according to case (ii) (b) of the above theorem, the system is consistent and the system has infinitely many solutions. ⎛ x⎞ ⎛ 1 0 −1 ⎞ ⎜ ⎟ ⎛ 0 ⎞ y =⎜ ⎟ Now the above system is equivalent to ⎜ ⎟ ⎝ 0 1 1⎠ ⎜ z ⎟ ⎝ 1 ⎠ ⎝ ⎠ i.e., x z=0 y+ z =1 Let z = k , then x So ( x, y, z ) = (k ,

k and y

k 1, where k is any arbitrary constant.

1, ) = (0, 1, 0) + kk(1, (1, 1, 1).

Hence, we have infinitely many solutions for the given system.

2.18

2.13 2.13.1

Engineering Mathematics-I

EIGEN VALUES AND EIGEN VECTORS Characteristic Polynomial and Characteristic Equation

Let us consider an n × n matrix A. Then the characteristic polynomial f A (λ ) is defined as a11 λ a21 det( A λ I n ) = … an1

a12 a22 − λ … an 2

… a1n … a2 n … … … ann − λ

where A = (aij ) n n . It is obvious from the definition that the characteristic polynomial f A (λ ) is of n th degree and is of the form fA λ

b0 λ n + b1λ n

where b0

1

b2 λ n −2 +

+ bn ,

b1 , b2 , ..., bn are constants.

Note: It can be easily shown that the constant term appears in the polynomial (i.e., the term bn ) is equal to det( A). The characteristic equation is defined as a 11 λ a21 det( A λ I n ) = … an1

a12 a22 − λ … an 2

… a 1n … a2 n = 0. … … … ann − λ

i.e., f A λ b0 λ n + b1λ n 1 b2 λ n −2 + + bn = 0. So the degree of the equation is n, which is same as the order of the matrix A. Example 20

Let us find the characteristic polynomial and characteristic equa⎛1 2 ⎞ tion of the 2 2 matrix A = ⎜ ⎟. ⎝3 5⎠ Here, characteristic polynomial is fA λ

A λ I2 ) =

1 λ 2 = λ2 3 5−λ

6λ − 1.

It is very important to note that in the polynomial ( 1) is the constant term which 1 2 is equal to det A = = 1. 3 5 The characteristic equation is f A (λ ) = 0. i.e., λ 2 λ 1 = 0. The degree of the equation is 2, which is same as order of the matrix A.

2.19

Matrix II

2.13.2

Cayley–Hamilton Theorem

If A be a square matrix of order n n then A satisfies its own characteristic equation. From the above theorem, we can say that if the characteristic equation of a square matrix A of order n n is −

b

+b

bn = 0,

then we have −

b A −2 + + where O is the null matrix of order n. For example, 1=

In the last example, the chracteristic equation is I

So by Cayley–Hamilton theorem we have

2.13.3

.

Determination of Inverse of a Matrix using Cayley– Hamilton Theorem

Let us consider a nonsingular square matrix A of order n n and its chracteristic equation is b +b bn = 0. Then by Cayley–Hamilton theorem, we have b An

2

On

n

Now since A is nonsingular, b

A)

...(1)

0 and so b

−1

exists.

So from the above equation (1), b An

2

+

2A

0

b− b A

−2

−

+

b

+b A

I

)

+

So, 2

−

+

= In

)

Let us find the characteristic equation of A

Example 21

using Cayley–Hamilton theorem, find A 1 . The characteristic equation of A is det( A 2 i.e., i.e.,

1 3− 0

0 1 l

0 −2 = 0 −2 − λ

4 l + 10 = 0.

In ) = 0

2 0 1

1 3 0

0 2 2

and

2.20

Engineering Mathematics-I

By Cayley–Hamilton theorem, A3

3 A2 3

i.e.,

4 A 10 I 3 = O3

3 A2 - 4 = 10 I 3

(

)

i.e., A A2 - 3A 3 A 4 I 3 = 10 I 3 i.e., A i.e., A

(

(

1 0⎞ ⎛ 2 3 −2 ⎟ ⋅ ⎜ 0 ⎟ ⎜ 0 −2 ⎠ ⎝ 1 Ê 4 3 A - 4 3 ) = Á -2 2 Á Ë 0

2

6 Ê -6 = Á -2 2 Á 3 Ë -3

1 0⎞ ⎛ 4 3 −2 ⎟ = ⎜ −2 ⎟ ⎜ 0 −2 ⎠ ⎝ 0 5 9 1

2ˆ

Ê6 2 - Á0 Á 4¯ Ë 3

5 2⎞ 9 −2 ⎟ ⎟ 1 4⎠ Ê 4 0 0ˆ 9 6 - Á0 4 0 Á 0 -6¯ Ë 0 0 4¯ 3

0ˆ

2 2ˆ 4 4 . 1 6¯

Hence, A

2.13.4

)

È 1 2 ˘ A - 3 A 4 I3 ˙ Î 10 ˚

1

⎛2 A2 = ⎜ 0 ⎜ ⎝1 So, (

)

È 1 2 ˘ A - 3 A 4I3 ˙ = I3 Î 10 ˚

1

Ê 6 2 -2ˆ È 1 2 ˘ 1 Á A - 3 A 4I3 ˙ = 2 4 -4 . Î 10 ˚ 15 Á Ë 3 1 -6¯

(

)

Eigen Values of a Matrix

Roots of the characteristic equation of a square matrix A are called the eigen values of A. Eigen values are also known as characteristic roots. For example, ⎛1 3⎞ let A = ⎜ ⎟. ⎝0 2⎠ Then the characteristic equation is det( A λ I 2 ) = 0. i.e.,

1 λ 3 =0 0 2−λ

i.e., (1 − λ )(2 )( − λ ) = 0 i.e., λ = 1, 2. So, 1 and 2 are the eigen values.

2.21

Matrix II

Theorem 2.8: 0 is always an eigen value for a singular matrix. Proof: Let A be any singular matrix of order n × n and its characteristic equation is b0 λ n b1λ n −11 + b2 λ n 2 bn 1λ + bn = 0. Now since A is singular, bn So, b0 λ n

b1λ n −1 + b2 λ n

i.e., λ ( 0 λ n

1

b1λ n

2

de A = 0. 2 2λ

bn −1λ = 0 n 3

bn −1 ) = 0

Therefore, λ = 0 is an eigen value. For example, ⎛1 0 1⎞ let A = ⎜ 2 2 3 ⎟ . ⎜ ⎟ ⎝0 0 0⎠ Clearly, det A = 0. So, A is singular. Then the characteristic equation is det( A λ I 3 ) = 0. i.e.,

1 λ 0 2 2−λ 0 0

i.e., λ (

1 3 =0 −λ

λ )(2 − λ ) = 0

i.e., λ = 0, 1, 2. So, 0 is an eigen value. Theorem 2.9: The diagonal elements are the eigen values for any diagonal matrix. ⎛ d1 0 ⎜0 d 2 Proof: Let us consider an n × n diagonal matrix, A = ⎜ … … ⎜⎜ ⎝0 0 d1 − λ 0 Then its characteristic equation is … 0 i.e., (d1 λ )(d )( 2

λ )… (d n λ ) = 0

i.e., λ = d1 , d 2 , , d n . Hence, the theorem is proved. For example, ⎛1 0 0⎞ let A = ⎜ 0 2 0 ⎟ . ⎜ ⎟ ⎝0 0 3⎠

0 d2 − λ … 0

… 0⎞ … 0⎟ ⎟. … …⎟ … d n ⎟⎠

… 0 … 0 =0 … … … dn − λ

2.22

Engineering Mathematics-I

The diagonal elements are 1, 2, and 3. Then the characteristic equation is det( A λ I 3 ) = 0. i.e.,

1 λ 0 0 0 2−λ 0 =0 0 0 3−λ

i.e., (1 − λ )(2 )( − λ ))(3 − λ ) = 0 i.e., λ = 1, 2, 3. So, the eigen values are 1, 2, and 3.

2.13.5

Eigen Vectors of a Matrix

Let us consider a square matrix A of order n × n. Now if an n-dimensional non-null vector x^ = ( x1 x2 , … , xn ) satisfies the equation Ax^ = λ x^, where λ is any scalar then x^ is called an eigen vector of the matrix A. Now if Ax^ = λ x^ holds, then (

λ n ) x^ = O. This is nothing but a homogeneous system of n equations with n unknowns. Since the system has a non-null solution, we have det( A λ I n ) = 0. This leads to the conclusion that the scalar λ is the eigen value corresponding to the eigen vector x^ . Theorem 2.10: There exists a unique eigen value corresponding to a eigen vector. Proof: Beyond the scope of the book. Theorem 2.11: If x^ 1 and x^ 2 are two eigen vectors corresponding to two distinct eigen values then x^ 1 and x^ 2 are independent. Proof: Beyond the scope of the book. Example 22

Let us find the eigen values and eigen vectors of the ⎛4 6⎞ matrix A = ⎜ ⎟. ⎝2 8⎠

The characteristic equation is i.e., λ 2

4 λ 6 =0 2 8−λ

λ + 20 = 0

i.e., (λ 10)( )(λ 2) = 0 So, the eigen values are λ = 10 and 2, say λ1 = 10 and λ 2 = 2. ^ ⎛x ⎞ Let X1 = ⎜ 1 ⎟ be the eigen vector corresponding to λ1 = 10. ⎝ x2 ⎠

Matrix II ^

2.23

^

Then AX1 = λ1 X1 ⎛x ⎞ ⎛ 4 6 ⎞ ⎛ x1 ⎞ i.e., ⎜ = 10 ⎜ 1 ⎟ ⎟ ⎜ ⎟ ⎝ 2 8 ⎠ ⎝ x2 ⎠ ⎝ x2 ⎠ i.e., 4 1 6 x2 = 10 x1 2 1 8 x2 = 10 x2 i.e., −6 1 + 6 x2 = 0 2 x1 2 x2 = 0 The above system is equivalent to x1 x2 = 0. Now if x2 = c, then x1 = c, where c is an arbitrary real number. ^ ⎛c⎞ ⎛1⎞ So the eigen vectors are X 1 = ⎜ ⎟ = c ⎜ ⎟ corresponding to the eigen value c⎠ ⎝ ⎝1⎠ λ1 = 10. ^ ⎛x ⎞ Let X 2 = ⎜ 1 ⎟ be the eigen vector corresponding to λ 2 = 2. ⎝ x2 ⎠ ^ ^ Then AX 2=λ2 X 2 ⎛ x1 ⎞ ⎛ 4 6 ⎞ ⎛ x1 ⎞ i.e., ⎜ ⎟ ⎜ x ⎟ = 2⎜ x ⎟ 2 8 ⎝ ⎠⎝ 2⎠ ⎝ 2⎠ i.e., 4

1

6 x2 = 2 x1

2

1

8 x2 = 2 x2

2

1

6 x2 = 0

2

1

6 x2 = 0

i.e.,

The above system is equivalent to x1 + 3 x2 = 0. Now if x2 = c, then x1

3c, where c is an arbitrary real number.

^ ⎛ −3 ⎞ ⎛ −3c ⎞ So the eigen vectors are X2 = ⎜ ⎟ = c ⎜ 1 ⎟ corresponding to the eigen value c ⎝ ⎠ ⎝ ⎠ λ2 = 2.

Theorem 2.12: Suppose l be an eigen value of an n × n square matrix A. Then the following hold: (1) l is also an eigen value of AT . (2) cl is also an eigen value of cA for any scalar c.

2.24

Engineering Mathematics-I

(3) ln is an eigen value of An . (4) l−1 is an eigen value of A−1 . Proof: (1) Since λ is an eigen value of an n × n square matrix A,

λ n )T = AT

we have (

λ I nT = AT

T

T

So, det[( A λ I n ) ] = det ( A

λ I n {since I nT

In }

λ In )

T

i.e., det( λ n ) = det ( A λ I n ). Since λ is an eigen value of an n × n square matrix A, det( A λ I n ) = 0. So, det( AT

λ I n ) = 0.

This proves the fact that λ is also an eigen value of AT . (2) Since λ is an eigen value of an n × n square matrix A, det( A λ I n ) = 0. Now det (

λ n ) = det [ ( A − I n ) ] = c n det (

λ n ) = 0.

This proves the fact that cλ is also an eigen value of cA for any scalar c. (3) Let us consider X be the eigen vector corresponding to the eigen value λ . Then AX = λ X . So, A2 X = A(λ X ) = λ ( AX ) = λ (λ X ) i.e., A2 X = λ 2 X . Therefore λ 2 is an eigen value of A2 . Again A3 X = A( A2 X ) = A(λ 2 X ) =

2

( AX A ) = λ 2 (λ X )

i.e., A3 X = λ 3 X . So, λ 3 is an eigen value of A3 . Proceeding in the similar manner we have An X = λ n X . This proves the fact that λ n is an eigen value of An . (4) Let us consider X be the eigen vector corresponding to the eigen value λ . Then AX = λ X . So, A 1 ( AX ) = A−11 (λ X ) ⇒ (

1

) X = λ ( A−1 X )

⇒ In X = λ( A 1 X ) ⇒ X = λ ( A−1 X ) ⇒ A−1 X = λ −1 X . This proves the fact that λ −1 is an eigen value of A−1 .

2.25

Matrix II

Theorem 2.13: For an idempotent matrix, the eigen values are either 0 or 1. Proof: Let λ be an idempotent matrix A and X be an eigen vector corresponding to λ . Then we have A2

A and AX = λ X .

So, A( AX ) = A(λ X ) ⇒ A2 X = λ ( AX ) ⇒ AX = λ (λ X ) ⇒ AX = λ 2 X . So, λ

λ2 X

⇒ (λ 2 − λ ) X = O ⇒ (λ 2 − λ ) = 0 ⇒ λ ( λ − 1) = 0 ⇒ λ = 0, 1. This proves the theorem. The following topic is included as advanced reading for interested students.

2.14

DIAGONALISATION OF A SQUARE MATRIX

2.14.1

Similar Matrices

Any matrix A of order n is said to be similar to another matrix B of the same order if there exists a nonsingular n × n matrix P such that B = P −1 AP. It is easy to prove that if A is similar to B then B also is similar to A and vice-versa. In this case, we say two matrices A and B of the same order are similar. Theorem 2.14: Two similar matrices have the same eigen values. Proof: Let A and B are two similar matrices of order n. Then for a n × n nonsingular matrix P, we have B = P −1 AP. Now the characteristic polynomial of B is det ( B

(

I n ) = det P 1 AP λ I n

Again P

1

(

In ) P

)

(

...(1)

)

P −1λ ( I n P ) = λ P P = λ I n .

So using the above result, from (1) det (

λ n)

det ⎡⎣ P 1 AP P

1

(

I n ) P ⎤⎦

2.26

Engineering Mathematics-I

= det ⎡⎣ P

1

= det

1

(A−

I n ) P ⎤⎦

( ) det ( A −

Therefore, det (

I n ) det P

(

)

det P 1 P ⋅ det ( A

λ n)

= det (

n

)

In )

det ( A − I n ) = det ( A − I n ) .

Since A and B have the same characteristic polynomial, they have the same characteristic equations and correspondingly they have the same eigen values. Note: The converse of the above theorem is not always true, i.e., the matrices having the same eigen values need not be always similar. Definition: A matrix A of order n × n is said to be diagonalisable if and only if A is similar to an n × n diagonal matrix. i.e., if D = P −1 AP, where P is an n × n nonsingular matrix and n × n diagonal matrix D is given by ⎛ λ1 0 ⎜0 λ 2 ⎜ D=⎜ 0 0 ⎜… … ⎜⎜ ⎝0 0

0 0 λ3 … 0

… 0 ⎞ … 0 ⎟ ⎟ … 0 ⎟. … …⎟ ⎟ … λ n ⎟⎠

Since A and P 1 AP have the same eigen values and also the eigen values of any diagonal matrix are its diagonal elements, we can say that λ 1 λ 2 , λ 3 , , λ n are the distinct eigen values of the matrix A. Theorem 2.15: A matrix A of order n × n is said to be diagonalisable if and only if there exist n eigen vectors of A which are linearly independent. Proof: Beyond the scope of the book. Theorem 2.16: If the eigen values of a matrix A of order n × n are all distinct and real, then A is diagonalisable. Proof: Beyond the scope of the book.

2.14.2

Steps for Diagonalisation of any Square Matrix

Here we consider a square matrix A of order 3 3 and which has distinct eigen values. Step (1) Find all three distinct eigen values of A. Suppose they are λ 1 λ 2 , λ 3 . Step (2) Find all three eigen vectors of A corresponding to λ 1 λ 2 , λ 3 . Suppose ⎛ x1 ⎞ ⎛ y1 ⎞ ⎛ z1 ⎞ ⎜ x ⎟, ⎜ y ⎟, ⎜ z ⎟ ⎜ 2⎟ ⎜ 2⎟ ⎜ 2⎟ ⎝ x3 ⎠ ⎝ y3 ⎠ ⎝ z3 ⎠

2.27

Matrix II

are the eigen vectors corresponding to λ 1 λ 2 , λ 3 respectively. ⎛ x1 Step (3) Form the nonsingular matrix P = ⎜ x2 ⎜ ⎝ x3

y1 y2 y3

z1 ⎞ z2 ⎟ . ⎟ z3 ⎠

⎛ λ1 ⎜ Step (4) The matrix P 1 AP is the diagonal matrix D = ⎜ 0 ⎜0 ⎝

0 λ2 0

0 ⎞ ⎟ 0 ⎟. λ 3 ⎟⎠

⎛ 1 1 −2 ⎞ Let us show that A = ⎜ −1 2 1 ⎟ is diagonalisable, and ⎜ ⎟ ⎝ 0 1 −1 ⎠ find P such that P 1 AP is a diagonal matrix. Example 23

The characteristic equation is det( A λ I 3 ) = 0. 1 λ 1 i.e., −1 2 − λ 0 1

−2 1 =0 −1 − λ

⇒ (1 − λ )(λ − 2)(λ + 1) = 0 ⇒ λ = 1, 2, − 1. So, the eigen values are 1, 2, and −1. ⎛ x1 ⎞ Let X = ⎜ x2 ⎟ be the eigen vector corresponding to λ = 1. ⎜ ⎟ ⎝ x3 ⎠ Then AX = λ X . ⎛ 1 1 −2 ⎞ ⎛ x1 ⎞ ⎛ x1 ⎞ 1 ⎟ ⎜ x2 ⎟ = 1 ⎜ x2 ⎟ ⇒ ⎜ −1 2 ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ 0 1 −1 ⎠ ⎝ x3 ⎠ ⎝ x3 ⎠ i.e., x1 + x2 − 2 x3 = x1 − x1 + 2 x2 + x3 = x2 x2 x3 = x3 i.e., x2 2 x3 = 0 − x1 + x2 + x3 = 0 x2 2 x3 = 0

2.28

Engineering Mathematics-I

So the system is equivalent to x2 2 x3 = 0 − x1 + x2 + x3 = 0 Now if we set x3 constant.

k1 , then x2

k1 and x1

k1 , where k1 is any arbitrary

⎛ 3k1 ⎞ ⎛3⎞ ⎜ ⎟ So the eigen vector corresponding to the eigen value λ = 1 is 2 k1 = k1 ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ k1 ⎠ ⎝1⎠ Similarly, the eigen vectors corresponding to the eigen value λ = 2 and λ = 1 are ⎛1⎞ ⎛1⎞ k2 ⎜ 3 ⎟ and k3 ⎜ 0 ⎟ respectively, where k2 , k3 are arbitrary constants. ⎜ ⎟ ⎜ ⎟ ⎝1⎠ ⎝1⎠ Since all the three eigen values of A are distinct, the eigen vectors are linearly independent and correspondingly A is diagonalisable. ⎛3 1 1⎞ So, we choose P = ⎜ 2 3 0 ⎟ , ⎜ ⎟ ⎝1 1 1⎠ Also det P = 6 ≠ 0, so P is nonsingular. ⎛ 3 Here adj ( P) = ⎜ 0 ⎜ ⎝ −3 So, P

−1

T

2 −1 ⎞ ⎛ 3 ⎟ 2 −2 = ⎜ −2 ⎜ ⎟ 2 7⎠ 1 ⎝ −1

0 −3 ⎞ 2 2⎟ ⎟ 2 7⎠

⎛ 3 0 −3 ⎞ 1⎜ = −2 2 2 ⎟ . 6 ⎜ −1 −2 7 ⎟ ⎝ ⎠

Now ⎛ 3 0 −3 ⎞ ⎛ 1 1 −2 ⎞ ⎛ 3 1 1 ⎞ 1 P 1 AP = ⎜ −2 2 2 ⎟ ⋅ ⎜ −1 2 1⎟ ⋅ ⎜ 2 3 0 ⎟ 6 ⎜ −1 −2 7 ⎟ ⎜ 0 1 −1⎟ ⎜ 1 1 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 1 0 0⎞ = ⎜0 2 0⎟ = D ⎜ ⎟ ⎝ 0 0 −1 ⎠ ⎛ 1 0 0⎞ where D = ⎜ 0 2 0 ⎟ , a diagonal matrix with the eigen values as its diagonal. ⎜ ⎟ ⎝ 0 0 −1 ⎠

2.29

Matrix II

WORKED-OUT EXAMPLES ⎛ −1 ⎜ 2 Find the rank of the matrix ⎜ ⎜⎜ 0 ⎝ 1

Example 2.1

2 −1 4 4 0 1 6 3

0⎞ 2⎟ ⎟ 5⎟ 2 ⎟⎠ [WBUT-2002, 2008].

⎛ −1 2 −1 0 ⎞ ⎜ 2 4 4 2⎟ Sol. Let A = ⎜ ⎟ 1 5⎟ ⎜⎜ 0 0 ⎟ ⎝ 1 6 3 2⎠ Applying elementary row operations on the matrix A, we have, ⎛ −1 ⎜ 2 A=⎜ ⎜⎜ 0 ⎝ 1

2 −1 4 4 0 1 6 3

0⎞ 2 ⎟ R2 2 R1 , R4 R1 ⎯⎯ → ⎟ ⎯⎯⎯⎯⎯⎯⎯ 5⎟ 2 ⎟⎠

⎛ −1 ⎜ 0 ⎜ ⎜⎜ 0 ⎝ 0

2 −1 8 2 0 1 8 2

0⎞ 2 ⎟ R4 +2 2 R2 → ⎟⎯ 5⎟ 2 ⎟⎠

⎛ −1 ⎜ 0 ⎜ ⎜⎜ 0 ⎝ 0

2 −1 8 2 0 1 0 0

0⎞ ( − ) R1 1 R2 2⎟ 2 → ⎟⎯ 5⎟ 0 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

2 4 0 0

1 1 1 0

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

2 4 0 0

0 0 1 0

5⎞ 1 R2 4⎟ ⎟ 4 5⎟ ⎯ → 0 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

2 1 0 0

0 0 1 0

5⎞ 1⎟ R1 + 2 R2 → ⎟⎯ 5⎟ 0 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

0 1 0 0

0 0 1 0

0⎞ 1⎟ R2 − R3 , R1 − R3 ⎟ ⎯⎯ ⎯⎯⎯⎯⎯→ 5⎟ 0 ⎟⎠

7⎞ 1⎟ ⎟=B 5⎟ 0 ⎟⎠

2.30

Engineering Mathematics-I

The row-reduced echelon matrix B has the 3 nonzero rows. So, rank of B = 3 and hence rank of A = 3.

Example 2.2

⎛ 1 3 4 3⎞ Find the rank of the matrix ⎜ 3 9 12 3 ⎟ ⎜ ⎟ ⎝ 1 3 4 1⎠

[WBUT-2005]

⎛ 1 3 4 3⎞ Let, A = ⎜ 3 9 12 3 ⎟ ⎜ ⎟ ⎝ 1 3 4 1⎠ Applying elementary row operations on the matrix A, we get,

Sol.

⎛ 1 3 4 3⎞ R2 3R 3R1 , R3 R1 A = ⎜ 3 9 12 3 ⎟ ⎯⎯⎯⎯⎯⎯⎯ ⎯⎯ → ⎜ ⎟ 1 3 4 1 ⎝ ⎠ ⎛1 3 4 ⎜0 0 0 ⎜ ⎝0 0 0

3 ⎞ −1 −1 R2 R3 6⎟ 6 2 → ⎟⎯ 2⎠

⎛ 1 3 4 3⎞ − R2 , 1 3R2 ⎜ 0 0 0 1 ⎟ ⎯R ⎯3 ⎯⎯⎯⎯⎯ ⎯⎯ → ⎜ ⎟ ⎝ 0 0 0 1⎠ ⎛1 3 4 0⎞ ⎜0 0 0 1⎟ = B ⎜ ⎟ ⎝0 0 0 0⎠ The row-reduced echelon matrix B has the 2 nonzero rows. So, rank of B = 2 and hence rank of A = 2.

Example 2.3

Sol.

⎛1 ⎜0 Let A = ⎜ ⎜⎜ 2 ⎝3

⎛1 ⎜0 Find the rank of the rectangular matrix ⎜ ⎜⎜ 2 ⎝3

3 0 6 9

2 4 1⎞ 2 2 0⎟ ⎟ 2 6 2⎟ 1 10 6 ⎟⎠

3 0 6 9

2 4 1⎞ 2 2 0⎟ ⎟ 2 6 2⎟ 1 10 6 ⎟⎠ [WBUT-2006]

Matrix II

Applying elementary row operations on the matrix A, we get, ⎛1 ⎜0 A=⎜ ⎜⎜ 2 ⎝3

3 0 6 9

2 4 1⎞ 2 2 0 ⎟ R3 2R 2 R1 , R4 3R1 ⎯ ⎯→ ⎟ ⎯⎯⎯⎯⎯⎯⎯ 2 6 2⎟ ⎟ 1 10 6 ⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

3 2 0 2 0 2 0 −5 5

4 2 2 2

1⎞ 1 ⎛ −1 ⎞ R , R 0⎟ 2 2 ⎝ 2 ⎠ 3 ⎯⎯ → ⎟ ⎯⎯⎯⎯⎯ ⎯ 0⎟ ⎟ 3⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

3 2 0 1 0 1 0 −5 5

4 1 1 2

1⎞ 0 ⎟ R3 − R2 →, ⎟⎯ 0⎟ 3 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

3 2 0 1 0 0 0 −5 5

4 1 0 2

1⎞ 0 ⎟ R34 ⎟⎯ → 0⎟ 3 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

3 2 0 1 0 −5 5 0 0

4 1 2 0

1⎞ 0 ⎟ R3 −5 5 R2 , R1 − 2 R2 →. ⎟ ⎯⎯ ⎯⎯⎯⎯⎯⎯ 3⎟ 0 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

3 0 0 0

0 1 0 0

2 1 3 0

1⎞ 1 0 ⎟ 3 R3 ⎟ → 3⎟ ⎯ ⎟ 0⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

3 0 0 0

0 1 0 0

2 1 1 0

1⎞ 0 ⎟ R1 −2 2 3 , R2 − R3 ⎯ ⎯⎯ →, ⎟ ⎯⎯⎯⎯⎯⎯ 1⎟ 0 ⎟⎠

⎛1 ⎜0 ⎜ ⎜⎜ 0 ⎝0

3 0 0 0

0 1 0 0

0 −1 ⎞ 0 −1 ⎟ ⎟=B 1 1⎟ 0 0 ⎟⎠

The row-reduced echelon matrix B has the 3 nonzero rows. So, rank of B = 3 and hence rank of A = 3.

2.31

2.32

Engineering Mathematics-I

Example 2.4

Using elementary row operations find the inverse of the matrix

⎛1 1 2⎞ A = ⎜ 2 4 4 ⎟. ⎜ ⎟ ⎝3 3 7⎠ Sol.

Let us apply elementary row operations on the matrix ( A | I 3 ) ⎛1 1 2 1 0 0⎞ R2 2 R1 , R3 3R1 4 4 0 1 0 ⎟ ⎯⎯⎯⎯⎯⎯⎯ ⎯ ⎯→ ⎟ 3 3 7 0 0 1 ⎝ ⎠

( A | I3 ) = ⎜⎜ 2

1 0 0⎞ 1 ⎛1 1 2 R2 ⎜ 0 2 0 −2 1 0 ⎟ ⎜ ⎟ ⎯2 → ⎝ 0 0 1 3 0 1⎠ ⎛1 1 2 ⎜ ⎜0 1 0 ⎜ ⎜0 0 1 ⎝

0 0⎞ ⎟ R −R 1 1 0⎟ ⎯ 1 2 → 2 ⎟ 3 0 1⎟⎠ 1

−1 ⎛ ⎞ 0⎟ ⎜1 0 2 2 2 ⎜ ⎟ 1 R − 2 R3 ⎜0 1 0 → 1 0⎟ ⎯ 1 ⎜ ⎟ 2 ⎜ 0 0 1 −3 0 1 ⎟ ⎜ ⎟ ⎝ ⎠ −1 ⎛ ⎞ −2 ⎟ ⎜1 0 0 8 2 ⎜ ⎟ ⎜ 0 1 0 −1 1 0 ⎟= ⎜ ⎟ 2 ⎜ 0 0 1 −3 0 1 ⎟ ⎜ ⎟ ⎝ ⎠ Therefore,

A−1

−1 ⎛ ⎞ −2 ⎟ ⎜8 2 ⎜ ⎟ 1 ⎜ = −1 0 ⎟ ⎜ ⎟ 2 ⎜ −3 0 1 ⎟ ⎜ ⎟ ⎝ ⎠

(

1 3

)

2.33

Matrix II

Example 2.5

Solve the system of equations by matrix inversion method:

x+ y−z =6 2 x 33yy + z = 1 2x 4 y + 2 = 1 Sol.

The above system of equations can be written as AX = B 1 −1 ⎞ ⎛ x ⎞ ⎛ 6 ⎞ 3 1⎟ ⎜ y ⎟ = ⎜ 1 ⎟ ⎟⎜ ⎟ ⎜ ⎟ 4 2⎠ ⎝ z ⎠ ⎝1⎠

⎛1 i.e., ⎜ 2 ⎜ ⎝2 Therefore,

1 −1 ⎞ ⎛ x⎞ ⎛6⎞ 3 1 ⎟ , X = ⎜ y ⎟ and B = ⎜ 1 ⎟ 3 ⎟ ⎜ ⎟ ⎜ ⎟ 4 2⎠ ⎝z⎠ ⎝1⎠

⎛1 A = ⎜2 ⎜ ⎝2 So,

1 −1 3 1 = −2 ≠ 0 4 2

1 det A = 2 2

Since det A ≠ 0, A−1 exists, and the system has a unique solution X = A−1 B. Now, ⎛ ⎜ ⎜ adj A = ⎜⎜ − ⎜ ⎜ ⎜ ⎝ 2 ⎛ −2 =⎜ 2 ⎜ 2 ⎝ −2

−3 −4 1 −4 1 −3

1 2 1 2 1 1

2 − 2 1 2 1 − 2

1 2 1 2 1 1

2 2 1 − 2 1 2

T

3⎞ 4⎟ ⎟ 1⎟ 4⎟ ⎟ 1⎟ 3 ⎟⎠

T

2 2⎞ 4 6⎟ ⎟ 3 −5 ⎠

⎛ −2 2 −2 ⎞ = ⎜ −2 4 −3 ⎟ ⎜ ⎟ ⎝ −2 6 −5 ⎠ Now,

A−1

⎛1 −1 ⎜ ⎛ −2 2 −2 ⎞ ⎜1 −2 adj A −1 ⎜ = = −2 4 −3 ⎟ = ⎜ det A 2 ⎜ −2 6 −5 ⎟ ⎜ ⎝ ⎠ ⎜1 −3 ⎜ ⎝

1⎞ ⎟ 3⎟ 2⎟ 5⎟ ⎟ 2⎟ ⎠

2.34

Engineering Mathematics-I

Therefore, A−1 B

X

⎛ 1 −1 1 ⎞ ⎛ ⎞ ⎜ ⎟ ⎜6⎟ 3 ⎟ ⎛6⎞ ⎜ ⎟ ⎛ x⎞ ⎜ 1 − 2 11 ⎜ y⎟ = ⎜ 2 ⎟ ⎜1⎟ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ 2 ⎟ ⎝ z ⎠ ⎜ 1 −3 5 ⎟ ⎝ 1 ⎠ ⎜ 11 1 ⎟ 2⎟ ⎜ ⎜ ⎟ ⎝ ⎠ ⎝2⎠ Therefore, the solution of the system of equations is x = 6; y =

11 11 ;z= . 2 2 ⎛1 If A = ⎜ 1 ⎜ ⎝3

Example 2.6

2 1⎞ ⎛2 ⎟ 4 1 and B = ⎜ 1 ⎟ ⎜ 0 −3 ⎠ ⎝2

1 1⎞ 1 0 ⎟ , show that AB = 6 I 3 . ⎟ 1 −1⎠

Utilize this result to solve the following system of equations 2x 2x Sol.

y

=5

x

y=0

y

=1

[WBUT-2009].

Here, ⎛1 AB = ⎜ 1 ⎜ ⎝3

2 1⎞ ⎛ 2 4 1⎟ ⎜ 1 ⎟⎜ 0 −3 ⎠ ⎝ 2

1 1⎞ 1 0⎟ ⎟ 1 −1⎠

⎛ 2 2 2 1 2 +1 1−1 ⎞ = ⎜ 2 4 + 2 1+ 4 1 1 1 ⎟ ⎜ ⎟ 3 3 3 3⎠ ⎝ 6 6 ⎛6 0 0⎞ ⎛1 0 0⎞ = ⎜ 0 6 0 ⎟ = 6 ⎜ 0 1 0 ⎟ = 6 I3 ⎜ ⎟ ⎜ ⎟ ⎝0 0 6⎠ ⎝0 0 1⎠ So, the first part is proved. Now, to solve the system 2x y =5 2x

x

y=0

y

=1

Matrix II

2.35

first we write the system in the following form ⎛2 ⎜1 ⎜ ⎝2

1 1⎞ ⎛ x ⎞ ⎛ 5 ⎞ 1 0⎟ ⎜ y ⎟ = ⎜0⎟ ⎟⎜ ⎟ ⎜ ⎟ 1 −1⎠ ⎝ z ⎠ ⎝ 1 ⎠ =C

i.e.,

...(1)

where, ⎛2 B=⎜1 ⎜ ⎝2

1 1⎞ ⎛ x⎞ ⎛5⎞ ⎟ ⎜ ⎟ 1 0 , X = y and C = ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ 1 −1⎠ ⎝z⎠ ⎝1⎠

Now from the above relation AB = 6 I 3 , we have ⎛1 ⎞ ⎜ A B = I3 ⎝6 ⎠ So, from the definition of inverse, we conclude that B −1 exists and it is given by 1 B 1 A 6 B

−1

1⎞ ⎛1 2 1⎜ = 1 −4 1⎟ . 6 ⎜ 3 0 −3 ⎟ ⎝ ⎠

Since B −1 exists, the solution of the system (1) is given by X

B −1C

⎛ x⎞ ⎛1 1 i.e., ⎜ y ⎟ = ⎜ 1 ⎜ ⎟ 6⎜ ⎝z⎠ ⎝3 =

2 1⎞ ⎛ 5 ⎞ 4 1⎟ ⎜ 0 ⎟ ⎟⎜ ⎟ 0 −3 ⎠ ⎝ 1 ⎠

⎛ 6⎞ ⎛1⎞ 1⎜ ⎟ ⎜ ⎟ 6 = 1 6 ⎜12 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠

Hence, the solution of the given system is x = 1, y = 1, z = 2. Example 2.7 2x

Solve by Cramer’s rule

y=3

3y 2z = 5 −2 + x = 4

[WBUT-2008]

2.36

Sol.

Engineering Mathematics-I

If we write the above equations in the form AX = B then ⎛2 the coefficient matrix A = ⎜ 0 ⎜ ⎝1

1 0⎞ ⎛ x⎞ ⎛ 3⎞ ⎟ ⎜ ⎟ 3 −2 , X = y and B = ⎜ 5 ⎟ . ⎟ ⎜ ⎟ ⎜ ⎟ 0 −2 ⎠ ⎝z⎠ ⎝ 4⎠ The determinant of the coefficient matrix is 2 det A = 0 1

1 0 3 −2 = 10 ≠ 0 0 −2

Therefore, the system of equations is consistent and the system has a unique solution. Now 3 det A1 = 5 4

1 0 3 −2 = 20 0 −2

2 3 0 det A2 = 0 5 −2 = 10 1 4 −2 2 det A3 = 0 1

1 3 3 5 = 10 0 4

So, the solution of the system of equations by Cramer’s rule is given by x=

det A1 −20 = = 1 det A −10

y=

det A2 −10 = =1 det A −10

z=

det A3 10 = = 1. det A −10

Example 2.8 3x

y

Solve by Cramer’s rule =4

x − y + 2z = 6 x + 2 y − z = −3

[WBUT-2009]

2.37

Matrix II

Sol.

If we write the above equations in the form AX = B then ⎛3 the coefficient matrix A = ⎜ 1 ⎜ ⎝1

1 1⎞ ⎛ x⎞ ⎛ 4⎞ 1 2 ⎟ , X = ⎜ y ⎟ and B = ⎜ 6 ⎟ . ⎟ ⎜ ⎟ ⎜ ⎟ 2 −1⎠ ⎝z⎠ ⎝ −3 ⎠ The determinant of the coefficient matrix is 3 det A = 1 1

1 1 1 2 = −3 ≠ 0 2 −1

Therefore, the system of equations is consistent and the system has a unique solution. Now, 4 det A1 = 6 −3

1 1 1 2 = 3 2 −1

3 det A2 = 1 1

4 1 6 2 =3 3 −1

3 det A3 = 1 1

1 4 1 6 = 6 2 −3

So, by Cramer’s rule, the solution of the system is given by x=

det A1 −3 = =1 det A −3

y=

det A2 3 = = 1 det A −3

z=

det A3 −6 = = 2. det A −3

Example 2.9

Investigate for what value of λ and μ the following equations

x+ y+z = 6 x + 2 y + 3z = 10 x + 2y + λz = μ have i) no solution, ii) a unique solution, and iii) an infinite number of solutions. [WBUT-2004]

2.38

Sol.

Engineering Mathematics-I

If we write the above equations in the form AX = B then the coefficient ⎛ x⎞ ⎛1 1 1 ⎞ ⎛6⎞ matrix A = ⎜1 2 3 ⎟ , X = ⎜ y ⎟ and B = ⎜10 ⎟ . ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝z⎠ ⎝1 2 λ ⎠ ⎝μ⎠

The determinant of the coefficient matrix is 1 1 1 det A = 1 2 3 1 2 λ = 1(2λ 6)) 1((λ − 3) + 1(2 2) =λ 3 Now, 6 1 1 det A1 = 10 2 3 μ 2 λ = 6(2λ 6)) 1(10 ( λ − 3μ ) + 1(20 2 μ ) = 2λ μ − 16 1 6 1 det A2 = 1 10 3 1 μ λ = 1(10λ 3μ ) 6(λ 3) 1( μ − 10) = 4λ 2 μ 8 1 1 6 det A3 = 1 2 10 1 2 μ = 1(2 μ 20) 1( μ − 10) = μ 10 [Note: The following cases will be discussed according to the observations of Section 3.8] Case (i): The system of equations have no solution when det A = 0 ⇒ λ = 3 and at least one of det A1 , det A2 , det A3 is nonzero, i.e., when λ = 3 and at least one of 2λ μ 16 ≠ 0, 4λ 2 μ + 8 ≠ 0, μ − 10 ≠ 0 i.e., when λ = 3 and μ ≠ 10.

2.39

Matrix II

Case (ii): The system of equations have a unique solution when det A ≠ 0, i.e., when λ ≠ 3. Case (iii): The system of equations have an infinite number of solutions when det A = 0 ⇒ λ = 3 and det A1 = det

2

det A3 = 0

i.e., when λ = 3 and 2λ μ 16 = 0, 4λ 2 μ + 8 = 0, μ − 10 = 0 i.e., when λ = 3 and μ = 10. Example 2.10

Determine the nature of the solution without solving the

homogeneous system of equations: x + y + 3z = 0 2x

y

=0

3x 2 y + 4 = 0 Sol.

1 1 3 The determinant of the coefficient matrix A is det A = 2 1 1 3 2 4 = 1(4 − 2) − 1(8 3) 3(4 − 3) = 2 5 3 = 0. Since the determinant of the coefficient matrix is zero for the given homogeneous system of equations, the system has infinitely many nontrivial solutions.

Example 2.11

Solve by the consistency of the following system of equations and

solve if possible x+ y+ z =1 2x

y 2z = 2

3x 2 y + 3 = 5 Sol.

[WBUT-2006, 2008]

The system of linear equations can be written in matrix form as AX = B ⎛1 1 1⎞ ⎛ x ⎞ ⎛1⎞ i.e., ⎜ 2 1 2 ⎟ ⎜ y ⎟ = ⎜ 2 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ 3 2 3⎠ ⎝ z ⎠ ⎝ 5⎠

The coefficient matrix of the system of equations is ⎛1 1 1⎞ A = ⎜ 2 1 2⎟ ⎜ ⎟ ⎝ 3 2 3⎠

2.40

Engineering Mathematics-I

and the augmented matrix is ⎛1 1 1 1⎞ A = ⎜ 2 1 2 2 ⎟. ⎜ ⎟ ⎝ 3 2 3 5⎠ Applying elementary row operations on the augmented matrix A, we have, ⎛1 1 1 1⎞ R2 2R 2 R1 , R3 3R1 A = ⎜ 2 1 2 2 ⎟ ⎯⎯⎯⎯⎯⎯⎯⎯ → ⎜ ⎟ ⎝ 3 2 3 5⎠ ⎛1 ⎜0 ⎜ ⎝0

1 1 1⎞ R3 − R2 , R1 + R2 1 0 0 ⎟ ⎯⎯⎯⎯⎯⎯⎯ → ⎟ 1 0 2⎠

⎛1 ⎜0 ⎜ ⎝0

1 0 1 1⎞ ( −1) 2 , R3 2 → 1 0 0 ⎟ ⎯⎯⎯⎯⎯⎯ ⎟ 0 0 2⎠

⎛1 0 1 1⎞ ⎜ 0 1 0 0 ⎟ ⎯R1 − R3 → ⎜ ⎟ ⎝0 0 0 1⎠ ⎛1 0 1 0⎞ ⎜0 1 0 0⎟ ⎜ ⎟ ⎝0 0 0 1⎠ Here, rank of A is 3 and rank of A is 2. Since rank of A ≠ rank of A, the given system of equations is inconsistent. In other words, the system does not have any solution. Example 2.12

For what value of k do the following equations

x+ y+ z =1 2x 4

y 4z = k y 10 z = k 2 have solutions? Solve them completely in each case. [WBUT-2003]

Sol.

The system of linear equations can be written in matrix form as AX = B ⎛ 1 1 1⎞ ⎛ x ⎞ ⎛ 1 ⎞ i.e., ⎜ 2 1 4 ⎟ ⎜ y ⎟ = ⎜⎜ k ⎟⎟ ⎜ ⎟⎜ ⎟ ⎝ 4 1 10 ⎠ ⎝ z ⎠ ⎜⎝ k 2 ⎟⎠

2.41

Matrix II

⎛ 1 1 1⎞ The coefficient matrix is A = ⎜ 2 1 4 ⎟ and the augmented matrix is ⎜ ⎟ ⎝ 4 1 10 ⎠ ⎛1 1 1 1 ⎞ A = ⎜⎜ 2 1 4 k ⎟⎟ . ⎜ 4 1 10 k 2 ⎟ ⎝ ⎠ Applying elementary row operations on the augmented matrix A, we have ⎛1 1 1 1 ⎞ R2 2R 2 R1 , R3 4 R1 → A = ⎜⎜ 2 1 4 k ⎟⎟ ⎯⎯⎯⎯⎯⎯⎯⎯ ⎜ 4 1 10 k 2 ⎟ ⎝ ⎠ ⎛1 ⎜0 ⎜ ⎜0 ⎝

1 1 1 ⎞ R −3 3 2 , R1 + R2 1 2 k 2 ⎟⎟ ⎯⎯3 ⎯⎯⎯⎯⎯ ⎯⎯ → 2 ⎟ 3 6 k 4⎠

⎛1 ⎜0 ⎜ ⎜0 ⎝

0 3 k 1 ⎞ 1 2 k 2 ⎟⎟ 0 0 k 2 3k + 2 ⎟⎠

⎛1 The coefficient matrix is equivalent to the matrix ⎜ 0 ⎜ ⎝0 Therefore, the rank of A is 2.

0 3⎞ 1 2 ⎟. ⎟ 0 0⎠

The system of equations have solutions if rank A = rank A. The matrix A has rank 2 if and only if k 2

k 2 = 0 ⇒ k = 1, 2.

For, k = 1, the system of equations becomes ⎛1 ⎜0 ⎜ ⎝0

0 3⎞ ⎛ x ⎞ ⎛ 0 ⎞ 1 2 ⎟ ⎜ y ⎟ = ⎜ −1⎟ ⎟⎜ ⎟ ⎜ ⎟ 0 0⎠ ⎝ z ⎠ ⎝ 0⎠

The above system is equivalent to x + 3z = 0 − y + 2 z = −1 Putting z = k1 , we get x constant.

k1 + 1, where k1 is an arbitrary

k1 and y

So the solution is ( x, y, z ) = ( 3k1 , 2 solutions is infinite.

1,

). In this case, the number of

2.42

Engineering Mathematics-I

For k

, the system of equations becomes

⎛ 1 0 3⎞ ⎛ x ⎞ ⎛ 1 ⎞ ⎜0 1 2⎟ ⎜ y ⎟ = ⎜0⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝0 0 0⎠ ⎝ z ⎠ ⎝0⎠ The above system is equivalent to x + 3z = 1 − y + 2z = 0 Putting z = k2 , we get x constant.

k2 and y

k2 , where k2 is an arbitrary

So the solution is ( x, y, z ) = ( 3k2 , 2k2 , k2 ). In this case also, the number of solutions is infinite. Determine the values of a and b so that the system of equations

Example 2.13 2 x 3y 3y + 4 = 9 x 2 y + az a =5 3x 4 y + 7 z = b

have i) a unique solution, ii) many solutions, and iii) no solution. Sol.

If we write the system of linear equations in the matrix form as AX = B then the coefficient matrix of the system of linear equations is 3 4⎞ ⎛2 A ⎜ a⎟ ⎜ ⎟ ⎝ 3 4 7⎠ and the augmented matrix is ⎛2 ⎜ ⎜ ⎝3

3 4 9⎞ a 5⎟ ⎟ 4 7 b⎠ The system of equations have unique solution when the determinant of the coefficient matrix is not equal to zero. A

2 det A 3

3 4 a 4 7

= 2( 14 4a)) 3(7 − 3 ) + 4(4 6) = a − 9

Matrix II

2.43

Therefore, for det A ≠ ⇒ a ≠ 9 the system of equations have a unique solution. When a = 9, the augmented matrix becomes 3 4 9⎞ ⎛2 A = ⎜ 1 −2 9 5 ⎟ ⎜ ⎟ ⎝ 3 4 7 b⎠ Applying elementary row operations on the matrix A, we have 3 4 9⎞ ⎛2 R A = ⎜ 1 −2 9 5 ⎟ ⎯ 12 → ⎜ ⎟ ⎝ 3 4 7 b⎠ ⎛ 1 −2 9 5 ⎞ R2 −2 2 1 , R3 + 3R1 ⎜2 3 4 9 ⎟ ⎯⎯⎯⎯⎯⎯⎯⎯ → ⎜ ⎟ 3 4 7 b ⎝ ⎠ 5 ⎞ ⎛1⎞ ⎛1 2 9 ⎜ 0 7 14 −1 ⎟ ⎜⎝ 7 ⎟⎠ R2 ⎜ ⎟⎯ → ⎝ 0 10 20 b 15 ⎠ 5 ⎞ ⎛1 2 9 ⎜ −1 ⎟ R1 +2 2 2 , R3 −10 R2 ⎯⎯ → ⎜0 1 2 ⎟ ⎯⎯⎯⎯⎯⎯⎯⎯ 7 ⎟ ⎜ ⎜ 0 10 20 b 15 ⎟ ⎝ ⎠ 33 ⎛ ⎞ ⎜ 1 0 13 ⎟ 7 ⎜ ⎟ −1 ⎜0 1 2 ⎟ ⎜ ⎟ 7 ⎜ 10 ⎟ ⎜ 0 0 0 b 15 + ⎟ 7⎠ ⎝ The system of equations is consistent when rank A = rank A and this is possible for b − 15 +

10 =0 7

i.e., b =

95 . 7

In this case, rank A = rank A = 2, which is less then the number of unknowns (= 3) and the system has infinitely many solutions. Again, if b − 15 +

10 95 ≠0⇒b≠ . 7 7

2.44

Engineering Mathematics-I

then rank A = 2 and rank A = 3, i.e., rank A ≠ rank A and so the system of equations is inconsistent and correspondingly, the system has no solution. Summarizing the above, we have the system of equations has a unique solutions when a ≠ 9 the system of equations has infinitely many solutions when a = 9 95 and b = 7 95 iii) the system of equations has no solution when a = 9 and b ≠ 7 i) ii)

Example 2.14

Solve the system of equations

x + 2 y + z 3u = 1 2x 4 y + 3

=3

3 x 6 y + 4 z 2u = 4 if possible . Sol.

The coefficient matrix of the system of equations is ⎛1 2 1 A = ⎜2 4 3 ⎜ ⎝3 6 4

3⎞ 1⎟ ⎟ 2⎠

and the augmented matrix is ⎛1 2 1 A = ⎜2 4 3 ⎜ ⎝3 6 4

3 1⎞ 1 3⎟ ⎟ 2 4⎠ Applying elemetary row operation on the augmented matrix A, we have ⎛1 2 1 A = ⎜2 4 3 ⎜ ⎝3 6 4 ⎛1 2 1 ⎜0 0 1 ⎜ ⎝0 0 1

3 1⎞ R 2R 2 R1 , R3 3R1 1 3 ⎟ ⎯⎯2 ⎯⎯⎯⎯⎯ ⎯ ⎯→ ⎟ 2 4⎠

3 1⎞ R3 − R2 , R1 − R2 7 1⎟ ⎯⎯⎯⎯⎯⎯ ⎯ ⎯→ ⎟ 7 1⎠

⎛ 1 2 0 10 0 ⎞ ⎜0 0 1 7 1⎟ ⎜ ⎟ 0 0 0 0 0⎠ ⎝ Here, the equivalent matrix has two nonzero rows and rank A = rank A = 2. So the system of equations is consistent.

2.45

Matrix II

Since, rank A = rank A = 2 < Number of unknowns ( = 4 ), the system of equations has infinitely many solutions. The equivalent system of equations becomes ⎛1 2 0 ⎜0 0 1 ⎜ ⎝0 0 0

⎛ x⎞ 10 ⎞ ⎜ ⎟ ⎛ 0 ⎞ y 7⎟ ⎜ ⎟ = ⎜1⎟ ⎟⎜z⎟ ⎜ ⎟ 0⎠ ⎜ ⎟ ⎝0⎠ ⎝u⎠

or, x 2 y − 10u = 0 z + 7u = 1 Taking, u = k1 , y k2 , we have x k2 are arbitrary constants.

k1 2k2 , z

7 k1 , where k1 and

Hence, the solution is given by ( x, y, z ) = (10 ( k1 − 2k2 , k2 , 1 − 7 k1 , k1 ) where k1 and k2 are arbitrary constants. ⎛2 2 1⎞ If A = ⎜ 1 3 1 ⎟ find all eigen values of A and obtain all the ⎜ ⎟ ⎝1 2 2⎠ eigen vectors corresponding to its eigen values. [WBUT-2004] Example 2.15

Sol.

The characteristic equation of the matrix A is det( A λ I 3 ) = 0 or,

2 λ 2 1 3−λ 1 2

1 1 =0 2−λ

or, (2 λ ){(3 ){( − λ ))(2 − λ ) 2} 2{(2 λ ) − 1} + {2 − (3 − λ ))} = 0 or, (2 λ ){6 ){ − 5λ λ 2

2} 2(1 − λ ) (−1 λ ) = 0

or, (2 λ )(λ 2 5λ + 4)) − 3 3λ = 0 or, 2λ 2 10λ + 8 λ 3 5λ 2 − 4λ 3 3λ = 0 or, λ 3 7λ 2 + 11λ − 5 = 0

2.46

Engineering Mathematics-I

or, (λ 1)( )(λ 2 6λ + 5) = 0 or, (λ 1)( )(λ 5)(λ − 1) = 0 Therefore, the eigen values are λ = 1, 1, 5. ⎛ x⎞ Let X1 = ⎜ y ⎟ be the eigen vector corresponding to the eigen value λ = 1. ⎜ ⎟ ⎝z⎠ Therefore, we have, AX1 = 1⋅ X 1 ⎛2 2 1⎞ ⎛ x ⎞ ⎛ x ⎞ or, ⎜ 1 3 1 ⎟ ⎜ y ⎟ = ⎜ y ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝1 2 2⎠ ⎝ z ⎠ ⎝ z ⎠ or, 2 x + 2 y z = x x + 3y + z = y x + 2 y + 2z

z

or, x 2 y + z = 0 x + 2y + z = 0 x + 2y + z = 0 So the above system is equivalent to x + 2 y + z = 0. Let y k1 and z k2 then x k1 − k2 where k1 and k2 are arbitrary constants. Therefore, the eigen vector corresponding to the eigen value λ = 1 is given by ⎛ x ⎞ ⎛ −2k1 k2 ⎞ ⎟ X1 = ⎜ y ⎟ = ⎜ k1 ⎜ ⎟ ⎜ ⎟ k2 ⎝z⎠ ⎝ ⎠ ⎛ −2 ⎞ ⎛ −1⎞ = k1 ⎜ 1⎟ k2 ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0⎠ ⎝ 1⎠ ⎛ x⎞ Let X 2 = ⎜ y ⎟ be the eigen vector corresponding to the eigen value λ = 5. ⎜ ⎟ ⎝z⎠

2.47

Matrix II

Therefore, we have, AX 2 = 5 X 2 ⎛2 2 1⎞ ⎛ x ⎞ ⎛ x⎞ or, ⎜ 1 3 1 ⎟ ⎜ y ⎟ = 5 ⎜ y ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝1 2 2⎠ ⎝ z ⎠ ⎝z⎠ 2 x + 2 y z = 5x

or,

x + 3y + z = 5y x + 2 y + 2 z = 5z 3 + 2y 3x

or,

=0

x 2y + z = 0 x + 2 y − 3z = 0 Here, the determinant of the coeficient matrix is Δ=

−3 1 1

2 1 2 1 =0 2 −3

Therefore, the system of homogeneous equations have nontrivial solutions. The solutions are x y z = = =k 4 4 4 Therefore, the eigen vector corresponding to λ = 5 is ⎛ x ⎞ ⎛ 4k ⎞ ⎛1⎞ X 2 = ⎜ y ⎟ = ⎜ 4k ⎟ = 4k ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 4k ⎠ ⎝1⎠ ⎛0 Verify that the matrix A = ⎜ 1 ⎜ ⎝2 −1 istic equation. If possible, find A . Example 2.16

Sol.

1 2⎞ 0 3 ⎟ satisfies its own character⎟ 3 0⎠ [WBUT-2002]

The characteristic equation of the matrix A is det( A λ I ) = 0 or,

−λ 1 2

−1 2 λ 3 =0 3 −λ

2.48

Engineering Mathematics-I

⇒ −λ (λ 2 − 9) + 1(−λ − 6) + 2(3 + 2λ ) = 0 ⇒ −λ 3 + λ − λ − + + λ = 0 ⇒ −λ 3 + ⇒ λ3 −

λ=0 λ = 01

Now, ⎛0 A2 = ⎜ 1 ⎜ ⎝2 A3

1 2⎞ ⎛ 0 0 3⎟ ⎜ 1 ⎟⎜ 3 0⎠ ⎝ 2

⎛3 A2 A = ⎜ 6 ⎜ ⎝3

⎛ 0 = ⎜ 12 ⎜ ⎝ 24 A3

6 −3 ⎞ 8 2⎟ ⎟ 2 13 ⎠

1 2⎞ ⎛ 3 0 3⎟ = ⎜ 6 ⎟ ⎜ 3 0⎠ ⎝ 3

6 −3 ⎞ ⎛ 0 8 2⎟ ⎜ 1 ⎟⎜ 2 13 ⎠ ⎝ 2

1 2⎞ 0 3⎟ ⎟ 3 0⎠

12 24 ⎞ 0 36 ⎟ ⎟ 36 0 ⎠ ⎛ 0 A = ⎜ 12 ⎜ ⎝ 24

12 24 ⎞ ⎛0 0 36 ⎟ − 12 ⎜ 1 ⎟ ⎜ 36 0 ⎠ ⎝2

1 2⎞ 0 3⎟ ⎟ 3 0⎠

⎛0 0 0⎞ = ⎜0 0 0⎟ = O ⎜ ⎟ ⎝0 0 0⎠ Since we have A3

A = O, the matrixA satifies its characteristic equation (1).

Again from (1), we have λ (λ 2 − 12) = 0 which implies λ = 0 is an eigenvalue, i.e., the matrix A is singular. Therefore, A−1 does not exist.

Example 2.17

⎛1 If A = ⎜ 0 ⎜ ⎝0

0 2⎞ 1 1⎟ then verify that A satisfies its own characteristic ⎟ 1 0⎠

equation. Hence, find A−1 and A9 . Sol.

The characteristic equation of the matrix A is det( A λ I ) = 0

[WBUT-2007, 2008]

2.49

Matrix II

1 λ or, 0 0

0 1− λ 1

2 1 =0 −λ

or, λ 3 2λ + 1 = 0. Now ⎛1 A = ⎜0 ⎜ ⎝0

0 2⎞ ⎛ 1 1 1⎟ ⎜ 0 ⎟⎜ 1 0⎠ ⎝0

2

⎛1 A A = ⎜0 ⎜ ⎝0

3

2

A

0 2⎞ ⎛ 1 1 1⎟ = ⎜ 0 ⎟ ⎜ 1 0⎠ ⎝0

2 2⎞ ⎛ 1 2 −1⎟ ⎜ 0 ⎟⎜ 1 1⎠ ⎝ 0

2 2⎞ 2 −1⎟ ⎟ 1 1⎠

0 2⎞ ⎛ 1 1 1⎟ = ⎜ 0 ⎟ ⎜ 1 0⎠ ⎝0

0 4⎞ 3 2⎟ ⎟ 2 −1⎠

Therefore, A3

A+ I

⎛1 = ⎜0 ⎜ ⎝0

0 4⎞ ⎛ 1 3 2⎟ − 2⎜ 0 ⎟ ⎜ 2 −1⎠ ⎝ 0

0 2⎞ ⎛1 0 0⎞ 1 1⎟ + ⎜ 0 1 0 ⎟ ⎟ ⎜ ⎟ 1 0⎠ ⎝0 0 1⎠

⎛0 0 0⎞ = ⎜0 0 0⎟ = O ⎜ ⎟ ⎝0 0 0⎠ Since A3 A I O the matrix A satisfies its own characteristic equation. Now, from (1), we have A3

A I

O

or, ( A2 − 2 ) = I i.e.,

⎡ ( A2 − 2 ) ⎤ = I ⎣ ⎦

So, from the definition of inverse, we have A

1

i.e., ⎛1 = ⎜0 ⎜ ⎝0

( A2 − 2 I ) 1

2 I − A2

⎛1 0 0⎞ ⎛ 1 = 2⎜ 0 1 0⎟ − ⎜ 0 ⎜ ⎟ ⎜ ⎝0 0 1⎠ ⎝0 2 −2 ⎞ 0 1⎟ ⎟ 1 1⎠

2 2⎞ 2 −1⎟ ⎟ 1 1⎠

...(1)

2.50

Engineering Mathematics-I

Again, from (1), we obtain A3

A I

O

or, A3 = 2 A − I or, A9 = (

3 3

)3 = 8A 8 A3 12 A2 + 6A 6A I

) = (2

= 8(2 A I ) 12 ⎛1 or, A9 = 22 ⎜ 0 ⎜ ⎝0 ⎛1 = ⎜0 ⎜ ⎝0

2

6 A − I = 22 A 12 A2 9 I

0 2⎞ ⎛1 1 1⎟ − 12 ⎜ 0 ⎟ ⎜ 1 0⎠ ⎝0 24 55 34

2 2⎞ ⎛1 0 0⎞ 2 −1⎟ − 9 ⎜ 0 1 0 ⎟ ⎟ ⎜ ⎟ 1 1⎠ ⎝ 0 0 1 ⎠

20 ⎞ 34 ⎟ ⎟ 21⎠

⎛ 0 0 1⎞ Show that the matrix A = ⎜ 3 1 0 ⎟ satisfies the Cayley– Example 2.18 ⎜ ⎟ ⎝ −2 1 4 ⎠ Hamilton theorem. [WBUT-2007] Sol.

The characteristic equation of the matrix is det( A λ I ) = 0 −λ 0 or, 3 1 − λ −2 1

1 0 =0 4−λ

or, (−λ ){(1 ){( − λ ))(4 − λ ))} + {3 + 2(1 λ )} ) =0 or, (−λ )(λ 2 − 5λ 4)) (5 ( 2λ ) = 0 or, λ 3 5λ 2 − 4λ − 2λ + 5 = 0 or, λ 3 5λ 2 + 6λ 5 = 0 Now, ⎛ 0 0 1⎞ ⎛ 0 0 1⎞ ⎛ −2 1 4 ⎞ A2 = ⎜ 3 1 0 ⎟ ⎜ 3 1 0 ⎟ = ⎜ 3 1 3 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ −2 1 4 ⎠ ⎝ −2 1 4 ⎠ ⎝ 5 5 14 ⎠ A3

⎛ −2 1 4 ⎞ ⎛ 0 0 1⎞ ⎛ −5 5 14 ⎞ A2 A = ⎜ 3 1 3 ⎟ ⎜ 3 1 0 ⎟ = ⎜ −3 4 12 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ 5 5 14 ⎠ ⎝ −2 1 4 ⎠ ⎝ −13 119 61⎠

2.51

Matrix II

So, A3 5 A2

6 A 5I

⎛ −5 5 14 ⎞ ⎛ −2 1 4 ⎞ ⎛ 0 0 1⎞ ⎛ 1 0 0 ⎞ = ⎜ −3 4 12 ⎟ − 5 ⎜ 3 1 3 ⎟ +6 ⎜ 3 1 0 ⎟ − 5 ⎜ 0 1 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −13 19 61⎠ ⎝ 5 5 14 ⎠ ⎝ −2 1 4 ⎠ ⎝ 0 0 1 ⎠ ⎛ 0 0 0⎞ = ⎜ 0 0 0⎟ = 0 ⎜⎝ 0 0 0⎟⎠ Therefore, the matrix A satisfies its own characteristic equation. Example 2.19 ⎛1 matrix ⎜ 2 ⎜ ⎝3 Sol.

Find the eigen values and corresponding eigen vectors of the

1 2⎞ 2 4⎟ ⎟ 3 6⎠

[WBUT-2008].

⎛1 Let A = ⎜ 2 ⎜ ⎝3

1 2⎞ 2 4⎟ ⎟ 3 6⎠

The characteristic equation of A is det ( A or

1 λ 2 3

I) = 0 1 2 2−λ 4 =0 3 6−λ

or, (1 λ ){(−2 − λ ))(6 − λ ) + 12} + 1{2(6 λ ) − 12} + 2{−6 3(2 + λ ))}} = 0 or, (1 λ ){ 12 4λ + λ 2 + 12} + {12 2λ − 12} + 2{−6 6 + 3λ} = 0 or, (1 λ )(λ 2

4λ ) 4λ = 0

or, λ{(1 {( λ )( ) λ − 4) + 4} = 0 or, λ{λ − λ 2

4λ 4 + 4} = 0

or, λ 2 (λ − 5) = 0 Therefore, the eigen values of the matrix A are λ = 0, 0, 5. ⎛ x⎞ Let X1 = ⎜ y ⎟ be the eigen vector corresponding to the eigen value λ = 0. ⎜ ⎟ ⎝z⎠

2.52

Engineering Mathematics-I

Therefore, we have AX = 0 ⋅ X ⎛1 or, ⎜ 2 ⎜ ⎝3

1 2⎞ ⎛ x ⎞ ⎛0⎞ 2 4⎟ ⎜ y ⎟ = ⎜0⎟ ⎟⎜ ⎟ ⎜ ⎟ 3 6⎠ ⎝ z ⎠ ⎝0⎠ y 2z = 0

or,

2x 2 y + 4 = 0 3 x 3y 3y + 6 = 0 The above system is equivalent to x − y + 2z = 0 Let y k1 and z = k2 , then x constants.

k1 2k2 where k1 and k2 are arbitrary

Therefore, the eigen vector corresponding to the eigen value λ = 0 ⎛ x ⎞ ⎛ k1 k2 ⎞ X = ⎜ y ⎟ = ⎜ k1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ k2 ⎠ ⎛1⎞ ⎛ −2 ⎞ = k1 ⎜ 1 ⎟ k2 ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝ 1⎠ ⎛ x⎞ Let X 2 = ⎜ y ⎟ be the eigen vector corresponding to the eigen value λ = 5. ⎜ ⎟ ⎝z⎠ Therefore, we have AX 2 = 5 X 2 ⎛1 or, ⎜ 2 ⎜ ⎝3 or,

1 2⎞⎛ x ⎞ ⎛ x⎞ ⎟ ⎜ ⎟ 2 4 y = 5⎜ y ⎟ ⎟⎜ ⎟ ⎜ ⎟ 3 6⎠⎝ z ⎠ ⎝z⎠ x − y + 2 z = 5x

2x 2 y + 4z = 5 y 3 x 3y 3 y + 6 = 5z or, 4 x − y + 2 = 0 2x 7 y + 4 = 0 3 x 3y 3y + z = 0

Matrix II

2.53

This is a homogeneous system and the determinant of the coeficient matrix is −4 −1 2 2 −7 4 = 0. 3 −3 1 Therefore, the system of homogeneous equations has nontrivial solutions and the solutions are x y z = = = k, 10 20 30 where k is any arbitrary constant. Therefore the eigen vector corresponding to the eigen value λ = 5 ⎛ x ⎞ ⎛ 10k ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ X 2 = y = 20k = 10k ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 30k ⎠ ⎝ 3⎠ ⎛5 4⎞ Determine the eigen vectors of A = ⎜ ⎟ and then diagonalise ⎝1 2⎠ A with the help of the basis of eigen vectors. [WBUT-2003] Example 2.20

Sol.

The characteristic equation of the matrix A is det ( A or,

5 λ 1

I) = 0 4 =0 2−λ

or, (5 − λ )(2 )( − λ ) 4 = 0 or, λ 2 7λ + 6 = 0 or, (λ 1)( )(λ 6) = 0 So, the eigen values of the matrix A are λ = 1, 6. ⎛ x⎞ Now let X1 = ⎜ ⎟ be an eigen vector corresponding to the eigen value λ = 1, ⎝ y⎠ then AX1 = 1⋅ X 1 ⎛5 4⎞⎛ x ⎞ ⎛ x ⎞ or, ⎜ ⎟⎜ ⎟ = ⎜ ⎟ ⎝1 2⎠⎝ y ⎠ ⎝ y ⎠ or, 4 x + 4 y = 0 x+ y = 0

2.54

Engineering Mathematics-I

The above system is equivalent to x+ y = 0 Taking x = k1 , we have y

k1 , where k1 is any arbitrary constant.

Therefore, the eigen vector corresponding to the eigen value λ = 1 ⎛ x⎞ ⎛ k ⎞ ⎛ 1⎞ X1 = ⎜ ⎟ = ⎜ 1 ⎟ = k1 ⎜ ⎟ . ⎝ y ⎠ ⎝ −k1 ⎠ ⎝ −1⎠ ⎛ x⎞ Again let X 2 = ⎜ ⎟ be an eigen vector corresponding to the eigen value ⎝ y⎠ λ = 6, then AX 2 = 6 X 2 ⎛ 5 4 ⎞ ⎛ x ⎞ ⎛ 6x ⎞ or, ⎜ ⎟⎜ ⎟ = ⎜ ⎟ ⎝1 2⎠⎝ y ⎠ ⎝6y ⎠ or,

x 4y = 0 x 4y = 0

The above is equivalent to x 4y = 0 Taking y k2 , we have x k2 , where k2 is any arbitrary constant. Therefore, the eigen vector corresponding to the eigen value λ = 6 ⎛ x ⎞ ⎛ 4k ⎞ ⎛ 4⎞ X 2 = ⎜ ⎟ = ⎜ 2 ⎟ = k2 ⎜ ⎟ . ⎝ y ⎠ ⎝ k2 ⎠ ⎝1⎠ Since both the two eigen values of A are distinct, the eigen vectors are linearly independent and correspondingly A is diagonalisable. ⎛ 1 4⎞ So, we choose P = ⎜ ⎟, ⎝ −1 1⎠ Also, det P = 5 ≠ 0, so P is nonsingular. T

⎛ 1 1⎞ ⎛1 Here, adj ( ) = ⎜ ⎟ = ⎜1 − 4 1 ⎝ ⎠ ⎝

4⎞ 1⎟⎠

1 ⎛1 −4 ⎞ So, P −1 = ⎜ ⎟. 5 ⎝1 1⎠ Now 1 ⎛1 −4 ⎞ ⎛ 5 4 ⎞ ⎛ 1 4 ⎞ P 1 AP = ⎜ ⎟⋅⎜ ⎟⋅⎜ ⎟ 5 ⎝1 1⎠ ⎝ 1 2 ⎠ ⎝ −1 1⎠ ⎛1 0⎞ =⎜ ⎟=D ⎝0 6⎠ ⎛1 0⎞ where D = ⎜ ⎟ , a diagonal matrix with the eigen values as its diagonal. ⎝0 6⎠

2.55

Matrix II

EXERCISES

Short and Long Answer Type Questions 1. Find the rank of the following matrices: ⎛ 5 4 5⎞ (a) ⎜ 4 5 7 ⎟ ⎜ ⎟ ⎝ 5 7 10 ⎠ [Ans : Rank is 2] 2 −1 1 2 2 3 1 1

⎛1 ⎜3 (b) ⎜ ⎜⎜ 2 ⎝1

3⎞ 1⎟ ⎟ 2⎟ 1⎟⎠ [Ans : Rank is 3]

⎛ 2 1 0 4⎞ (c) ⎜ 1 3 2 1⎟ ⎜ ⎟ ⎝ −1 2 4 3 ⎠ ⎛0 ⎜1 (d) ⎜ ⎜⎜ 2 ⎝3

0 3 6 9

1 1 4 4

[Ans : Rank is 3]

2 1⎞ 0 3⎟ ⎟ 2 8⎟ 2 10 ⎟⎠ [Ans : Rank is 3]

⎛ 1 3 4 3⎞ (e) ⎜ 3 9 12 3 ⎟ ⎜ ⎟ ⎝ 1 3 4 1⎠

[WBUT-2005] [Ans : Rank is 2]

⎛ 6 1 3 8⎞ ⎜ 4 2 6 1⎟ (f) ⎜ ⎟ ⎜⎜10 3 9 7 ⎟⎟ ⎝16 4 12 15 ⎠ ⎛0 ⎜1 (g) ⎜ ⎜⎜ 3 ⎝1

1 −3 3 0 1 1 0 1 2

[Ans : Rank is 2]

1⎞ 1⎟ ⎟ 2⎟ 0 ⎟⎠ [Ans : Rank is 2]

2.56

Engineering Mathematics-I

⎛1 ⎜2 (h) ⎜ ⎜⎜ 1 ⎝1

1 2 3 1 3 −1 7 −4

0 5 0 1

4⎞ 2⎟ ⎟ 3⎟ 1⎟⎠ [Ans : Rank is 4]

⎛ 3 (i) ⎜ −6 ⎜ ⎝ −3

1 2⎞ 2 −4 ⎟ ⎟ 1 −2 ⎠ [Ans : Rank is 1]

2. Find all values of μ for which the rank of the following matrix is 2. 1 ⎞ ⎛1 2 3 ⎜2 5 3 μ ⎟ ⎜ ⎟ ⎝ 1 1 6 μ + 1⎠ [Ans : μ = 1]. 3. Using elementary row operations, find the inverse of the matrix ⎛ 2 0 0⎞ ⎜ 4 3 0⎟ ⎜ ⎟ ⎝6 4 1⎠

⎡ ⎛ 1 ⎢ ⎜ 2 ⎢ ⎜ ⎢ Ans : ⎜ −2 ⎢ ⎜ 3 ⎢ ⎜ −1 ⎢ ⎜ ⎝ 3 ⎣

⎞⎤ 0 ⎟⎥ ⎟⎥ 1 0 ⎟⎥ ⎟⎥ 3 ⎟⎥ −4 1 ⎟⎥ 3 ⎠⎦ 0

4. Using elementary row operations, find the matrix A if A

−1

⎛ 3 −1 1⎞ = ⎜ 1 −2 3 ⎟ . ⎜ ⎟ ⎝ 3 −3 4 ⎠

⎡ ⎛ 1 1 −1⎞ ⎤ ⎢ Ans : ⎜ 5 9 −8 ⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎝ 3 6 −5 ⎠ ⎦ ⎣

5. Find the eigen values of the following matrices: ⎛ 1 (a) ⎜ ⎝ −5 ⎛5 4⎞ (b) ⎜ ⎟ ⎝1 2⎠

2⎞ 4 ⎟⎠

[Ans : 6, −1 ] [Ans : 6, 1].

2.57

Matrix II

6. Find the eigen values of the matrix ⎛ −6 0 ⎜ 0 −2 ⎜ ⎜⎜ 0 0 ⎝ 0 0

0 0 7 0

0⎞ 0⎟ ⎟ 0⎟ 1⎟⎠

[Ans : −6, −2, 7, 1]

⎛ 2⎞ ⎛3 1 4⎞ 7. Prove that the vector ⎜ 0 ⎟ is an eigen vector of the matrix ⎜ 0 2 0 ⎟ . Mention ⎜ ⎟ ⎜ ⎟ ⎝1⎠ ⎝0 0 5⎠ the corresponding eigen value. [Ans : Eigen value is 5] ⎛ 2⎞ ⎛ 8 8. Prove that the vector ⎜ 1⎟ is an eigen vector of the matrix ⎜ −6 ⎜ ⎟ ⎜ ⎝ −2 ⎠ ⎝ 2 corresponding to the eigen value 3.

6 2⎞ 7 −4 ⎟ ⎟ 4 3⎠

9. Verify the Cayley–Hamilton theorem for the following matrices. ⎛ 2 1 3⎞ a) ⎜ −1 3 −7 ⎟ ⎜ ⎟ 1⎠ ⎝ 1 0

⎛2 b) ⎜ 2 ⎜ ⎝2

1 1⎞ 1 −1⎟ ⎟ 2 1⎠

⎛ 1 2 4⎞ c) ⎜ −1 4 −8 ⎟ ⎜ ⎟ ⎝ 5 1 8⎠

10. If a matrix A is invertible and its eigen values are λ1 λ 2 , … λ n and B = A−1 , show that the eigen values of B are ⎛1 ⎜0 11. Find the rank of the matrix ⎜ ⎜⎜ 2 ⎝3

3 0 6 9

1 1 , , λ1 λ 2 2 4 1⎞ 2 2 0⎟ ⎟ 2 6 2⎟ 1 10 6 ⎟⎠

,

1 . λn

[WBUT-2006]

[WBUT-2006] [Ans : Rank is 3]

⎛λ 1 1 1 ⎞ ⎜1 λ 1 1⎟ 12. Find all values of λ for which the rank of the matrix ⎜ ⎟ is less than 4. ⎜⎜ 1 1 λ 1 ⎟⎟ ⎝1 1 1 λ⎠ −1 ⎤ ⎡ ⎢ Ans : λ = 1, 3 ⎥ ⎦ ⎣ c −b a ′ ⎞ ⎛ a ⎜ −c a b′ ⎟ 0 13. Find the rank of the matrix ⎜ ⎟ where aa ′ + bb′ + cc′ = 0. a 0 c′ ⎟ ⎜⎜ b ⎟ ⎝ − a ′ b′ c ′ 0 ⎠

2.58

Engineering Mathematics-I

14. Solve the following system of equations by matrix inversion method: a)

x y z=4 x + 2 y + 2z = 5 3x y 4 z = 6 [Ans : x = 1, y = 1, z = 1]

b)

x+ y+ z =1 x + 2 y + 3 z = 16 x + 3 y + 4 z = 22 [Ans : x = 1, y = 3, z = 3 ]

c)

x + 2 y + 3z = 6 2x 4 y + z = 7 3 x 2 y + 9 = 14 [Ans : x = 1, y = 1, z = 1]

15. Solve the following system of equations by Cramer’s rule: a)

x+ y−z = 6 2 x 33yy + z = 1 3 x 4 y + 2 = −1 [Ans : x

b)

3 y

2 z = 1]

−x + y + z = 2 2 y 3z = 4 3x 2 y − 6 = 1 ⎡ ⎢ Ans : x ⎣

9 59 33 ⎤ y= ,z= ⎥ 7 28 28 ⎦

16. Examine the consistency of the following system of equations and if possible, solve: a) 2 x y =1 x + y + 2 z = −1 3x 2 y − z = 4 [Ans : Consistent and unique solution x = 1, y = 0, z = 1 ] b)

4x 2 y + 6 = 8 x + y − 3 z = −1 15 x 3 y 9 z = 21 [Ans : Consistent and infinitely many solutions x = 1, y = 3k 2, z = k ]

2.59

Matrix II

c)

x − y + 2z = 4 3 y 4z = 6 x+ y+z = 2 [Ans : Inconsistent]

d)

x 4 y + 7z = 8 3 x 8y 8y − 2 = 6 7 x 8y 8 y + 26 z = 31 [Ans : Inconsistent]

e)

x 4y − z = 3 3x y 2 z = 7 2 x 33yy + z = 10 ⎡ ⎢ Ans : Consistent and unique solution x ⎣

f)

62 −5 31 ⎤ y= ,z= ⎥ 17 17 17 ⎦

2x y 4z = 4 x 3y − z = 5 3 x 2 y + 2 = −1 −8 x + 3 y − 8 = −2 [Ans : Consistent and unique solution x = 1, y = 2, z = 0 ]

17. Examine whether the following homogeneous system of equations have nontrivial solutions and find them if they exist. a)

x + 2 y + 3z = 0 2 x 33yy + z = 0 x + y + 2z = 0 [Ans : Only trivial solution]

b)

x + 2 y + 3z = 0 3x 4 y + 5 = 0 2 x 33yy + 4 = 0 [Ans : Nontrivial solution x = k , y

c)

k, z = k ]

x 2y + z w = 0 x + y − 2 z + 3w = 0 4 x y 5z + 8 = 0 ⎡ ⎢ Ans : Nontrivial solution x ⎣

5 k1 − k 3

y = k1

4 k2 , z 3

k1 w

⎤ k2 ⎥ ⎦

2.60

Engineering Mathematics-I

18. Find the values of a and b so that the following system of equations have (i) a unique solution, ii) no solution, and iii) infinitely many solutions. a) 2 x 3y 3y + 5 = 9 7 x 3y 3y − 2 = 8 2 x 3y 3 y + az = b [Ans: i) Unique solution for a ≠ 5, b = k (any constant) ii) No solution for a = 5, b 9 iii) Infinitely many solutions for a = 5, b = 9 ] b)

x+ y+z b 2 x y 3z = b 1 5 x 2 y + az = b 2 [Ans : i) unique solution for a ≠ , b = k (any constant) ii) no solution for a = 8, b 3, 1 iii) infinitely many solutions for a = 8, b = 3, − 1 ]

c) 3 x 2 y + z = b 5 x 88yy + 9 = 3 2 x y az = −1 ⎡ ⎤ ⎢ Ans : i) Unique solution for a ≠ , b = k ( anyy constant ) ⎥ ⎢ ⎥ 1 ⎢ ⎥ ,b≠ ii) No sollution for a ⎢ ⎥ 3 ⎢ ⎥ 1 ⎢ ⎥ iii) Infinitely y many solutions for a = 3, b = 3 ⎣ ⎦ 19. Find the eigen values and corresponding eigen vectors of the following matrices: ⎛ 2 1 1⎞ a) ⎜ −1 2 −1⎟ ⎜ ⎟ ⎝ 1 1 2⎠ ⎡ Ans : Eigen values : λ = 1, 1, 4., ⎤ ⎢ ⎥ ⎛ 1⎞ ⎛1⎞ ⎛ 1⎞ ⎥ ⎢ ⎢ Eigen vectors : for λ = 1, k1 ⎜ 0 ⎟ + k2 ⎜ 1 ⎟ and for λ = 1, k3 ⎜ −1⎟ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ ⎢ ⎝ −1⎠ ⎝0⎠ ⎝ 1⎠ ⎦ ⎣ ⎛ 2 b) ⎜ 2 ⎜ 1 ⎝ −1

1 3 1

1⎞ 4⎟ ⎟ 2⎠

⎡ Ans : Eigen values : λ = 1, 1, 3, ⎤ ⎢ ⎥ ⎛ 1⎞ ⎛ 0⎞ ⎛ −2 ⎞ ⎥ ⎢ ⎢ Eigen vecto s : for λ = 1, k1 ⎜ −1⎟ , for λ = −1, k2 ⎜ 1⎟ , for λ = 3, k3 ⎜ −3 ⎟ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ ⎢ ⎝ 0⎠ ⎝ −1⎠ ⎝ 1⎠ ⎦ ⎣

2.61

Matrix II

⎛ 2 20. Verify Cayley–Hamilton theorem for the matrix A = ⎜ −1 ⎜ and A4 . ⎝ 1 ⎛0 ⎜0 21. If A = ⎜ ⎜⎜ 0 ⎝1

1 0 0 0

0 1 0 0

1 1⎞ 2 −1⎟ and find A−1 ⎟ 1 2⎠

0⎞ 0⎟ ⎟ , by Cayley–Hamilton theorem show that A 1⎟ 0 ⎟⎠

1

= A3 .

22. Find the inverse of the following matrices by finding the characteristic equation (using Cayley–Hamilton theorem): ⎛2 (a) ⎜ ⎝ −1

1⎞ 2 ⎟⎠

⎛ 1 (b) ⎜ −1 ⎜ ⎝ 2

2 1⎞ 0 3⎟ ⎟ 1 1⎠

⎡ ⎛2 ⎢ ⎜ −1 ⎢ Ans : A = ⎜ 3 ⎢ ⎜⎜ 1 ⎢⎣ ⎝3

⎡ ⎛1 ⎢ ⎜6 ⎢ ⎜ ⎢ Ans : A−1 = ⎜ 7 ⎢ ⎜ 18 ⎢ ⎜1 ⎢ ⎜ ⎝9 ⎣ ⎛1 (c) ⎜ 1 ⎜ ⎝2

⎛ 1 (d) ⎜ −1 ⎜ ⎝ 0

2 1⎞ 1 1⎟ ⎟ 3 −1⎠

2 −2 ⎞ 3 0⎟ ⎟ 2 1⎠

1 ⎞⎤ 3 ⎟ ⎥⎥ ⎟ 2 ⎟⎥ ⎟ 3 ⎠ ⎥⎦

−1 1 ⎞ ⎤ 6 3 ⎟ ⎥⎥ ⎟ −1 2 ⎟ ⎥ 18 9 ⎟ ⎥ 5 1 ⎟⎥ ⎟⎥ 18 9 ⎠ ⎦

[WBUT-2005] ⎡ ⎛ −2 ⎢ ⎜ 9 ⎢ ⎜ ⎢ Ans : A−1 = ⎜ 1 ⎢ ⎜ 3 ⎢ ⎜ 5 ⎢ ⎜ ⎝ 9 ⎣

5 9 −1 3 1 9

⎞⎤ ⎟⎥ ⎟⎥ 0 ⎟⎥ ⎟⎥ −1 ⎟ ⎥ ⎟⎥ 3 ⎠⎦ 1 3

⎡ ⎛ 3 2 6 ⎞⎤ ⎢ Ans : A−1 = ⎜ 1 1 2 ⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎝ 2 2 5 ⎠⎦ ⎣

2.62

Engineering Mathematics-I

23. Determine eigen vectors of A and then diagonalise A with the help of the basis of eigen vectors: ⎛ 2 0⎞ (a) A = ⎜ ⎟ ⎝ 2 3⎠ ⎛5 4⎞ (b) A = ⎜ ⎟ ⎝1 2⎠ 24. Find the matrices P so that P 1 AP is a diagonal matrix (i.e., find P which diagonalises the following matrices): ⎛ 1 2 2⎞ (i) A = ⎜ 1 2 −1⎟ ⎜ ⎟ ⎝ −1 1 4 ⎠ ⎡ ⎛ 1 1 2 ⎞⎤ ⎢ Ans : P = ⎜ 1 0 −1⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎝ 0 0 1⎠ ⎦ ⎣ ⎛ 4 2 2⎞ (ii) A = ⎜ 2 4 2 ⎟ ⎜ ⎟ ⎝ 2 2 4⎠

⎛1 2 3⎞ (iii) A = ⎜ 0 1 0 ⎟ ⎜ ⎟ ⎝ 2 1 2⎠

⎡ ⎛1 ⎢ Ans : P = ⎜1 ⎢ ⎜ ⎝1 ⎣

1 0 1

0 ⎞⎤ 1⎟ ⎥ ⎟⎥ 1⎠ ⎦

⎡ ⎛1 ⎢ Ans : P = ⎜ 0 ⎢ ⎜ ⎝1 ⎣

3 1⎞ ⎤ 0 −6 ⎟ ⎥ ⎟⎥ 2 4 ⎠⎦

Multiple-Choice Questions ⎛2 1⎞ 1. The rank of the matrix A = ⎜ ⎟ is ⎝ 3 4⎠ a) 2

b) 3

c) 4

d) none of these

2. For what value of λ does the system of equations x + y + z = 1; x + 2 y − z = 2; 5 x 7 y + λ z = 4 have a unique solution? a) λ ≠ 2

b) λ ≠ 1

c) λ ≠ 3

d) λ ≠ 4

2.63

Matrix II

⎛ 2 0 1⎞ 3. The value of a for which rank of the matrix ⎜ 5 3 ⎟ is less than 3? ⎜ ⎟ ⎝ 0 3 1⎠ a)

3 4

b)

3 5

c)

3 2

d) 1.

⎛ 1 1 1⎞ 4. The value of k for which the rank of the matrix ⎜ 1 k 1⎟ is 2 is ⎜ ⎟ ⎝10 1 0 ⎠ a) 1 b) 0 c) −1 d) 2 5. The system of equations x + 2 y − z = 2; 4 x 8y 8 y − 4 = 8 has a) infinite many solutions c) a unique solution

b) no solution d) none of these

1⎞ ⎛ 6. The rank of the matrix ⎜ is 6 6 3 ⎟⎠ ⎝ a) 2 b) 3 c) 1 d) none of these 7. The equation x y = 0 has a) no solution b) exactly one solution c) exactly two solutions d) infinite number of solutions 8. All the eigen values of any nilpotent matrix are a) 0 b) 1 c) 2

d) none of these

⎛ 1 1 3⎞ 9. The sum of the eigen values of A = ⎜ 1 5 1 ⎟ is ⎜ ⎟ ⎝3 1 1⎠ a) 6 b) 5 c) 4 d) 7 [Hint : The trace of any square matrix is equal to the sum of the eigen values.] 10. The system of equations x + y − 3 z = 0; 3 a) a nontrivial solution c) no solution

y

= 0; 2 x

y 4 z = 0 has

b) a trivial solution d) none of these

⎛3 0 0⎞ 11. The eigen values of A = ⎜ 0 2 0 ⎟ are ⎜ ⎟ ⎝0 0 4⎠ a) 3, 2, 4 b) 5, 4, 6 c) 4, 3, 5

d) 7, 2, 9

⎛ 2 2⎞ 12. The eigen values of A = ⎜ ⎟ are ⎝ 2 2⎠ a) 2, 4 b) 0, 4 c) 0, 2

d) 0, 0.

2.64

Engineering Mathematics-I

⎛ 3 2 2⎞ 13. One of the eigen values of A = ⎜ 2 1 3 ⎟ is ⎜ ⎟ ⎝ 3 2 2⎠ a) 16 b) 15 c) 0

d) 14.

14. λ be an eigen value of an n × n square matrix A then λ −1 is an eigen value of a) A2

b) 2A

d) A−1

c) AT

15. λ be an eigen value of an n × n square matrix A then 2λ is an eigen value of a) A2

b) 2A

d) A−1

c) AT

16. λ be an eigen value of an n × n square matrix A, then λ is also an eigen value of a) A2 b) 2A c) AT d) A−1 17. λ be an eigen value of an n × n square matrix A then λ 2 is an eigen value of a) A2 b) 2A c) AT d) A−1 18. If the sum of the eigen values of the matrix A a) 7

b) 1

c) 8

⎛4 0 9⎞ ⎜ a 0 ⎟ is 7, then a is ⎜ ⎟ 7 0 2⎠ ⎝ d) 2

19. If trace of a 3 3 matrix is 12, and two of its eigen values are 4, 6 then the third eigen value is a) 2 b) 1 c) 8 d) 3 20. If A2 A then its eigen values are either a) 0 or 2 b) 0 or 1 c) 2 or 1

d) none of these.

Answers: 1. (a) 9. (a) 17. (a)

2. (b) 10. (a) 18. (b)

3. (c) 11. (a) 19. (a)

4. (a) 12. (b) 20. (b)

5. (a) 13. (c)

6. (c) 14. (d)

7. (d) 15. (b)

8. (a) 16. (c)

CHAPTER

3 Successive Differentiation 3.1

INTRODUCTION

Suppose we have a differentiable function y = f ( x) defined over an interval I. Then its first-order derivative is denoted by dy d , f ′ x), ( f ( x) ) , y ′ y1. dx dx Now suppose the first-order derivative is again differentiable on a certain interval. Then the second-order derivative is denoted by d2 y

d2

( f ( x) ) , y ′′ y2. dx 2 dx 2 In this way we can find the higher-order derivatives, differentiating the functions again and again, if they exist. Basically, this leads to the formation of the present chapter named as successive differentiation. 3.2

, f ′′ x),

SUCCESSIVE DIFFERENTIATION

Successive differentiation of a function means differentiation of a function successively or repeatedly. Suppose any function is given and you are to find its 100-th derivative, if it exists. Now the question is whether you can find it without having prior knowledge of the 1st, 2nd, ..., 99-th derivatives of the function. This means when you are finding derivatives of higher orders, you should know all of its previous-order derivatives. This is a very laborious and time-consuming job.

3.2

Engineering Mathematics-I

So, to meet the above difficulty, if we can find a general formula for the n-th order derivative of a particular function, if it exists, then by putting simply the value of n, we can get the derivative of any order as we require. Keeping similarity with the notations as before, we denote the n-th order derivative of a function y = f ( x) by dny dx

n

dn

, f ( n) x),

3.3

dx

n

x) x ) , y ( n ) yn .

(f

n -TH DERIVATIVE OF SOME IMPORTANT FUNCTIONS

(a) Let us consider y ax b )m , m is any number. Then y1 = m (ax + b) m 1 a ) (ax b) m −2 ⋅ a 2

y2 = m(m

= (m − 2 1) (ax (ax + b) m

2

a2

y3 = m(m 1))(m 2) (ax + b) m

3

a3

m 3 = (m −1)( 1)(m 3 1)) (ax b) m− ⋅ a3 ..................................

yn = m(m 1)(m 2)

(m n − 1) ⋅ (ax (ax b) m

n

an ,

if n < m, m is any number. Especially when m is any +ve integer and n < m, yn = m(m 1))(m 2) =

m! (

)!

(m n − 1) ⋅ (ax (ax b) m

n

an

)m−n .

⋅ an (

When n = m, yn = m(m 1))(m 2)

1 ⋅(ax (ax b) m

m

am

= ! am = ! an . When n > m and m is any +ve integer, yn = 0. Example 1

If y

(2 x 3)6 , let us find y4 , y6 and y7 .

Here, m = 6, a +ve integer. In the first case, n = 4. So n < m. Then m! yn = ⋅ a n (ax b) m − n . (m n)!

Successive Differentiation

6! ⋅ 2 4 (2 x 3)6 − 4 (6 − 4)! 6! = ⋅ 2 4 (2 3)2 2!

y4 =

In the second case, n = 6. So n = m. Then y n = n! ⋅ a n . y6 = 6! 26. In the third case, n = 7. So n > m. Then y7 = 0. x −2 , let us find y4 .

If y

Example 2

Here, m = 2, a −ve integer and n = 4. (m n − 1) ⋅ x m

yn = m(m 1))(m 2)

y4 = (−2)( 2)( 2 −1)( 1)( 2 − 2)( 2 − 3) x −22

n

4

= 120 ⋅ x −6 3 x4 ,

If y

Example 3

let us find y3 .

3 , a fraction and n = 3. 4 yn = m(m 1))(m 2) (m n − 1) ⋅ x m

Here, m =

n

3

⎛ 3 ⎞⎛ 3 ⎞⎛ 3 ⎞ −3 y4 = ⎜ ⎟ ⎜ − 1 ⎟ ⎜ − 2 ⎟ x 4 ⎝ 4 ⎠⎝ 4 ⎠⎝ 4 ⎠ (b) Let us consider y = Then y1

1 = (ax + b )−1 ax + b

(ax b) −2 ⋅ a b) −11 1 ⋅ a

= (−1) ⋅ (

y2 = (−1) 1)(( 22) (ax b)−33 ⋅a 2 1 b) −2 −1 ⋅ a 2

= (−1)2 ⋅ 2! ⋅ (

y3 = (−1) 11)((−22)(−33) ⋅(ax (ax b)−44 ⋅a 3 = (−1)3 ⋅ 3! ⋅ ( b) −3−1 ⋅ a 3 .................................. ( 1) n n! (ax + b)

yn =

( 1) n (

! an !⋅ b) n +1

.

n 1

an

3.3

3.4

Engineering Mathematics-I

If y =

Example 4

1 3x 5

then ( 1) n n! ⋅ 3n

yn =

(3 x + 5) n +1 ( 1) 4 4! 34

y4 =

(3 x + 5)5

.

If y =

Example 5

1 2x 3

then ( 1) n n! ⋅ 2 n

yn = y3 =

(2 x − 3) n +1 ( 1)3 3! 23

.

(2 x − 3) 4

(c) Let us consider y

ax b)

Then y1 =

1 ⋅a ax + b

(ax b) −1 ⋅ a

(ax b) −2 ⋅ a 2

y2

b) −2 ⋅ a 2

= (−1) ⋅ 1! ⋅ (

= (−1)2 1 ⋅ (2 −1)! 1)! (ax + b) b) −2 ⋅ a 2 y3 = (−1)( 1)( 2) (ax b) −3 ⋅ a 3 b) −3 ⋅ a 3

= (−1)2 ⋅ 2! ⋅ (

= (−1)3 1 ⋅ (3 −1)! 1)! (ax + b) b) −3 ⋅ a 3 .................................. yn = (−1) n −1 ⋅ (n 1)! ⋅ (ax b) =

( 1) n

1

(n − 1)! ⋅ a n

(

b) n

Alternative Method y

(ax a b)

so, y1 =

1 ⋅a ax + b

.

n

an

Successive Differentiation

(n − 1)th derivative of y1

yn

⎛ 1 ⎞ = (n − 1)th derivative of ⎜ ⋅a⎟ ⎝ ax + b ⎠ ⎧ ⎛ 1 ⎞⎫ = ⎨(n − 1)th derivative of ⎜ ⎟⎬ ⎝ ax + b ⎠ ⎭ ⎩ i.e., yn = a ⋅ =

( 1) n

(n 1)! a n −1 )n

(

( 1) n

1

(n − 1)! ⋅ a n )n

(

If y = log

Example 6 then y

1

x3 2x 1

3 log x log log(2 x + 1)

yn = 3 ⋅ y6 = 3 ⋅

( 1) n

(n − 1)! (−1) n −1 ⋅ (n 1)! ⋅ 2 n − x (2 x + 1) n

1

n

( 1)6

1

(6 1)! ( 1)6 1 (6 1)! ⋅ 26 − x (2 x + 1)6 6

⎛ 26 3 − = 5! ⎜ ⎜ (2 x + 1)6 x 6 ⎝ (d) Let us consider y

⎞ ⎟⎟ . ⎠ e ( ax

b)

Then e( ax

y1

b)

( ax b )

⋅a ⋅ a2

y2

e

y3

e( ax b) ⋅ a 3 ............

yn

e( ax

b)

Example 7

⋅ an If y

yn = e(2 x + 3) 2 n y5 = e(2 x + 3) 25 = 25 ⋅ e(2

3)

.

e(2 x + 3) then

3.5

3.6

Engineering Mathematics-I

(e) Let us consider y Then y1 ax b) ⋅ a

ax b)

⎛π ⎞ ax b ⎟ ⎝2 ⎠ π ⎛ ⎞ y2 a 2 ⋅ sin ⎜ + ax + b ⎟ ⎝2 ⎠ ⎛ π ⎞ = a 2 cos 2 ⋅ + ax b ⎟ ⎝ 2 ⎠ ⎛ π ⎞ y3 = a 3 si 2 ⋅ + ax + b ⎟ 2 ⎝ ⎠ ⎛ π ⎞ 3 = cos 3 ⋅ + ax b ⎟ ⎝ 2 ⎠ ..................... = a cos

yn

⎛ π ⎞ a n ⋅ cos ⎜ n + ax a b ⎟. ⎝ 2 ⎠

Example 8

If y

2

(3 x 4), find yn and y4 .

Here, 2

y

(3 x 4)

1 [1 cos 2(3 x + 4)] 2 1 = [1 (6 8) ] 2 1 yn = [1 (6 (6xx 8) ]n 2 6n ⎛ π = − ⋅ cos ⎜ n ⋅ + 6 2 ⎝ 2 =

y4 = − =−

⎞ 8⎟. ⎠

64 ⎛ π ⎞ ⋅ cos ⎜ 4 ⋅ + 6 x 8 ⎟ 2 2 ⎝ ⎠ 64 cos(6 x + 8) 2

(f) Let us consider y Then y1 ax b) b) ⋅ a =a s

⎛π ⎝2

⎞ ax b ⎟ ⎠

ax b)

[WBUT 2003]

Successive Differentiation

3.7

⎛π ⎞ + ax + b ⎟ ⎝2 ⎠ ⎛ π ⎞ = a 2 si 2 ⋅ + ax b ⎟ 2 ⎝ ⎠ ⎛ π ⎞ 3 y3 = a cos 2 ⋅ + ax + b ⎟ ⎝ 2 ⎠ π ⎛ ⎞ = 3 si 3 ⋅ + ax b ⎟ ⎝ 2 ⎠ ..................... ⎛ π ⎞ yn a n ⋅ sin ⎜ n + ax a b ⎟. ⎝ 2 ⎠ y2

a 2 ⋅ cos

Example 9 Here, y

If y

(2 x 3)

(2 x 3)

(2 x 3), find yn and y8 . (2x

(2x (2 x 3)

1 sin 2(2 3) 2 1 = sin (4 x + 6) 2 1 yn [sin(4 x + 6)]n 2 4n ⎛ π = ⋅ sin ⎜ n + 4 2 ⎝ 2 =

y8 = =

3.4

⎞ 6 ⎟. ⎠

48 ⎛ π ⎞ ⋅ sin ⎜ 8 ⋅ + 4x 4x 6 ⎟ 2 ⎝ 2 ⎠ 48 ⋅si (4 x + 6) 2

n -TH ORDER DERIVATIVE OF PRODUCT OF TWO FUNCTIONS OF SAME VARIABLE

First, we recall the multiplication rule of the differentiation for finding 1st order derivative of the product of two functions. Suppose u and v are two functions of x and 1st order derivative exists for them then 1st order derivative of y u ⋅ v is given by y1

u1 v + u v1 .

Also, we can find the higher order derivatives of the product u v, differentiating repeatedly if they exist.

3.8

Engineering Mathematics-I

Now the question is whether we can have any general formula for finding directly the n-th order derivative of the product of two functions of the same variable. The answer is yes and it is given by the following theorem: Leibnitz’s Theorem Suppose u and v are two functions of x and the n-th order derivative exists for both of them. Then the n -th order derivative of the product y u.v is given by ( u ⋅ v )n = un v

yn

n

C1 un 1v1 + n C 2 un − 2 v2

n

C r un − r vr +

+ uvn .

Proof: Beyond the scope of the book. Note: If we put n = 1 in the above formula, we obtain y1 (u ⋅ v)1 = u1 ⋅ v + u ⋅ v1 , which is the multiplication rule of differentiation for finding 1st order derivative of the product of two functions. So it is clear that Leibnitz’s Theorem is nothing but the generalised multiplication rule for finding higher-order derivatives. Selection of u and v In general, we can choose any function of the product as u or v, but if we carefully see formula, the order of the derivative of v increases term by term. So, among the two functions, which ever has more priority of vanishing at the higher order derivatives should be taken as v. Basically, the reason behind it is to make the calculation easier and, of course, to save time too.

log x and v = x 2 .

Here, we set u Then un =

( 1) n

x 2 log x then let us find yn .

If y

Example 10

1

x

(n − 1)! n

and v1

2 x, v2 = 2, v3 = v4 =

=0

i.e., vn = 0, for n > 2 Here, it is obvious from the above that x 2 has the more priority over log x of vanishing at the higher order derivatives and due to that factor we have choosen x 2 as v. Now using Leibnitz’s theorem, we find the n-th derivative of y u ⋅ v log x ⋅ x 2 as

( )

n

(u ⋅ v) n = un v

yn

i.e., yn =

( 1)

C1un 1v1 + n C2 un −2 v2 +

+ uvn

{log x ⋅ ( x )} 2

n

n 1

(n − 1)!

xn

⋅(

2

) + n C1

( 1) n

2

(n − 2)!

x n −1

⋅ (2 x) + n C2

( 1) n −−3 ⋅ ( − 3)! x n −2

⋅ (2).

3.9

Successive Differentiation

If y = e x sin x then let us find yn .

Example 11

e x and v = sin x.

Here, we set u Then

⎛ nπ ⎞ e x and vn = sin ⎜ + x⎟ ⎝ 2 ⎠ Here, it is obvious from the above that no function has the priority over another of vanishing at the higher order derivatives and due to that we can choose u and v arbitrarily. Now using Leibnitz’s theorem, we find the n-th derivative of y u ⋅ v e x sin x as un

yn

n

(u ⋅ v) n = un v

{

C1un 1v1 + n C2 un −2 v2 +

+ uvn

}

i.e., yn = e x ⋅

x

n

⎛π ⎞ = e x ⋅ si x + n C1 e x ⋅ sin ⎜ + x ⎝2 ⎠

n

C2 ⋅ e x ⋅ si

⎛ 2π ⎞ + x⎟ + ⎝ 2 ⎠

⎛ nπ ⎞ + e x ⋅ sin i ⎜ + x⎟ ⎝ 2 ⎠

WORKED-OUT EXAMPLES If y =

Example 3.1 Sol.

x2 , find yn . ( x 1)( x − 2)( x 3)

[WBUT 2001]

Let us consider y=

x2 A B C + + = ( x 1)( x − 2)( x 3) ((xx 1) ( x − 2) ( x 3)

x2 A( x 2)( x 3) B(x B( x 1)( x − 3) + C ( x 1)( x − 2) = ( x 1)( x − 2)( x 3) ( x 1)( 2)( x − 3) ⇒

2

= A( − 2)( x − 3) + B( x − 1)( x − 3) + C ( x − 1)( x − 2)

Substituting x = 1, 2, 3, we have respectively 12 = A(1 − 2)(1 − 3) i.e., A =

1 2

22 = B(2 − 1)(2 − 3) i.e., B = 4. and 32 = C (3 − 1)(3 − 2) i.e., C =

9 2

3.10

Engineering Mathematics-I

Therefore, y

x

x 2) x

yn =

(ax b)

4 )

when y

n +1

+

x

9 2 x

)

1 we get from the above, ax + b

1 n! n! 9 n! − ⋅ + . n n n + 1 + 1 2 ( x 1) 2 ( x 3) + ( x 2)

Example 3.2

e ax

If n

Sol.

x

n a

since yn

y

1 2

x2

a

prove that

e ax cos bx + n tan −1

b . a

Here, we have e ax os bx Now differentiating w.r.t x y

e ax

a⋅

x

a cos Consider a r2

x)

r os

and

r in

then

b tan −1 a

a 2 + b2 and

So, e

r

bx r sin sin bx) bx + )

Again differentiating w.r.t x we have y

r⋅

ax

⋅

bx +

ax 2

ax

bx + ) b bx

⋅ {cos

r sin

+

+

bx + 2 ) Proceeding similarly as above r

bx n ) n

y

a

ax

⋅ cos bx + n tan

1

b . a

sin

}

3.11

Successive Differentiation

If y

Example 3.3 yn Sol.

(a

2

n 2 2 +b

)

b⎞ ⎛ e ax sin ⎜ bx + n tan −1 ⎟ . a⎠ ⎝

Follow Example 3.2. If y

Example 3.4 yn = Sol.

e ax sin bx, prove that

x n −1 log x, prove that

(n 1)! . x Here, we have x n −1 log x

y

Now differentiating w.r.t. x, y1

(n − 1) ⋅ x n −2 ⋅ log x + x n −1 ⋅

1 x

1) x n −1 ⋅ log x + x n −1

i.e., xy1 = (

i.e., xy1 = (n 1) y + x n −1 Now applying Leibnitz’s rule, we differentiate ( [ y1 x] x]n

1

= [(n 1) ⋅ y] y ]n

1) times,

[ x n −1 ]n −1

1

i.e., {y1}n 1 ⋅ x + n −1 C1 {y1}n

2

1 = (n (n 1) yn −1 + (

1)!

i.e., yn ⋅ x + (n 1) yn− n 1 = ( n − 1) yn− n 1 + ( n − 1)! i.e., yn x = (n − 1)! Hence (n 1)! yn = . x Example 3.5 Sol.

If y

2 cos xx((

It is given that y

2 cos x( x(

x

x)

= 2 cos sin x − 2 cos2 x = sin 2

cos 22x 1 cos

x

x) then show that ( y10 )0 = 210 .

3.12

Engineering Mathematics-I

Differentiating n times, we have yn

{sin 2 x}n {

2 x}n

⎛ π ⎞ ⎛ π ⎞ = 2 n sin ⎜ n + 2 x 2 n cos n + 2 x ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ ⎧ ⎛ π ⎞ ⎛ π ⎞⎫ = 2 n ⎨sin ⎜ n + 2 x ⎟ − cos n + 2 x ⎟ ⎬ ⎠ ⎝ 2 ⎠⎭ ⎩ ⎝ 2 Now putting x = 0, we have ⎧ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎫ ( yn )0 = 2 n ⎨sin cos ⎜ ⎟⎬ ⎝ 2 ⎠⎭ ⎩ ⎝ 2 ⎠ So ⎧ ⎛ 10π ⎞ ⎛ 10π ⎞ ⎫ ( y10 )0 = 210 ⎨sin cos ⎜ ⎟⎬ ⎝ 2 ⎠⎭ ⎩ ⎝ 2 ⎠ = 210 {

5π

cos 5 }

10

5

= 2 {0 − ( 1) } Hence ( y10 )0 = 210 .

If y =

Example 3.6 Sol.

1 x2

a

, show that yn = 2

Here, 1

y= =

x

2

a

2

=

1 ( x a)(x ( x a)

1 ⎡ 1 1 ⎤ − ⎢ 2a ⎣ ( ) ( ) ⎥⎦

So differentiating n times, we have yn =

=

1 ⎡⎧ 1 ⎫ ⎧ 1 ⎫ ⎤ ⎢⎨ ⎬ ⎥ ⎬ −⎨ 2 a ⎣ ⎩ ( x a ) ⎭n ⎩ ( x a ) ⎭n ⎦ 1 ⎡ ( 1) n ! ( 1) n ! ⎤ − ⎢ ⎥ 2a ⎣ ( ) n +1 ( ) n +1 ⎦

Hence, yn =

⎤ ( 1) n n! ⎡ 1 1 − ⎢ ⎥. n n + 1 + 1 2 a ⎣ ( x a) ( x a) ⎦

⎤ ( 1) n n! ⎡ 1 1 − ⎢ ⎥ n n + 1 + 1 2 a ⎣ ( x a) ( x a) ⎦

3.13

Successive Differentiation

Sol.

1

If y =

Example 3.7

x2 + a

⎤ ( 1) n n! ⎡ 1 1 − ⎢ ⎥ n n + 1 + 1 2ia ⎣ ( x ia) ( x ia) ⎦

, show that yn = 2

Here, 1

y= =

x2 + a2

=

1 ( x ia)(x ( x ia)

1 ⎡ 1 1 ⎤ − ⎢ 2ia ⎣ ( ) ( ) ⎥⎦

So differentiating n times, we have yn =

=

1 ⎡⎧ 1 ⎫ ⎧ 1 ⎫ ⎤ ⎢⎨ ⎬ ⎥ ⎬ −⎨ 2ia ⎣ ⎩ ( x ia) ⎭n ⎩ ( x ia) ⎭n ⎦ 1 ⎡ ( 1) n ! ( 1) n ! ⎤ − ⎢ ⎥ n + 1 2ia ⎣ ( ) ( ) n +1 ⎦

Hence, yn =

⎤ ( 1) n n! ⎡ 1 1 − ⎢ ⎥. 2ia ⎣ ( x ia) n +1 ( x ia) n +1 ⎦

Example 3.8

If f ( x) = tan x and n is a +ve integer, prove with the help of

Leibnitz’s theorem that f n

Sol.

n

C2 f ( n

n

2)

C4 f ( n

4)

(0)

⎛ nπ ⎞ = sin ⎜ ⎟ ⎝ 2 ⎠ [WBUT 2001]

We have f x) = tan x =

sin x cos x

or, f ( x) ⋅ cos x = sin x. Applying Leibnitz’s theorem, we differentiate n times w.r.t x, { f ( x)

}n = {sin x}n

f n ( x) cos x

n

C1 f ( n

1)

( x)(

x)

n

C2 f ( n

2)

( x)(

x)

+ n C3 f ( n −3) ( x) sin x + n C4 f ( n − 4) ( x) cos x + Now, putting x = 0 in the above, we get fn

n

C2 f ( n

2)

n

C4 f ( n

4)

(0)

⎛ nπ = sin ⎜ ⎝ 2

⎞ ⎟. ⎠

⎛ nπ ⎞ = sin ⎜ + x⎟ ⎝ 2 ⎠

3.14

Engineering Mathematics-I

If x + y = 1, prove that the n-th derivative of x n y n is

Example 3.9

{

n! y n −

( C) n

2

(

yn 1x

1

n

)

C2

2

y n 2 x2 −

(C) n

2

3

}

y n −3 x 3 +

( 1) n x n . [WBUT 2002]

x n and v

Let u

Sol.

y n = (1 − x) n

Then un = n!, vn = ( 1)) n n!. Also ur =

n! xn r , r < n (n r )!

So, ) n = un v + n C1un 1v1

(

( x y ) = {x n n

n

n

x

n

n

n

n! ⋅ x n(1 − x) n 1 (−1) 1!

n! 2 ⋅ x ⋅ n(n − 1)(1 − x) n −2 (−1)2 + 2!

{

( )

= ! (1 x) n − So,

(x y ) n n

n

{

−

( )y (1 + ) y

(i) 1 (ii)

2

1

2

If y

n

1

( )

= n! y n −

Example 3.10

+ uvn

}

= n!(1 − x) n + n C1 + n C2

C2 un −2 v2 +

2

n

(C)

2

2

(

)n 1 x

(1

(1 − x) n −3 x3 +

( C) n

2

yn 1x

(

y n −3 x 3 +

n

n

C2

(−1) n x n

C

)

2

(1 − x) n −2 x 2

}

}

(−1) n x n .

tan −1 x then show that

=1

n +1

+ 2 nxy x n + n(

If we differentiate y y1 =

2

y n −2 x2

1) yn

1

=0

Find also the value of yn at x = 0. Sol.

)

+ x n (−1) n ⋅ n!

1 1 + x2

tan −1 x w.r.t x, then

[WBUT 2003, 2005]

Successive Differentiation

(

)

i.e., y1 1 x 2 = 1

3.15

...(1)

Applying Leibnitz’s theorem, we differentiate n times w.r.t x,

(

)

⎡ y1 1 x 2 ⎤ = [1]n ⎣ ⎦n ⇒ { 1}n ⋅

fi

(

)+

+

( + )y

n +1

n

1 { 1}n −1 ⋅ (2

) + n C2 { 1}n −2 ⋅ (2) = 0

+ 2 nxy x n + n( - 1) yn -1 = 0

...(2)

Now we will find ( yn )0. From (1), y1 =

1 1 + x2

i.e., ( y1 )0 = 1 Also, from (1), y2 = −

2x

(1 + x ) 2

2

i.e., ( y2 )0 = 0. Now putting x = 0 in (2), ( yn +1 )0 + n(n 1)( yn −1 )0 = 0 i.e., ( yn +1 )0 = n(n 1)( yn −1 )0

...(3)

If we put n = 3, 5, 7, … in the above then ( y4 )0 = 3 ⋅ 2 ( y2 )0 = 0, since ( y2 )0 = 0 ( y6 )0 = 5 ⋅ 4 ( y4 )0 = 0, since ( y4 )0 = 0 ( y8 )0 = −7 7 6 ( y6 )0 = 0, since( y6 )0 = 0 and so on. Therefore, ( yn )0 = 0, when n is even. Again putting n = 2, 4, 6, … in the relation (3), we have ( y3 )0 = −2 2 1 ( y1 )0 = ( 1) 2!, since ( y1 )0 = 1 ( y5 )0 = 4 ⋅ 3 ( y3 )0 = ( 1)2 4!, since ( y3 )0 = (−1) ⋅ 2! ( y7 )0 = 6 ⋅ 5 ( y5 )0 = ( 1)3 6!, since( y5 )0 = (−1)2 ⋅ 4! Therefore, ( yn )0 = ( -1)

n -1 2

◊(

1) , when n is odd.

3.16

Engineering Mathematics-I

Hence, we can say

( yn )0 = ( -1)

n -1 2

◊(

1)!, when n is odd

= 0, when n is even. Example 3.11

Show that

d n ⎛ log x ⎞ n! ⎛ 1 1 = (−1) n n +1 ⎜ log x − 1 − − − n ⎜ x ⎟ 2 3 dx ⎝ x ⎝ ⎠ Sol.

1⎞ − ⎟. n⎠

[WBUT 2003, 2008].

Here we are to find d n ⎛ log x ⎞ d n ⎛ 1 ⎞ ⎜ ⎟= ⎜ ⋅ log x ⎟ dx n ⎝ x ⎠ dx n ⎝ x ⎠ ⎛1 ⎞ = ⎜ ⋅ log x ⎟ ⎝x ⎠n We set u =

if v

1 ( 1) n n! , then un = (using (b) of Art. 3.3) and x x n +1

log g xx, then vn =

( 1) n

1

(n − 1)! n

x Now using Leibnitz’s theorem, we have (

) n = un v + n C1un 1v1

n

C2 un −2 v2 +

( 1) n ! ⎧1 ⎫ i.e., ⎨ ⋅ log x ⎬ = ⋅ log x x n +1 ⎩x ⎭n + n C2

n

C1

(using (c) of Art. 3.3)

+ uvn ( 1) n

1

(n − 1)! 1 ⋅ x x n

(−1) n −2 ⋅ (n − 2)! ⎛ 1 ⋅⎜ − 2 x n −1 ⎝ x

⎞ ⎟+ ⎠

1 (−1) n −1 ⋅ (n − 1)! + ⋅ x xn

( 1) n ! ( 1) n ! ( 1) n ! 1 ⎧1 ⎫ i.e., ⎨ ⋅ log x ⎬ = log − − ⋅ x 2 x n+1 x n +1 x n +1 ⎩x ⎭n −

−

(−1) n ⋅ ! 1 ⋅ . n x n +1

1 1 n! ⎛ ⎧1 ⎫ i.e., ⎨ ⋅ log x ⎬ = (−1) n n +1 ⎜ log x − 1 − − − 2 3 x ⎝ ⎩x ⎭n

1⎞ − ⎟. n⎠

d n ⎛ log x ⎞ n! ⎛ 1 1 = (−1) n n +1 ⎜ log x − 1 − − − n ⎜ x ⎟ 2 3 dx ⎝ x ⎝ ⎠

1⎞ − ⎟. n⎠

i.e.,

Successive Differentiation

3.17

Show that if u = sin ax + cos ax then

Example 3.12

{

}

D n u = a n 1 + ( 1) n

2 ax

1 2

, where D ≡

d . dx

[WBUT 2003].

It is given that u = sin ax + cos ax

Sol.

Now differentiating n times, we have un = {sin ax} ax}n {cos ax}n Using (e) and (f ) of Art. 3.3, ⎛ nπ ⎞ ⎛ nπ ⎞ a n sin ⎜ + ax ⎟ + a n cos + ax ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

un

⎧ ⎛ nπ ⎞ ⎛ nπ ⎞⎫ = a n ⎨sin ⎜ + ax ⎟ + cos + ax ⎟ ⎬ ⎠ ⎝ 2 ⎠⎭ ⎩ ⎝ 2 Squaring both sides, we have ⎧ ⎛ nπ ⎞ ⎛ nπ ⎞⎫ a 2 n ⎨sin ⎜ + ax ⎟ + cos + ax ⎟ ⎬ ⎠ ⎝ 2 ⎠⎭ ⎩ ⎝ 2

( u n )2

2

⎧ ⎛ nπ ⎞ ⎛ nπ ⎞⎫ = a 2 n ⎨1 + 2 sin + ax ⎟ ⋅ cos + ax ⎟ ⎬ ⎝ 2 ⎠ ⎝ 2 ⎠⎭ ⎩ ⎧ ⎛ nπ ⎞⎫ = a 2 n ⎨1 + si 2 + ax ⎟ ⎬ ⎝ 2 ⎠⎭ ⎩ = a 2 n {1 + si (

2 ax)}

{

}

= a 2 n 1 + ( 1) n

2 ax

Hence,

{

}

un = a n 1 + ( 1) n

Example 3.13

(1 − ) y 2

n+2

− (2

Also, find ( yn )0.

If y

2 ax

1 2

.

(

)

cos m sin −1 x then prove that

1) xyn +1 +

(

2

−

2

)y

n

= 0. [WBUT 2004]

3.18

Sol.

Engineering Mathematics-I

We have m sin −1 x

y

1

i.e.

=

1

sin

x

Now differentiating w.r.t x 1

1 1

1 − y2

1 x2

Squaring both sides, we have y

2

= m2

x

y2

...(1)

Differentiating (1) w.r.t x 2

2

−

2

= m2 ( 2

−2

1

1

)

Cancelling 2 y1 from both sides, xy1 + m2 y = 0.

x

...(2)

Now Applying Leibnitz’s rule, we differentiate (2), n times x2

y

n

−

n

+ m2 y

n

= 0. C2

i.e., −

⋅x+

i.e., y i.e., 1

x 2

Now we will find (

−

n

y

1

+

−

− yn = 0

)0. m sin −1 x

We have from (1), by putting x =0

Again from (2), by putting x i.e.,

=0

−2

yn + 2

Putting x = 0 in y

i.e.,

⋅1 +

2

0

.

Now putting x = 0 in (3), m −n

0

( 2)

we get ( )0 = 1.

n +1

n

2 n

n

...(3)

3.19

Successive Differentiation

i.e.,

yn ) 0

=

If we put n 1, 3, 5,

...(4)

in the above then

=

2

m2

= 0,

(y

0

=

2

m2

= 0,

(y

0

=

2

m2

= 0,

(y

0

and so on. )0 = 0, when n is odd.

Therefore, (

Again putting n = 2, 4 , … in the relation (4), we have =

2

m2

2

m2

=

2

m2

2

m2

Similarly,

2

=

,

m2

=

since

− m2

y

m2 ⋅

2

m2

2

m2

and so on. Therefore,

0

2

=

− m2

2

− m2

)

.

when n is even Hence we can say (

)0 = 0, when n is odd 2

=

−

2

2

−

2

…

,

when n is even. Example 3.14

If y

x2 Sol.

x

1

−n n

n

) yn = 0

We have y

x

1

n

Now differentiating w.r.t x n

n −1

then prove that

x

[WBUT 2006]

3.20

Engineering Mathematics-I

i.e., x 2

x

2

i.e., x 2

n

1

= 2 nxy

Again differentiating w.r.t x, x2

= 2 y xy1

i.e., x 2

y

2

y=0

Now Applying Leibnitz’s rule, we differentiate n times 2

y

−1

− n y1

n 2

i.e., +

−1

C

−

{

+ −

i.e., y +

= 0.

+ nx

nn

−n x −n n

If yn =

dn

x

dx n

) yn = 0 then prove that [WBUT 2007]

Here, dn

yn =

dx n

dn dx

n−1

dn dx So, yn = n i.e.

1

1

n −1

dn

(

x

nx n −1 log

1

dx n −1 d n −1 dx

xn d x dx

⋅

n 1

( nx (x

} 0

n − 1)! Sol.

n

}

i.e., x 2

Example 3.15

nyn = 0

−1

1

(2)

n −1

n 1

) x

)

x +

)

1 x d n −1 dx n

x + n − 1) n − 1)!.

1

(x ) n −1

3.21

Successive Differentiation

Example 3.16

If

a

x 2 yn + 2

y

n +1

n

log x

b

og x), prove that

=0

[WBUT 2007]

Here

Sol.

a

(log x

b

og x)

Now differentiating w.r.t x 1 ⋅ + x i.e., xy1

a

x

x ⋅ b

1 x

x)

Again differentiating w.r.t x, y

a

i.e., y

=

(log x) 1 x

1 b x

{a

1 x

x

x}

x +b

So, xy1 +

−y

i.e., x 2 y

y=0

Now Applying Leibnitz’s rule, we differentiate n times. y

x2

n

+

x

n

x

i.e.,

n

{

= 0.

C1

⋅x+ i.e.,

n

x −

x

2 nx

n

y

y}

⋅1 + (n

2)

=0 yn

n +1

n

n

n

=0

Hence x 2 yn + 2 Example 3.17

n +1

=0

With the help of result obtained by differentiating n times x 2 n in

two different ways, show that 1+

n2 2

1

+

n2

2

+

− 2)2

n ⋅ 2

⋅3

+

=

(2 n)! ( !)

.

3.22

Sol.

Engineering Mathematics-I

x 2 n then

Let y yn =

(2 n)! 2 n x (2 n n)!

n

=

(2 n)! n x . n!

...(1)

Now again we can write x2 n = x n x n

y

Now applying Leibnitz’s rule, we differentiate n times. ⎡ xn ⋅ xn ⎤ ⎣ ⎦n

yn

{ }

= xn

n

{ }

⋅ x n + n C1 x n

+ n C3

{ }

n −3

n −1

⋅ n xn

1

{ }

+ n C2 x n

n −2

⋅ n(n − 1) ⋅ x n −2

⋅ n(n − 1)(n − 2) ⋅ x n −3 +

n! n(n 1) n! 2 ⋅ x ⋅ n ⋅ x n −1 + ⋅ ⋅ x n(n n(n 1) x n −2 1! 2 2! n(n − 1)(n − 2) n! 3 + ⋅ ⋅ x ⋅ n(n − 1)(n − 2) ⋅ x n −3 + 3! 3!

i.e. yn = n! ⋅ x n + n ⋅

⎧ n2 n2 ⋅ (n (n 1)2 n2 ⋅ (n (n 1)2 (n − 2)2 i.e. yn = n! ⋅ x n ⎨1 + 2 + + + 12 22 12 22 ⋅ 32 ⎩ 1

⎫ ⎬ ⎭

...(2)

From (1) and (2), we have ⎧ n2 n2 ⋅ (n (n 1)2 n2 ⋅ (n (n 1)2 (n − 2)2 n! x n ⎨1 + 2 + + + 12 22 12 22 ⋅ 32 ⎩ 1 Hence, 1+

n2 2

1

+

n2 ⋅ ( 2

1 2

1)2 2

+

n2 ⋅ (

1)2 (n − 2)2 2

2

2

1 2 ⋅3

+

=

⎫ (2 n)! n x ⎬= n! ⎭

(2 n)! ( !)2

.

EXERCISES

Short and Long Answer Type Questions 1. Find the n-th derivative yn of the following functions: 1

(i) y

(a − bx) 2

−1

(ii) y

⎡ ⎧1 3 ⋅ 5 (2 ⎢ Ans : ( 1) n b n ⎨ 2n ⎩ ⎢⎣

3) ⎫ ⎬⋅( ⎭

)

−n+

1 2

⎤ ⎥ ⎥⎦

x2 ⎡ ⎧1 3 ⋅ 5 (2 ⎢ Ans : ( 1) n ⎨ 2n ⎩ ⎢⎣

1 1) ⎫ − n − 2 ⎤ ⎥ ⎬⋅ x ⎭ ⎥⎦

3.23

Successive Differentiation 3

(iii) y

x4 3 4n ⎤ ⎡ 3 1 ⋅ 5 9… (4 n − 7) 4 ⎥ ⎢ Ans : ( 1) n −1 x ⋅ 4n ⎢⎣ ⎥⎦

(iv) y =

1 (5 x − 7) ⎡ ( 1) n ! ⋅ 5n ⎤ ⎢ Ans : ⎥ (5 x − 7) n +1 ⎦ ⎣

(v) y

4x

a3

⎡ Ans : a 3 ⋅ ( 1) n 4 n ( g a) n ⋅ a ⎣ (vi) y

4x ⎤

⎦

(2 x 9) ⎡ ( 1) n 1 (n −1)! 1)! 2 n ⎤ ⎢ Ans : ⎥ (2 x + 9) n ⎣ ⎦

(vii) y = log

a+x a x ⎡ ⎢ Ans : ( ⎢⎣

(viii) y

⎧ 1 (−1) n ⎫⎤ )! ⎨ − ⎬⎥ n (a + x) n ⎭⎥⎦ ⎩ ( a x)

3 i xx.

⎡ 1⎧ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎫⎤ + x ⎟ − 3n sin + 3 x ⎟ ⎬⎥ ⎢ Ans : ⎨3 sin ⎜ 4⎩ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎭⎦ ⎣ (ix) y = e2 x sin x si 2 x n n ⎡ 1 ⎧ nπ ⎞ ⎢ Ans : e x ⎪⎨2 2 cos ⎛⎜ + x ⎟ +10 10 2 cos 3 2 ⎪ ⎢ ⎝ 4 ⎠ ⎩ ⎣

(

(x) y =

⎫⎪⎤ 3 ⎬⎥ ⎪⎭⎥⎦

)

1 3

( x 1) ( x − 2) ⎡ n ⎢ Ans : ( 1) ⎣

(xi) y =

1

1

⎧ ( 2)(n + 1) ( 1) 1 !⎨ + + + n +3 n+2 2( x − 1) ( 1) 2( x − 1)n +1 ( ⎩

x2 ( x 1)( x − 2)( x 3)

1 2)

⎫⎤

n +1 ⎬ ⎥

⎭⎦

[WBUT 2001] ⎡ 1 ( 1) n ! ( 1) n ! 9 ( 1) n ! ⎤ − ⋅ + 4 ⎢ Ans : ⎥ 2 ( 1) n +1 ( 2) n +1 2 ( 3) n++1 ⎦ ⎣

3.24

Engineering Mathematics-I

1

(xii) y

[WBUT 2001]

x 2 + 16

⎡

!⎡

An

1

8i

4i)

n +1

⎤

1

−

4i )

n +1

2. Find the n-th derivative yn of the following functions using Leibnitz’s theorem: e og x

(i)

⎡ ⎢ Ans : e

1 x

x

1 x2

+

⋅

x3

(n 1)! ⎤

)

+

xn

x3 cos x

(ii) y

⎡ ⎢ An

x3 cos

n + 2

⋅ cos

⎢ +

(n

⋅ co

)

( n − 1) 2

+

2

n

x

⎤

x2

+x C3 cos

(n

)

+x

2

6

xne x

(iii) y

Ans : e x x n + xn

(iv)

−

⋅n n − x

x

!

x

Ans : n

n

x

2

C

−x

2

C

⋅

2

x

n 2

+

+ xn

n

+x

⎤⎤

e x os x

(v) ⎡

An

(vi) y

e

⎧

x

+ n C2 cos

2

⎧ 2 n 1

{n − +n (vii) y

+x +

+ cos

⎤

x 2 tan 1x An

3. If

+

x log

x

x 1 x +1

x

⎡

x

Ans :

then show that (

)! sin n

2

− 2) sin n cos − 2) si

cos

) sin (n − 2)

n

)0 = 210 .

)!

x

x n ⎫⎤

x n ( x 1)

( x + 1) n ⎭

[WBUT 2001]

3.25

Successive Differentiation

e tan

4. If

1

x

, prove that

(i) 1

=0

(ii) 1 +

2

yn + 2

=0

x = tan x and n is a +ve integer, prove with the help of Leibnitz’s theorem

5. If

that f

C4 f (

C f

6. If y

1

m sin

(i) 1

2

y

(ii) 1

2

yn + 2 1

7. If y (i) 1 −

2

= n

n 2

[WBUT 2001]

0 y

2

+1

2

yn = 0

2

x , prove that 1

−2 = 0

yn + 2 1

8. If y

(0)

x , prove that m2 y

y2

2

(ii) 1

4)

xy

n 2 yn = 0

+1

x then show that

(i) 1

=1

(ii) 1 +

2

+

yn +1

=0

Find also the value of yn at x = 1

[WBUT 2003, 2005]

1

= 2 x prove that

9. If 2

(i)

x

(ii)

x2

10. If x n yn

m2 y

y yn

2

n

0 n −m

y

yn = 0

y = 1, prove that the n-th derivative of x n y n is n

C

2

yn x

C2

2

yn x

n

C

2

y n −3 x 3

n n

}

x . [WBUT 2002]

11. If y =

1+ x prove that 1 1− x

{2( −

−

−

)( −

n 2

=0

3.26

Engineering Mathematics-I

12. If y

(x

l

)

(

)

1 + x 2 , prove that x 2 1 yn + 2 (2n (2 n 1) xxyn +1 n2 yn = 0

d n ⎛ log x ⎞ n! ⎛ 1 1 = (−1) n n +1 ⎜ log x − 1 − − − n ⎜ x ⎟ 2 3 dx ⎝ x ⎝ ⎠

13. Show that

1⎞ − ⎟. n⎠ [WBUT 2003, 2008]

x3

14. If y =

⎧ 0, if n is even ⎫ , prove that ( yn )0 = ⎨ ⎬ for n > 1. ⎩−n!, if n is odd ⎭ x −1

15. If y

(x

2

2

1

)

n

(

)

then prove that x 2 1 yn

2

2 xyn

1

− n(n 1) yn = 0 [WBUT 2006, 2009]

−1

sin x

16. If y =

17. If y

1− x e m cos

18. If yn =

dn dx n

−1x

(

2

(

2

, prove that 1 −

2

, prove that 1 −

(x

n

)y

)y

n+2

− (2

3) yn +1 (n + 1)2 yn = 0

n+2

− (2

1) xyn +1 −

)

log x then prove that yn

nyn

1

(

2

+

(n − 1)!

2

)y

n

=0

[WBUT 2007]

n

⎛ y⎞ ⎛x⎞ 19. If cos −1 ⎜ ⎟ = log ⎜ ⎟ , prove that x 2 yn + 2 + (2 n + 1) xxyn +1 + 2 n2 yn = 0 ⎝b⎠ ⎝n⎠ 20. If y = a cos(log x) b n (log x), prove that

(

)

x 2 yn + 2 + (2 n + 1) xy x n +1 + n2 + 1 yn = 0

[WBUT 2007]

Multiple-Choice Questions 1. If y

e −3 x then yn is given by

a) e −3 x

b) ( 3) n e −3 x

c) ( 3) n

d) none.

2. If y = 3−5 x then yn is given by a) ( 1) n 5n ( g 3) n 3−5 x

b) ( 1) n (log 3) n 3−5 x

c) 5n (log 3) n 3−5 x

d) none

3. The n-th derivative of ( a) a10

b) 10!a10

)10 when n > 10 is c) 0

d) 10! [WBUT 2007]

3.27

Successive Differentiation

4. If y = x cos x then yn is given by ⎛ π ⎞ a) x cos n + x ⎟ ⎝ 2 ⎠ π ⎛ π ⎞ ⎛ ⎞ c) cos ⎜ n + x ⎟ + n cos n − 1 + x ⎟ 2 ⎝ 2 ⎠ ⎝ ⎠

d) none

i 4 x + cos 4 x then ( yn )0 is given by

5. If y

a) ( 1) n 4 n 2

6. If y a) 1 7. If y

π ⎛ π ⎞ ⎛ ⎞ b) x cos n + x + n cos n − 1 + x ⎟ 2 ⎝ 2 ⎠ ⎝ ⎠

b) 4 n

c) 0

d) none

4 x then ( y4 )0 is given by b) 4

c) 0

d) none

e ax + b then ( y6 )0 is given by

a) aeb

b) a 6 eb

c) a 6 e a

b

d) none

1+ x then ( yn )0 is given by 1− x

8. If y = log

a) 0 b) 1, when n is even and −1, when n is odd c) −1, when n is even and 0, when n is odd d) none 9. If y

(

)

a 2 − x 2 then yy2 + y12 is

a) 2

b) 1

c) 0

d) none

c) ( !)

d) none

n

x then ( yn )0 is x −1 n a) −( !) b) ( − )

10. If y =

11. If y = e3 x si 4 x then yn is given by 4⎞ ⎛ 44xx n tan −1 ⎟ 3⎠ ⎝ 4⎞ ⎛ c) sin 4 x n tan −1 ⎟ e3 x 3⎠ ⎝ a) 5n si

12. If y

2

b) 5n si d) none

4 x + x 4 then ( yn )0 is given by

a) 120

b) 4 n si

c) 0

d) none

13. If y = a) 3

4⎞ ⎛ 4x 4 x n tan −1 ⎟ e3 x 3⎠ ⎝

4⎞ ⎛ 4 n tan −1 ⎟ 3⎠ ⎝

x cos 3 x then ( y5 )0 is given by b) 15

c) 0

d) none

3.28

Engineering Mathematics-I

(2 3 x) then ( y6 )0 is given by

14. If y a)

−(

36

15. If y =

a)

) ⋅ 26

−(

) ⋅ 36 26

c) 1

d) none

1 then ( y4 )0 is given by (3 x + 5)

34

b)

5

5

Answers: 1. b 10. a

b)

2. a 11. b

−(

3. c 12. d

) ⋅ 34 5

5

4. b 13. c

c)

4! ⋅ 34

5. b 14. b

d) none

55

6. d 15. c

7. b

8. c

9. d

CHAPTER

4 Mean Value Theorems and Expansion of Functions 4.1

INTRODUCTION

There are some real-valued functions being continuous and derivable on a certain interval, which possess some special properties at any point lying in between boundary points of that interval. Mean-value theorems are such theorems which involve some particular results as stated above. Basically, in this chapter we discuss the very well-known three mean-value theorems, namely, Rolle’s, Lagrange’s and Cauchy’s mean-value theorems along with their wide range of applications in various fields. In this chapter, we also deal with some series expansion theorems and formulas, namely, Taylor’s and Maclaurin’s series expansion and their application towards x og(1 + x), etc. some standard functions like e , sin x, log(

4.2

ROLLE’S THEOREM

4.2.1 Statement Let f I → R be a real-valued function where I lowing conditions: i) f is continious in the closed interval [a , b]

[a , b] and f satisfies the fol-

4.2

Engineering Mathematics-I

ii) f is derivable in the open interval ( , b ), i.e., f ′ x ) exists for x ∈ (a , b ) and iii) f a ) = f b ) Then there exists at least one value of x (say c), c ∈ (a , b ), i.e., a < c < b such that, f ′ c ) = 0. [WBUT-2003] Proof: Since the function f is continuous in the closed interval [ a, b], it is also bounded there. Let us consider that m and M are the greatest lower bound ( g.l.b) and least upper bound (l.u.b) respectively for the function f . Then there exists two points c and d in [ a, b] such that f c) = m and f d ) = M. Now two cases may arise: Case i) m M In this case, the function f x ) = m is constant for all [x ∈ a, b] and correspondingly, the derivative f ′ x ) = 0 for all [x ∈ a, b]. Hence, the result is proved. Case ii) m ≠ M Since in this case f a ) = f b) and m ≠ M , at least one of m and M is different a or f b) b. from f a) Suppose m And m

f (a ); then f (c) ≠ f (a) ⇒ c ≠ a

f (b); then f c) ≠ f b) ⇒ c ≠ b

So c is neither a nor b. Therefore, a < c < b By hypothesis, f is derivable in the open interval (a, b), i.e., f ′ c) exists for a < c < b. Now it remains to prove f ′ c) = 0. For this purpose, first consider f ′ c) < 0. Then there exists an interval ( , c + h1 ), h1 > 0, for every point x ∈ (c, c h1 ), f x ) < f c) = m, which contradicts the fact that m is the greatest lower bound (g · l · b) Next we consider f ′ c) > 0. Then there exists an interval ( h 2 , c), h 2 > 0, for every point x ∈ (c − h 2 , c), f x ) < f c) = m, which is again a contradiction as before. So the only possiblity is that f ′ c) = 0 for a < c < b. Hence, the theorem is proved. Note: The theorem asserts the existence of at least one value c, where f ′ c) = 0. So, there may be more than one value of c for which the derivative vanishes.

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4.3

Mean Value Theorems and Expansion of Functions

4.2.2 Geomertical Interpretation y

A

f(a)

y = f(x)

B

C M

N

f(b)

x x=a

x=c

x=b

Figure 4.1

Rolle’s Theorem

From Fig. 4.1, it is clear that at the points A (a, f ( )) and B (b, f (b)), the ordinates are same for the continuous graph y f ( x ), i.e., the values at the points x = a and x b (which are f aa) and f bb) respectively) are equal. Since the function f is derivable in the open interval (a, b), a tangent exists at each point of the graph except the extreme points A (a, f ( )) and B (b, f ( )). Now, we can see in the graph that there exists a point C (c, f ( )) in between two extreme points A and B, at which the tangent MN is parallel to the x -axis, i.e., gradient of the tangent at C (c, f ( )) is zero. It implies that f ′ c) = 0. Correspondingly, we have a point x c in between the points x a and x = b, such that f ′ c) = 0.

4.2.3 Important Observation The conditions of Rolle’s theorem are only sufficient, they are not neccessary. This will be followed by three important examples: Example 1 Sol.

Verify Rolle’s theorem for the function f x ) = 1 − x 2 for −1 ≤ x ≤ 1.

Here, we are to examine three conditions. i) Since, f x) x is a polynomial in x and all polynomials in x are continious functions for all values of x R, f x ) = 1 − x 2 is continuous for all x, where −1 ≤ x ≤ 1.

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4.4

Engineering Mathematics-I

ii) Due to the same reason as above, f x) x is also derivable for all x, where −1 ≤ x ≤ 1. Moreover, f ′ x ) = −2 x, which exists for all values of x in −1 < x < 1. iii) f f (1) = 0. Thus, all the conditions of Rolle’s theorem are satisfied by f x ) = 1 − x 2 . Now f ′ c) = 0 ⇒ −2c = 0 ⇒ c = 0. Definitely, c = 0 lies between −1 and 1, i.e., −1 < c < 1. Hence, we can conclude that f x) x satisfies all the conditions and as such there exists c ∈ (−1, 1) such that f ′ c) = 0. Therefore, Rolle’s theorem holds good. Verify Rolle’s theorem for the function f x ) = x , −1 ≤ x ≤ 1.

Example 2

[WBUT-2003, 2009] Sol. Here, f xx) is defined as f x ) = x, for 1 =

x < 0,

0 ≤ x < 1.

f

i) f x) x is continious for all x in −1 ≤ x ≤ 1 except at x = 0. Now, lim f x ) = lim x = 0 x →0 +

x →0 +

lim f x ) =

x →0 −

x →0 −

(− x ) = 0

and f (0) = 0 so, f x) x is also continious at x = 0. Therefore, f xx) is continuous for all x in −1 ≤ x ≤ 1. ii) Here, f ′ x ) = −1, f

− 1 < x < 0,

= 1 for 0 < x < 1. So, f x) x is derivable for all x in −1 < x < 1 except at x = 0. Now we check the derivability of f x) x at x = 0. Since, f ′(0 +) = lim

h →0 +

Skar2_04_001-023.indd 4

f h) − f (0) =1 h

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Mean Value Theorems and Expansion of Functions

and f ′(0 −) = lim

h →0 −

4.5

f h) − f (0) = −1 h

i.e., f ′(0 ) ≠ f ′(0 −) so, f x) x is not derivable at x = 0. Therefore, f xx) is not derivable for all x in −1 < x < 1. iii) f

f (1) = 1.

Since all the conditions of Rolle’s theorem is not satisfied by f x) x , Rolle’s theorem is not applicable here. Here, we can observe that there exists no such c, −1 < c < 1 for which f ′ c) = 0, i.e., Rolle’s theorem does not hold since the condition (ii) is violated. Example 3

Verify Rolle’s theorem for the function f x ) =

1 1 + , i 0 ≤ x ≤ 2. x 2 x

Sol. (i) f xx) is not continious for all x in 0 0 < x < 2 ).

2 (since it is continuous for

(ii) f xx) is derivable in 0 < x < 2, since f ′ x) =

1 (2 − x )

2

−

1 x2

exists for 0 < x < 2.

(iii) f (0) and f (2) are not defined. So, f

f (2).

All the conditions are not satisfied by the function f x) x , so Rolle’s theorem is not applicable here. But it is interesting to see that 1 1 f ′ c) = − 2 =0 2 (2 − c) c ⇒ c = 1, hi h li

b

0 and 2

i.e., there exists c where 0 < c < 2 such that f ′ c) = 0. So, the result of Rolle’s theorem is still true, though all the conditions are not satisfied.

Conclusion from Examples 1 to 3: From Example 1 We have observed that all the conditions are satisfied, so Rolle’s theorem holds good.

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4.6

Engineering Mathematics-I

From Example 2 We have observed that some of the conditions are violated, so Rolle’s theorem does not hold. From Example 3 We have observed that some of the conditions are violated but Rolle’s theorem is still true. Summing up the above, we conclude that if all the conditions are satisfied by f x) x in [ a, b] then the result f ′ c) = 0, where a < c < b surely occurs. But if any of the conditions are violated by f x) x , the result f ′ c) = 0, where a < c < b may still be true but not in at all times. In the latter case, we can say that Rolle’s theorem is not applicable. So it is clear that the conditions of Rolle’s theorem are only sufficient, by no way they are neccessary.

4.3

LAGRANGE’S MEAN VALUE THEOREM (LAGRANGE’S MVT)

4.3.1 Statement Let f I → R be a real-valued function where I [a , b] and f satisfies the following conditions: i) f is continuous in the closed interval [a , b] ii) f is derivable in the open interval (a , b ), i.e., f ′( x ) exists for x ∈ (a , b ). Then there exists at least one value of x (say c), c ∈ (a , b ), i.e., a < c < b such that f b) − f a ) = f ′ c) b a

a < c < b.

[WBUT 2002, 2004]

Proof: Let us construct a function

φ ( ) as

φ ( x ) = f ( x ) k ⋅ x for x ∈ a, b]

(1)

where the constant k is to be determined such that φ (a ) = φ (b). So,

φ ( a ) = φ (b ) ⇒k

−

f (a) k

f b) − f a ) . b−a

= f (b ) + k b (2)

Now since f x) x is continuous in [ a, b], φ ( ) is continuous there and also since f xx) is derivable in ( a, b ) , φ ( ) is derivable there. Also φ (a ) = φ (b). Therefore, φ ( ) satisfies all the conditions of Rolle’s theorem in [ a, b]. So, there exists a value x = c, a < c < b such that φ ′( ) = 0.

4.7

Mean Value Theorems and Expansion of Functions

Therefore, by (1),

φ ′(c) = f ′(c) + k = 0 ⇒ f ′(c) = −k Using (2) in the above, we have f ′ c) =

f b) − f a ) for a < c < b. b−a

Hence, the theorem is proved. Note: If we consider f a ) = f b) then from the above f ′ c) = 0 for a < c < b. So, Lagrange’s MVT becomes Rolle’s theorem.

4.3.2

Geomertical Interpretation y y = f(x)

B

A N C M f(b) f(a) x x=a

x=c

x=b

Figure 4.2 Lagrange’s MVT Figure 4.2 represents a curve y f ( x ) which is continuous in [ a, b] and derivable in (a, b). Now we consider the chord AB joining the two points A (a, f ( )) and B (b, f (b)) of the curve. So, gradient of the chord AB is BN f b) − f a ) = . AN b a Again since the function is derivable everywhere in (a, b), a tangent exists at every point between the extreme points A and B. Now we draw a tangent MN which is parallel to the chord AB and touches the curve at the point C c, f (c)). Here the point C lies between the points A and B and correspondingly x c lies between x a and x = b.

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4.8

Engineering Mathematics-I

Then gradient of the tangent MN is f ′ c) (Since here at C, x c ). Since the chord AB and the tangent MN are parallel, their gradients are same. So, we have f b) − f a ) = f ′ c) h a < c < b. b a

4.3.3 Other Forms of Lagrange’s Mean-Value Theorem (1) If in the statement of Section 4.3.1 we consider b = a + h, h > 0, then the point c where a < c < a h can be repesented as c = a + θ h, h, 0 < θ < 1 and correspondingly Lagrange’s MVT in the interval [ , a + h] can be written as f a h

f (a ) hf h ′ a + θ h), ),

0 < θ < 1.

(2) Now if we put a = 0 and h x in the above form then Lagrange’s MVT in the interval [0, ] can be written as f x) = f x f ′ θ x ), 0 < θ < 1. Example 4 1 Sol.

x

Verify Lagrange’s MVT for the function f x ) = x 2

3 x + 2 for

2. Here, we are to examine two conditions. i) Since f x) x is a polynomial in x and all polynomials in x are continuous functions for all values of x R, f x ) = x 2 3 x + 2 is continuous for all x, where 1 x 2.

ii) Due to the same reason as above, f x) x is also derivable for all x, where 1 x 2. Moreover, f ′ x ) = 2 x + 3, which exists for all values of x in 1 x 2. Since all the conditions of Lagrange’s MVT are satisfied by f x ) = x 2 in 1 x 2,

3x + 2

there should exist c ∈ (1, 2) such that f (1) = f ′ c). 2 1 Now the above implies f

12 − 6 = 2 c + 3. 1 3 i.e., c = . 2

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Mean Value Theorems and Expansion of Functions

4.9

3 lies between 1 and 2. 2 Hence Lagrange’s MVT is verified for the given function.

Here, c =

Example 5 where 0 ≤ x ≤ Sol.

Verify Lagrange’s mean-value theorem for f x ) = cos x,

π 2

Here we are to examine two conditions. i) f x ) = cos x is continuous for all values of x, 0 ≤ x ≤

π . 2

π . 2 Since all the conditions of Lagrange’s MVT are satisfied by f x ) = cos x in π 0≤x≤ , 2 ⎛ π⎞ there should exist c ∈ ⎜ 0, ⎟ such that ⎝ 2⎠

ii) f ′ x ) = − sin x exists for all values of x, 0 < x
0 for x1 < c < x2 . Therefore, f x2 ) − f x1 ) > 0 ⇒ f x2 ) > f x1 ) x2 x1 So, x1 < x2

f x1 ) < f x2 ) and hence the function is increasing.

In the 2nd case, since f ′ x ) < 0 in [ a, b], also f ′ x ) < 0 in [ 1 , x2 ] and so, f ′ c) < 0 for x1 < c < x2 . Therefore, f x2 ) − f x1 ) < 0 ⇒ f x2 ) < f x1 ) x2 x1 So, x1 < x2

4.4 4.4.1

f x2 ) < f x1 ) and hence the function is decreasing.

CAUCHY’S MEAN-VALUE THEOREM (CAUCHY’S MVT) Statement

Let f I → R and g I → R be two real-valued functions where I f and g satisfy the following conditions, i) f and g are both continuous in the closed interval a , b]; ii) f and g are both derivable in the open interval (a , b ); iii) g ′( x ) ≠ 0 for all values of x in a < x < b;

[a , b] and

4.16

Engineering Mathematics-I

Then there exists at least one value of x (say c) c ∈ (a , b ), i.e., a < c < b such that, f b) − f a ) f ′ c ) = for a < c < b. g b) − g(a ) g ′(c )

[WBUT 2001, 2003]

Proof: Let us construct a function φ ( ) as

φ ( x) = f ( x))

g ( x)

, b]

(1)

where the constant k is to be determined such that φ (a ) = φ (b). So, φ ( a ) = φ (b ) f (a ) k g((a a ) = f (b ) k ⋅ g (b ) ⇒k

−

f b) − f a ) . g b) − g( a )

(2)

Here, g b) ≠ g(a ), otherwise it satisfies the conditions of Rolle’s theorem which results g ′( x ) = 0, a < x < b, which contradicts the condition (iii) of the theorem. So k is finite. Now since f x) x and g( x ) both are continuous in [ a, b], φ ( ) is continuous there x and g( x ) are derivable in ( a, b ) , φ ( ) is derivable there. and also since f x) Also, φ (a ) = φ (b). Therefore, φ ( ) satisfies all the conditions of Rolle’s theorem in [ a, b]. So, there exists a value x = c, a < c < b such that φ ′( ) = 0. Therefore by (1), f ′(c) φ ′(c) = f ′(c) + k ⋅ g ′(c) = 0 ⇒ = −k g ′(c) Using (2) in the above, we have f ′ c) =

f b) − f a ) for a < c < b. g b) − g( a )

Hence the theorem is proved. Note: When, g( x ) = x, Cauchy’s mean-value theorem takes the form of Lagrange’s mean-value theorem f b) − f a ) = f ′ c) f a < c < b b a

4.4.2 Other Forms of Cauchy’s Mean-Value Theorem 1) Let b = a + h; then the point c where a < c < a h is repesented as c = a + θ h, h,

Skar2_04_001-023.indd 16

0 1) is a positive integer π 2 0

∫

cos n x dx , where here n (> 1) is a positive integer [WBUT 2008]

(a) Let us consider In

∫ cos x dx = ∫ cosn −1 cos cos x dx n

Integrating by parts, we have In

cosn

1

⎧d ( n x ∫ cos x dx − ∫ ⎨ dx ⎩

1

x)

⎫

∫ cos xdx ⎬ dx ⎭

= cosn −1 ssin in x − ∫ −(n 1) cosn −2

sin x sin xdx

= cosn −1 ⋅ ssin i x + (n −1) 1) ∫ cosn −2

sin 2 xdx

= cosn −1 ssin in x + (n (n − 1) ∫ cosn −2 (1 cos2 x)dx = cosn −1 ssin in x + (n (n − 1) ∫ cosn −2 x − ( i.e., I n

cosn

i.e., n I n = cos

1

1) ∫ cosn xdx

x sin x + (n − 1) I n −2 (n 1) I n

n −1

x sin x + (n 1) I n −2

5.7

Reduction Formula

⇒ In =

cos n−1 x sin x ( n − 1) + I n− 2 n n

...(1)

Therefore, (1) represents the reduction formula of I n positive integer. (b) Let us consider

n

x dx, where n is any

π

π

∫0

Jn

∫ cos

n

xdx = [ I n ]02

Now taking limits on both sides of (1), we have π I n 02

[ ]

π

⎡ cos n −1x sin x ⎤ 2 (n − 1) =⎢ [ In ⎥ + n n ⎣ ⎦0

i.e., J n =

(

1) n

π

]02

⋅ J n −2

...(2) π

Therefore, (2) represents the reduction formula for J n

∫0

n

x dx d .

Calculation of the value of the Definite Integral π

Jn

∫0

n

xd x , where n(> 1) is a positive integer. xdx xdx,

It is very interesting to see that π

Jn

∫0

π

n

⎛π ⎞ x dx d = ∫ 2 cosn ⎜ − x dx 0 ⎝2 ⎠

π

∫0

s

So, the values of two definite integrations same.

n

x dx π 2 0

∫

π 2 0

∫

cosn x dx and

sin n x dx are

π

Hence from the last section 5.2, we have the value of J n following

Case (i) When n is even. Jn =

( n 11)) ( n − 3) ( n 5) n ( n 2) ( n 4)

3 1 π . 4 2 2

Case (ii) When n is odd. Jn =

( n 1) ( n − 3) ( n 5) n ( n 2) ( n 4)

4 2 . 5 3

∫0

n

xdx as the

5.8

Engineering Mathematics-I

Alternative method of finding Reduction Formula for π 2 0

∫

cos n d , where n (> 1) is any positive integer.

Let π

∫0

Jn

n

xdx

π

= ∫ 2 cosn −1 x cos x dx 0

Using integration by parts, p

È Î

Jn

n -1

x cos xdx ˘ 2 ˙˚ 0

Ú

Ú

p 2

0

¸ Ï d( n 1 x) c xdx ˝dx cos Ì dx ˛ Ó

Ú

π

= 0 + ∫ 2 ( − 1) cosn −2 ⋅ sin x ⋅ sin x ⋅ dx 0

π

= (n − 1) ∫ 2 cosn −2 x ⋅ sin 2 xdx 0

π

n 2 = (n − 1) ∫ 2 cosn− x(1 − cos2 x)dx d 0

π

i.e., J n = (n − 1) ∫ 2

n −2

0

π

x (n 1) ∫ 2 cosn xdx 0

i.e., J n = (n −1) 1) J n −2 (n 1) J n i.e., J n =

(n 1) J n −2 n π

Therefore the reduction formula of J n = ∫ 2 cosn x dx, where n(> 1) is any positive 0 integer is given by Jn =

(n 1) J n −2 . n

Example 4 Sol.

Using reduction formula, find

If we consider In

∫ cos

n

xdx

then the reduction formula is n −1

x sin x (n 1) + I n −2 n n We are to calculate In =

cos

I4

∫ cos

4

x dx.

∫ cos

4

x dx.

5.9

Reduction Formula

Here cos3 x sin x 3 + I2 4 4 1 cos x s in x I 2 ∫ cos2 xdx = + I0 2 2 I 0 ∫ dx x Therefore, I4

∫ cos

I4

4 ∫ cos xdx =

4

xdx =

cos3 x sin x 3 + I2 4 4

=

cos3 ssin in x 3 ⎧ cos sin x 1 ⎫ + ⎨ + I0 ⎬ 4 4⎩ 2 2 ⎭

=

cos3 ssin in x 3 ⎧ cos sin x 1 ⎫ + ⎨ + x⎬ 4 4⎩ 2 2 ⎭

Example 5

Using reduction formula, find

π

Sol.

If we consider π

Jn

∫0

n

xdx

Then the reduction formula is Jn =

(n 1) ⋅ J n −2 n

We are to calculate π

J6

∫0

6

x dx

Here, J6

5 J4 6

Again J4

3 J2 4

and J2

1 J0 2

But π

J0

∫0

dx =

π 2

∫02 cos

6

d .

5.10

Engineering Mathematics-I

Therefore, 5 3 1 π 5π ⋅ ⋅ ⋅ = 6 4 2 2 32

J6 =

5.4

REDUCTION FORMULA FOR

∫ sin

m

(a)

cosn xdx, where m (> 1) and (

) are positive

integers π 2 0

∫

(b)

sinm

cos n x dx , where m (> 1) and a d (

positive integer s

) are [WBUT 2008]

(a) Let I m, n = ∫ sin m x cosn x dx = ∫ cosn

1

((sin (ssin n m cos x) dx

Integrating by parts, we have I m, n = cosn

1

⎫ ⎧ d ( n 1 x) x ∫ sin m x cos xdx − ∫ ⎨ sin m x cos xdx ⎬ dx d ∫ dx ⎭ ⎩

= cosn −1 x

sin m +1 x − ∫ −( m +1

= cosn −1 x

sin m +1 (n − 1) + sin m cosn −2 x sin 2 xdx 1) ∫ m +1 (

= cosn −1 x

sin m +1 x ( + ( m +1

so, I m, n = cosn −1 x ⇒

1) cosn −2 sin x

1) sin m cosn −2 x(1 − cos2 x)dx 1) ∫

sin m +1 x (n 1) (n 1) sin m x cosn −2 xdx − + sin m cosn xdx ∫ (m 1) (m + 1) ∫ m +1

sin m +1 x ( − 1) ( + ) I m, n = cos n −1 x + I m,, ( + 1) ( + 1) m +1

⇒ I m, n =

sin m +1 x dx m +1

cos nn−1 x s ( m + n)

+1

If we write I m, n = ∫ sin m x cosn x dx

x

+

( n − 1) I m, n − 2 ( m + n)

−2

...(1)

Reduction Formula

= ∫ sin m

5.11

((cos cosn sin x) dx

1

Then we can get sin m

Im,n = −

1

n x ( m + n)

1

x

+

( m 1) Im ( m n)

2, n

...(2)

Therefore, the reduction formula of I m, n = ∫ sin m x cosn x dx is given by both the formulas (1) and (2). (b) Let us consider π 2 0

J m, n = ∫ sin x cos x dx m

n

π 2

⎡⎣ I m, n ⎤⎦ . 0

Now taking limits on both sides of (1), we have π ⎡⎣ I m, n ⎤⎦ 2 0

⎡ cosn 1 x sin m =⎢ (m n) ⎣

i.e., J m, n =

π

1

π x ⎤ 2 (n 1) ⎡⎣ I m, n −2 ⎤⎦ 2 ⎥ + 0 ⎦ 0 (m n)

(n 1) J m, n − 2 (m n)

...(3)

Again taking limits on both sides of (2), we have π ⎡⎣ I m, n ⎤⎦ 2 0

π

π ⎡ sin m −1 x cosn +1 x ⎤ 2 (m 1) ⎡⎣ I m −2, n ⎤⎦ 2 = ⎢− ⎥ + 0 (m n) ⎣ ⎦ 0 (m n)

i.e., J m, n =

(m 1) Jm (m n)

2, n

...(4)

Therefore, both the formulas (3) and (4) represent the reduction formula of π

I m, n = ∫ 2 sin m x cosn x dx d . 0

Observation: π

J m, n = ∫ 2 sin m x cosn xdx 0

π

= ∫ 2 sin m 0

⎛π ⎝2

⎞ ⎛π ⎞ cosn ⎜ − x ⎟ dx ⎠ ⎝2 ⎠

= J n, m

Calculation of the Value of the Definite Integral π

J m, n = ∫ 2 sin m x cosn x dx d . 0

5.12

Engineering Mathematics-I

Let us consider the reduction formula (4), i.e., J m, n =

(m 1) Jm ( m n)

2, n

(m 1) Jm (n m)

=

...(5)

2, n

Now replacing m by m − 2, we get from (5) Jm

2, n

=

(m − 3) Jm (n + m − 2)

...(6)

4, n

From (5) and (6), we obtain J m, n =

(m 1) (m 3) ⋅ ⋅ Jm (n m) (n m 2)

...(7)

4, n

Again replacing m by m − 2, we get from (6) Jm

4, n

=

(m − 5) Jm (n + m − 4)

...(8)

6, n

From (7) and (8), we have J m, n =

(m 1) (m 3) (m 5) ⋅ Jm ⋅ (n m) (n m 2) (n m 4)

...(9)

6, n

Similarly proceeding, we have the following cases:

Case (i) m is odd and n is any (odd or even) integer J m, n =

(m 1) (m 3) (m 5) ⋅ ⋅ (n m) (n m 2) (n m 4)

4

⋅

2

(n 5) (n 3)

J1, n

Again π

∫0

J1, n

i x cosn x dx π

⎡ cosn +1 x ⎤ 2 1 = −⎢ . ⎥ = + 1 ( ) ( 1) ⎣ ⎦0 Hence ( m 1) ( m 3) ( m 5) J m, n = ( n m) ( n m 2) ( n m 4) =

2 4 i 6 ( m 3) i ( ( 1) (n ( n + 3)… (

4

2

1 ( n 5) (n ( n 3) ( n + 1)

1) . )

Case (ii) m and n both are even integers J m, n =

(m 1) (m 3) (m 5) ⋅ ⋅ (n m) (n m 2) (n m 4)

3

⋅

1

(n 4) (n 2)

J 0, n

5.13

Reduction Formula

Again π

J 0, n = ∫ 2 cosn x dx 0

Since n is even we have from section 5.3, J 0, n =

3 1 π ⋅ ⋅ . 4 2 2

(n 1) (n − 3) (n 5) ⋅ ⋅ n (n 2) (n 4)

Hence 3 ⎤ ⎡ ( n 1) ( n − 3) ⎢ ⎥ ( 2) ( n 4) ((n n 2) ⎦ ⎣ n

( m 3) ⎡ ( m 1) J m, n = ⎢ ⎣ ( n m) ( n m 2) =

[1

3i5

3

1)][1 i 3 5 … (

( m 3) i (

2 4 6 i

3) ( n − 1)] π i 2

( n + m)

Using reduction formula, find ∫ sin 3

Example 6 Sol.

1

cos2 x dx

The reduction formula of I m, n = ∫ sin m x cosn x dx is given by I m, n =

cosn −1 x sin m +1 x (n 1) + I m, n ( m n) ( m n)

2

Here, m = 3, n = 2 Now I 3, 2 =

cos x sin 4 x 1 + I 3, 0 5 5

and I 3, 0

∫ si n = ∫ sin

3

x dx d = ∫ (1 − cos2 x) sin x dx d

∫ cos

2

x ⋅ sin x dx

3

= − cos x +

cos x 3

Therefore, cos x sin 4 x 1 + I 3, 0 5 5 4 cos sin x 1 ⎧ cos3 x ⎫ + ⎨− cos x + = ⎬ 5 5⎩ 3 ⎭

I 3, 2 =

Example 7 Sol.

Using reduction formula, find

π 2 0

∫

sin 3 cos2 x dx

Since π

J m, n = ∫ 2 sin m x cosn xdx = 0

2 4 ⋅ 6 (m − 1) (n 1)(n + 3)(n 5) 5) (n + m)

3 1 π⎤ . 4 2 2 ⎥⎦

5.14

Engineering Mathematics-I

is odd and n may be odd or even integers.

when

Here, m = 3 (odd), n = 2 (even) Therefore, π

J 3, 2 = ∫ 2 sin 3 x cos2 x dx 0

=

2 2 = 3.5 15

Example 8 Sol.

Using reduction formula, find

π 2 0

∫

sin 2 cos2 x dx

Since π

1 3 ⋅ 5...(m − 1) ⋅ 1 3 ⋅ 5...(n − 1) π 2 2 4 ⋅ 6...(m + n) when both m, n are even integers. J m, n = ∫ 2 sin m x cosn xdx = 0

Here, m = 2 (even), n = 2 (even). Therefore, π

J 2, 2 = ∫ 2 sin 2 x cos2 x dx 0

11 π π = = 2 4 2 16

5.5 (a)

REDUCTION FORMULA FOR

∫ cos

m

x sin si nx dx , where here m and n are

positive integers. (b)

π 2 0

∫

cosm x sin nx dx , where he e m and

are

positive integers. (a) Let I m, n = ∫ cosm x sin nx dx Integrating by parts, we have

(

)

⎧ d cosm x ⎫ ⎪ ⎪ I m, n cos x ∫ sin nx dx − ∫ ⎨ sin nx dx ⎬ dx ∫ dx ⎪⎩ ⎪⎭ ⎛ − cos nx ⎞ m m −1 = cosm x ⎜ ssin in x ⋅ cos nx dx ⎟ − n ∫ cos n ⎝ ⎠ m

5.15

Reduction Formula

Since sin nx cos

sin I

cos nx n

x

− cosm cos nx n − cos

cos nx n

⇒

+ n

⇒ Im

n

Im

cos nx n

os nx si

m cos m −1 x n∫

nx

m n −

∫ cos

m [ n nx

+

m Im n

cosm x nx m + Im m+n m+n

− cos m x m n

n

x

+

x} dx

s

− xdx

sin

sin(n 1)

1]

Therefore, the reduction formula of I m integers is given by Im

nx cos

x sin n

sin xdx −

−

− cosm x n

n

m

s

1

n −1

n

m Im m n

= cosm x sin nx dx where m n are positive

...(1)

1

(b) Let us consider Jm

n

= ∫ 2 cosm x sin nx dx = I m

2 n 0

Now taking limits on both sides of (1), we have Im i.e.,

2 n 0

Jm

n

⎡−

nx ⎤ 2

m

m n

+

0

m 1 + ⋅ Jm m n m n

m I m n

2 −1 0

...(2)

1

Therefore, (2) represents the reduction formula for

∫ 2 cos

Alternative Method of Finding Reduction Formula for

∫ 2 co

m

s n x dx

Let Jm

n

=

cosm x sin nx dx

= cosm

sin nxd

2 0

∫

d cosm x 2

dx

∫ sin nxdx

dx

sin x dx.

5.16

Engineering Mathematics-I π

π ⎤ ⎡ − cosm cos nx ⎤ 2 m ⎡ π m s nxddx − ∫ 2 cos m −1 sin ( − 1) x dx ⎥ =⎢ ⎥ − ⎢ ∫02 cos x sin 0 n ⎥⎦ ⎣ ⎦ 0 n ⎢⎣ π

⎡ − cosm x cos nx ⎤ 2 m ( + ) ⇒ J m, n = ⎢ ⎥ + J m −1, 1, n −1 n n ⎣ ⎦0 n π

⎡ − cosm x cos nx ⎤ 2 m =⎢ J m −1, n −1 ⎥ + m+n ⎣ ⎦0 m + n

⇒ J m, n

⇒ J m, n =

m 1 + J m−1, n −1 m+n m+n π

Therefore, the reduction formula of J m, n = ∫ 2 cosm x sin nx dx where m, n are 0 positive integers is given by J m, n =

m 1 Jm + m+n m+n

1, n 1

π

Example 9

Using reduction formula, find π

Sol.

∫02 cos

3

sin 2 x dx

The reduction formula of J m, n = ∫ 2 cosm x sin nx dx where m, n are positive 0 integers is J m, n =

m 1 Jm + m+n m+n

1, n 1

Here, m = 3 and n = 2. Now, J 3, 2 = J 2, 1 =

1 3 2

+

3 3 2

J 2, 1 =

1 2 1 2 J1, 0 = + J1, 0 + 2 +1 2 +1 3 3 π

and J1, 0

∫0

dx = 0

Therefore, J 3, 2

1 3 + J 2, 1 , 5 5

1 3 J 2, 1 5 5 1 3 1 2 = + ⋅ = 5 5 3 5

5.17

Reduction Formula

5.6

REDUCTION FORMULA FOR dx , where n(> 1) is a positive integer a 2 )n

∫ (x 2

(a)

∞

∫0

(b)

dx (x

2

, where n (> 1) is a positive integer

a 2 )n

(a) Let us consider dx 1 ⋅ 1 dx In = ∫ 2 =∫ 2 2 n (x a ) ( x a 2 )n Now integrating by parts, we have In = = = = =

⎡d ⎧ ⎤ ⎫ 1 ⋅ ∫ 1 dx − ∫ ⎢ ⎨ 2 ⎬ ⋅ ∫ 1 dx ⎥ dx n 2 a ) ⎣ dx ⎩ ( x a ) ⎭ ⎦

1 (x

2 n

2

1 (

2

(

2

2 n

)

⋅( ) + ∫

2 n

)

+ 2n 2n ⋅ ∫

2 n

+ 2n 2n ⋅ ∫

x

x (

2

(

2

)

n 2x (

2

(

2

x 2 n

2

(x + a )

⇒ 2 ⋅ a2 ⋅

n +1

=

2 n +1

2 n +1

2

(

dx

dx

)

1 2 n

2

)

dx − a 2 ⋅ ∫

1 (

2

2 n +1

)

⎤ dx ⎥ ⎦

+ 2 n ⋅ ⎡⎣ I n − a 2 ⋅ I n +1 ⎤⎦ x

(

⋅ ( x) dx

)

x2 + a2 − a2

2 n

⇒ In =

)

x2

⎡ + 2n 2n ⋅ ⎢ ∫ ) ⎣ (

x

2 n +1

2

+

2 n

)

+ (2 n − 1) ⋅ I n

Now replacing n by n −1, we have from above 2 ( n 1) a2 i I n =

x (

2

2 n −1

)

+ (2n (2 n 3) I n−1

So (1) represents the reduction formula for dx In = 2 , h n(> 1) is a positive integer. ( x a2 )n (b) Let us consider Jn = ∫

∞

0

dx (x

2

a2 )n

...(1)

5.18

Engineering Mathematics-I

[ I n ]∞0 .

so, J n

Now taking limits on both sides of (1), we have ⎡ ∞ 2 (n 1) a 2 ⋅ [ I n ]0 = ⎢ ⎣(

x 2

⎡ 2(n 1) a 2 ⋅ J n = ⎢ lim ⎣ x →∞ ( Again x lim

x →∞

x 2 n −1

2

)

−0

⎤ ⎦

(2

is an indeterminate form of

2 n −1

2

∞

⎤ + (2n (2 n 3) [ I n 2 n −1 ⎥ ) ⎦0

(x + a )

∞ 1 0

]

3) J n −1

...(2)

∞ . ∞

So using L’Hospital’s rule, we have x lim

x →∞ ( x

2

+ a 2 ) n −1

x →∞

d ⎡ 2 ( x + a 2 ) n −1 ⎤⎦ ⎣ dx

1

= lim

1) ( x + a 2 ) n 2

x →∞ (

i.e.,

= lim

d ( x) dx

x lim

x →∞

( x + a 2 ) n −1 2

2

2x

= 0.

...(3)

Using the result (3) in (2), we have 2 (n 1) a 2 ⋅ J n = (2 Jn =

i.e.,

3) ⋅ J n −1

1 (2 n 3) i i J n −1 a2 (2 n − 2)

...(4)

Hence (4) is the reduction formula for Jn = ∫

∞

0

dx (x

2

a2 )n

, where n(> 1)) is a positive integer.

Calculation of the Value of the Definite Integral Jn = ∫

∞

0

dx (x

2

a2 )n

,

n (> 1) is a positive integer.

From (4), we have the reduction formula for above as Jn =

1 (2 n 3) ⋅ ⋅ J n −1 a 2 (2 n − 2)

Now replacing n by n −1, we get from (5)

...(5)

Reduction Formula

1 (2 n − 5) ⋅ ⋅ J n −2 a 2 (2 n − 4) From (5) and (6), we obtain J n −1 =

Jn =

5.19

...(6)

1 (2 n 3) 1 (2 n 5) ⋅ ⋅ ⋅ ⋅ J n −2 a 2 (2 n − 2) a 2 (2 n − 4) ⎞ (2 n − 3) (2 n − 5) ⎟ ⋅ (2 n − 2) ⋅ (2 n − 4) ⋅ J n −2 ⎠

⎛ 1 1 i.e., J n = ⎜ 2 ⋅ 2 ⎝a a

...(7)

Again replacing n by n −1, we get from (6) J n −2 =

1 (2 n − 7) ⋅ ⋅ J n −3 a 2 (2 n − 6)

...(8)

From (7) and (8), we have ⎛ 1 1 1 ⎞ (2 n 3) (2 n − 5) (2 n − 7) Jn = ⎜ 2 ⋅ 2 ⋅ 2 ⎟ ⋅ ⋅ J n −3 ⋅ ⋅ ⎝ a a a ⎠ (2 n − 2) (2 n − 4) (2 n − 6) Similarly proceeding as above ⎛ ⎞ ⎜ 1 1 1 ⎟ (2 n 3) (2 n − 5) (2 n − 7) 5 3 1 Jn = ⎜ 2 ⋅ 2 ⋅ ⋅ ⋅ ⋅ ⋅ J1 ⎟⋅ a 2 ⎟ (2 n − 2) (2 n − 4) (2n − 6) 6 4 2 ⎜a a ⎝ ( n 1) terms ⎠ 1 (2 3) (2 n − 5) (2 n − 7) 5 3 1 = 2( n−1) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ J1 (2 n − 2) (2 n − 4) (2 n − 6) 6 4 2 a Again J1 = ∫

∞

0

dx ( x2

a2 ) ∞

x⎤ ⎡1 = ⎢ ⋅ tan −1 ⎥ a ⎦0 ⎣a

1 π ⋅ . a 2 So putting the value of J1 in (10), we have =

Jn =

(2n 3) (2n − 5) (2n − 7) (2n − 2) (2n − 4) (2n − 6)

1 a

2( n −1)

a

2n 1

(2 3) (2n − 5) (2n − 7) (2n − 2) (2n − 4) (2n − 6)

1

=

5 3 1 ⎛1 π ⎞ 6 4 2 ⎜⎝ a 2 ⎟⎠ 5 3 1 π 6 4 2 2

Alternative Method for Computation In = ∫

∞

0

dx (x

2

a2 )n

,

n (> 1) is a positive integer.

...(9)

...(10)

5.20

Engineering Mathematics-I

Let x

a tan θ then dx = a sec 2θ d θ

Putting the above values in integration, we have π

a sec 2 θ dθ

0

(a 2 tan 2 θ + a 2 ) n

In = ∫ 2 =

π 2 0

1 a

∫

2n 1

cos2 n 2 θ dθ π

∫0

We have from Section 5.3 if J n Jn =

cosn x dx d , then

3 1 π ⋅ ⋅ ., when n is even 4 2 2

(n 1) (n − 3) (n 5) ⋅ ⋅ n (n 2) (n 4)

Since 2 n 2 is always even, using the result of Section 5.3, we have In = =

π 2 a2n 1 0

a

1

∫

1

⎡ (2 3) (2 n − 5) (2 n − 7) ⎢ (2 n − 2) ⋅ (2 n − 4) ⋅ (2n − 6) ⎣

2n 1

Example 10 Sol.

cos2 n

2

θ dθ

∫ ( x2

Find

dx

3 1 π⎤ ⋅ ⋅ 4 2 2 ⎥⎦

.

a 2 )2

We have the reduction formula for dx In = 2 , n (> 1) is a positive integer. ( x a2 )n as 2 (n 1) a 2 ⋅ I n =

x (

+ (2n (2 n 3) I n −1 .

2 n −1

2

)

Here we are to find I 2 . By putting n = 2 in the above formula we get 2(2 1)

2 2

x

=

i.e., 2 a 2 I 2 =

2

(

2 2 1

)

x (

2

2

)

+ I1 .

Again I1 =

dx (x

2

2

a )

=

+ (2 ⋅ 2 3)

1 tan −1 a. a

2 1.

5.21

Reduction Formula

Hence I2 =

1 2a

⋅

2

x 2

2

(x + a )

+

1

⋅ tan −1a.

2a3

WORKED-OUT EXAMPLES π

If I n = ∫ 2 sin 2 n

Example 5.1 In =

1

0

θ d θ , where n is a positive integer, show that

2n I n −1 . 2n 1 π

Use this to evaluate ∫ 2 sin 7 θ dθ .

[WBUT 2001]

0

Sol.

Here π

I n = ∫ 2 sin 2 n

π

1

d

0

= ∫ 2 sin 2 nθ sin i θ dθ 0

Integrating by parts, taking sin 2nθ as the first function and sin θ as the second function, we have π π ⎫ ⎧ d ( 2n ) = sin 2 nθ ∫ 2 sin i θ dθ − ∫ 2 ⎨ sin θ d θ ⎬dθ ∫ 0 0 dθ ⎭ ⎩ π θ ]02

= [ − sin nθ

π 2

2 sin ∫ 2n

2n 1

θ

2

θ dθ

0

π 2

= 0 + 2 n ∫ sin 2 n 1 θ (1

) dθ

2

0

π 2

= 2n ∫ s

2

π 2

1

d

0

= 2 nI n

2n ∫ s

2

1

θ dθ

0

1

− 2 nI n

⇒ (2 n +1) + 1) I n = 2nI 2 nI n −1 2n ⇒ In = I n −1 . (2 n + 1) π 2

To evaluate

∫ sin 0

7

θ dθ by the above reduction formula, we have

5.22

Engineering Mathematics-I π 2

∫s

I3

7

θ dθ =

2 3 I2 (2 ⋅3 3 1)

6 I2 7

5

θ dθ =

2 2 I1 (2 ⋅ 2 1)

4 I1 5

3

θ dθ =

2 1 I0 (2 ⋅1 1 1)

2 I0 3

0

Similarly, π 2

∫s

I2

0

π 2

∫s

I1

0

π 2

∫s

I0

θ ]π0 /2 = 1

θ dθ = [

0

Therefore, from the above relations, we have π 2

∫s

I3

7

θ dθ =

0

Example 5.2 1

∫ 0

x6 1− x

Sol.

2

dx =

6 I2 7

Show that 5 π 32

Let x = sin θ then dx = cos θ dθ Then 1

x6

∫

dx

1 − x2

0

π 2

=

sin 6θ cos θ ∫ cos θ dθ 0

π 2

= ∫ sin 6θ dθ 0

64 642 16 I1 = I0 = 75 753 35

[WBUT 2003]

5.23

Reduction Formula π 2

Now from Section 5.2, we have if J n =

∫

n

x dx, then

0

Jn =

3 1 π ⋅ ⋅ , n is even. 4 2 2

(n 1) (n − 3) (n 5) ⋅ ⋅ n (n 2) (n 4)

Since here n = 6, using the result of Section 5.2, π 2

5 3 1 π

5π

∫ sin θ dθ = 6 ⋅ 4 ⋅ 2 ⋅ 2 = 32 . 6

0

Hence 1

∫ 0

x6 1− x

2

dx =

5 π. 32 1

Example 5.3

∫x

Prove that if un

n

tan −1 x dx then

0

(

Sol.

1)

n

(n 1) un −2 =

π 1 − . 2 n

[WBUT 2002]

Here 1

1

n −1 ∫ x tan xdx

un

∫

0

1

x ⋅ x n dx

0

Performing integration by parts, we have 1

⎡ −1 x n +1 ⎤ 1 1 x n +1 dx ⎢ tan x ⎥ − n + 1 ⎦ 0 ∫0 1 + x 2 n + 1 ⎣

un

1

=

1 π 1 x2 x n −1 dx ⋅ − 4 n + 1 n + 1 ∫0 1 + x 2

=

1 ⎛ π 1 1 ⋅ − 1− ⎜ ∫ 4 n + 1 n + 1 0 ⎝ 1 + x2

1

⎞ n −1 ⎟ x dx ⎠

So, 1

un =

1 π 1 ⋅ − xn 4 n + 1 n + 1 ∫0

1

1

dx +

1 1 xn n + 1 ∫0 1 + x 2

1

dx

5.24

Engineering Mathematics-I 1

un =

1 ⎡ xn ⎤ 1 ⎧⎪ ⎡ n −11 π 1 ⋅ − ⎢ ⎥ + ⎨ x tan 4 n +1 n +1 ⎣ n ⎦ n + 1 ⎩⎪ ⎣ 0

1

1

1

x ⎤⎦ − ∫ (n (n 1) x n −2 tan 0

0

1

⎫⎪ xdx ⎬ ⎭⎪

Hence from above 1

un =

1 1 1 π n − 1 n −22 π 1 ⋅ − + − x tan 4 n + 1 n + 1 n n + 1 4 n + 1 ∫0

1

xdx

1

i.e., un =

1 ⎛ π 1 ⎞ n 1 n −22 x tan − − n + 1 ⎜⎝ 2 n ⎟⎠ n + 1 ∫0

i.e., un =

1 ⎛π 1 ⎞ n 1 un − 2 − − n + 1 ⎜⎝ 2 n ⎟⎠ n + 1

1

x dx

Therefore, (

1)

(n 1) un −2 =

n

π 4

∫

If un

Example 5.4

nn

π 1 − . 2 n

dθ , prove that

0

n(un Sol.

un 1 ) = 1.

[WBUT 2003]

Here π 4

∫

un

n

0

π 4

θ dθ = ∫ tan n −2 θ tan 2 θ dθ 0

π /4

=

∫

tan n

2

1) dθ

((sec (se secc 2

0

π 4

=

∫

n 2

0

π 4

d ((tan θ ) − ∫ tan n −2 θ dθ 0

From above, we can write π

⎡ tan n −1 θ ⎤ 4 un = ⎢ ⎥ − un − 2 ⎣ n − 1 ⎦0 i.e., un =

1 − u n −2 n −1

Replacing n by n +1 in the above expression, we have

Reduction Formula

un +1 =

5.25

1 − un −1 n

i.e., n(un

un 1 ) = 1.

Hence the result is proved. Example 5.5

(n − 1) θ .

1)( I n − I n −2 ) = 2

( Sol.

sin nθ dθ , show that sin θ

If I n =

[WBUT 2004]

We have In =

sin nθ dθ sin θ

So, I n−2 = ∫

sin ( n −

)θ

sin θ

dθ

Now In

I n −2 = ∫ =∫ =∫

sin ( n − )θ sin nθ dθ − ∫ dθ sin θ sin θ sin θ sin ( n − 2 )θ sin θ 2

(

dθ

1)θ sin i θ dθ sin θ

= 2 ∫ cos (n 1) θ dθ Therefore In

I n −2 = 2

sin (n − 1) θ n −1

i.e., (n 1)(

n 2)

n

= 2 sin (

1) .

Hence the result is proved. Example 5.6 (

If I n = ∫

1) ( I n + I n −2 ) = 2

cos nθ dθ , show that cos θ

(n − 1) θ

Hence evaluate

∫ (4 cos

2

θ 3) dθ

[WBUT 2005]

5.26

Sol.

Engineering Mathematics-I

We have In = ∫

cos nθ dθ cos θ

...(1)

So, I n−2 = ∫

cos ( n −

)θ

dθ

cos θ

Now In

I n −2 = ∫

cos ( n − ) θ cos n θ d +∫ dθ cos θ cos θ cos θ + cos ( n − 2 ) θ

=∫

cos θ 2

=∫

(

dθ

1)θ cosθ dθ cosθ

= 2 ∫ cos(n 1) θ dθ Therefore In

I n −2 = 2

i.e., (n 1)(

sin (n − 1) θ n −1 n 2)

n

= 2 sin (

1) .

...(2)

Hence, the result is proved. Now we are to evaluate

∫ (4 cos θ 2

3) dθ

using the above result. For this purpose, if we put n = 3 in (1), then I3 = ∫

cos 3θ (4 cos3θ 3 dθ = ∫ cos θ cos θ

= ∫ (4

3) dθ

2

Basically, we are to find I 3 . Putting n = 3 in (2), we have (3 −1)( 1)( i.e., (

3

3

3 2) 1)

= 2 sin (3 − 1)θ

= si 2θ

)

dθ

Reduction Formula

5.27

cos θ

∫ cos θ dθ ∫ dθ = θ .

Again I1 So,

I 3 = si 2θ θ

∫

i.e.,

(4 cos2 θ 3) dθ = si 2θ θ . π 2

If I n

Example 5.7

∫x

n

sin x dx d , (n > 1) then prove that

0

In

⎛π ⎞ n (n 1) I n −2 = n ⎜ ⎟ ⎝2⎠

Sol.

Here we have π 2

In

∫x

n

n −1

.

[WBUT 2008]

sin x dx

0

Integrating by parts, we obtain

In

⎡ ⎣

π 2

π x n cos x ⎤⎦ 2 0

n ∫ x n −1 cos xdx 0

⎡ π ⎢ = 0 − n ⎢ ⎡⎣ − x n −1 sin x ⎤⎦ 2 − ( 0 ⎢ ⎢⎣ =n

⎛π ⎞ ⎝2⎠

π 2

1) ∫

n 2

0

⎤ ⎥ ( sin x) dx ⎥ (− ⎥ ⎥⎦

π 2

n −1

1) ∫ x n −2 sin x dx

n(

0

Therefore, In

⎛π ⎞ n⎜ ⎟ ⎝2⎠

n −1

− n(n 1) I n −2

⎛π ⎞ n(n 1) I n −2 = n ⎜ ⎟ ⎝2⎠ Hence the result is proved.

n −1

i.e., I n

π 2

Example 5.8

If I m, n = ∫ cosm x cos nx dx d , then prove that 0

5.28

Engineering Mathematics-I

I m, n =

m Im m+n

1, n 1 .

Hence show that π 2

π

∫ cos nx cos nx dx = 2n+1 . 0

Sol.

Here π 2

I m, n = ∫ cosm x cos nx dx

...(1)

0

π x sin nx ⎤ 2

⎡ cos =⎢ n ⎣ m

π 2

m −1 x ( sin x) ⎥ − ∫ m cos ⎦0 0

sin nx dx n

π

m2 = ∫ cosm −1 x (sin nx sin x)dx n 0

i.e., I m, n

Again cos( 1) = cos(

) = cos nx cos x + sin

...(2)

sin x

Then from (2) π

I m, n

m2 = ∫ cosm −1 x [ cos(n − 1) x cos nx cos x ] dx n 0 π

π

m2 m2 = ∫ cosm −1 x cos(n − 1)x 1) x dx − ∫ cosm x n 0 n 0 i.e., I m, n =

m Im n

⎛ m⎞ i.e., ⎜ 1 n⎠ ⎝ Hence

m, n

1, n 1

=

−

m I m, n n

m I m −11, n

1

m I m 1, n 1 . m+n So the first part is proved. To prove the second part, let I m, n =

π 2

Jn

∫ 0

n

x cos nx dx.

nx dx

...(3)

Reduction Formula

So it is clear from (1) that J n

5.29

I n, n .

Therefore, from (3), we have Jn

I n, n =

n I n −11, n n+n

1

1 J n −1 2

Now from (4) Jn

1 1 1 J n −1 = ⋅ J n −2 2 2 2 1 1 1 = ⋅ ⋅ J n −3 2 2 2 ..............

Jn =

1 1 1 ⋅ ⋅ 2 2 2

1 Jn 2

n

n terms

1

i.e., J n

2n

J0

Again π 2

∫

J0

0

0

π 2

x cos 0 dx = ∫ dx = 0

π . 2

Hence 1

Jn

2n

J0 =

π 1 π ⋅ = . 2 n 2 2 n +1 1

If I n = ∫ (si

Example 5.9

1

x) n dx then prove that

0

n

In Sol.

⎛π ⎞ n(n 1) I n −2 = ⎜ ⎟ . ⎝2⎠ Let Jn

∫( i

1

)

n

x dx

Putting x = sin v, i.e., dx = cos vdv, we have from the above, Jn

∫v

n

cos v dv

= v n sin v

∫ nv

n −1

sin vdv

= v n sin v − n ⎡⎣v n −1 ( − cos v ) (n − 1) ∫ v n −2 ( cos v ) dv ⎤⎦ Jn

v n sin v + nv n −1 cos v − n(n − 1) ∫ v n −2 cos vdv

...(4)

5.30

Engineering Mathematics-I

v n sin v + nv n −1 cos v − n(n − 1) J n −2

i.e., J n

(

= sin −1

)

n

(

)

⋅ x + n sin −1

n −1

⋅ 1 − x 2 − n(

1) J n −2

Now 1

∫ ( si

In

1

x

0

(

⎡ = ⎢ sin −1 ⎣

)

)

n

dx = [ J n ]0

n

⎤ ⎡ ⋅ x ⎥ + ⎢ n sin −1 ⎦0 ⎣

1

1

(

)

n −1

1

⎤ ⋅ 1 − x 2 ⎥ − n( ⎦0

1) [ J n −2 ]0 1

n

⎛π ⎞ =⎜ ⎟ +0 ⎝2⎠ Hence

(n − 1) I n −2

n

In

⎛π ⎞ n(n 1) I n −2 = ⎜ ⎟ . ⎝2⎠ π 2

If I m, n = ∫ sin m x cos nxdx and J m, n

Example 5.10

0

π 2

∫

i

m

x sin nxdx then prove

0

that ( Sol.

)I ) I m, n

mJ m

11,

1

= sin

nπ . 2

Here we have π 2

I m, n = ∫ sin m x cos nxdx 0

π

π

⎡ sin mx sin nx ⎤ 2 m 2 m −1 =⎢ x((cos x) sin nxdx ⎥ − ∫ sin n ⎣ ⎦0 n 0 π 2

=

nπ ⎞ m 1⎛ sin sin m −1 x(( i ⎜ ∫ n⎝ 2 ⎠ n 0

) dx

Again sin (

1) = sin (

) = sin nx cos x − cos

sin x

So, π 2

I m, n =

nπ ⎞ m 1⎛ sin sin m −1 x ( sin((n 1) x + cos nx sin x ) dx ⎜ ∫ n⎝ 2 ⎠ n 0

5.31

Reduction Formula π 2

=

nπ ⎞ m 1⎛ sin sin m −1 x sin ( n ⎜⎝ 2 ⎠ n ∫0

1) xdx −

m sin m x cos nxdx n ∫0

Therefore, I m, n =

nπ 1⎛ sin n ⎜⎝ 2

⎞ m ⎟ − n Jm ⎠

nπ ⎛ i.e., nI m, n = ⎜ sin 2 ⎝ Hence we have )I ) I m, n

(

mJ m

⎞ ⎟ − mJ m ⎠

11,

= sin

1

1

∫ (1

If I n

Example 5.11

11, n 1

)

−

m I m, n n

11, n 1

− mI m, n

nπ . 2

n

x 2 dx then prove that

0

(2 n 1) I n = 2nI 2 nI n −1 . Here

Sol.

1

In

∫ (1

x2

0

(

⎡ = 1− ⎣

2

1

)

)

n

n

⋅ 1dx

(

= 0 − 2n ∫ 1 − 0

1

(

= 2n ∫ 1 0

i.e., (2

(

)

n −1

n −1

(

2 nI n n 1)

n

2

)

n −1

(−2 ) ⋅ x dx

) ⋅ ( − ) dx 2

(

⎡ 1− ⎣

)

n

2 n ∫ 1 x 2 dx 0

i.e., I n

2

2

1

i.e., I n

1

1

⎤ ⋅ x ⎥ − n∫ 1 ⎦0 0

2

) − 1⎤⎦ dx 1

(

n ∫ 1 − x2 0

)

n −1

dx

2 nI n −1

=2

n −1 .

Hence the result is proved. 1

Example 5.12

If I m, n = ∫ x m (1 x) n dx then prove that 0

I m, n

1

=

n −1 I m, n 2 . m+n

π 2

5.32

Sol.

Engineering Mathematics-I

Here we have 1

I m, n

n −1 m ∫ x (1 − x ) dx

1

0

1

= ∫ ⎡⎣(1 0

) n −1 ⎤⎦ ⋅ ⎡⎣ x m ⎤⎦ dx

Integrating by parts, 1

I m, n

1

1 ⎡ x m +1 ⎤ x m +1 = ⎢(1 x) n −1 ⋅ (n 1) ∫ (1 x) n −2 ( 1) ⋅ dx ⎥ − (n m + 1 ⎦0 m +11 ⎣ 0 1

=0+

n −1 (1 − x) n −22 ⋅ x m 1 dx m + 1 ∫0 1

=−

n −1 (1 − x)) n m + 1 ∫0

=−

n −1 (1 − x)) n ∫ m +1 0

=−

n −1 n −1 (1 − x) n −1 ⋅ x m + (1 − x) n −2 ⋅ x m ∫ m +1 0 m + 1 ∫0

2

m

(− x) dx

2

m

⎡⎣(1 −

1

1

) − 1⎤⎦ dx 1

Therefore, I m, n

1

=−

n −1 I m, n m +1

1

+

n −1 I m, n m +1

n −1 ⎞ n −1 ⎛ 1+ ⎟ I m,, 1 = m + 1 I m,, m + 1 ⎝ ⎠ n −1 ⎛m+n⎞ i.e., ⎜ I m, n−1 = I m, n−2 ⎟ m +1 ⎝ m +1 ⎠ i.e.,

2

2

n −1 I m, n 2 . m+n Hence the result is proved. i.e., I m, n

1

=

π

Example 5.13 prove that In Sol.

1 − cos nθ dθ , where n is a positive integer or zero then 1 − cos θ 0

If I n = ∫

I n = 2 I n+1 . Here, we have π

1 − cos nθ dθ 1 − cos θ 0

In = ∫

π

d so I n + 2 = ∫ 0

1 cos ( n 2 )θ 1 − cos θ

dθ .

5.33

Reduction Formula

Therefore, π

In = ∫

In

1 − cos ( n +

)θ + 1 − cos nθ

1 − cos θ

0

dθ

os nθ ⎤⎦ 2 ⎡⎣cos ( 2 )θ + ccos dθ . 1 − cos θ 0

π

=∫ Since

⎛ A B⎞ ⎛ A B⎞ cos ⎜ ⎟, ⎝ 2 ⎠ ⎝ 2 ⎠ we obtain from above cos A cos B = 2 cos

π

In = ∫

In

2 − 2 cos ( n + )θ ⋅ cos θ 1 − cos θ

0

π

= 2∫

1 ⎡⎣(1 −

) − 1⎤⎦

π

(1 −

) ⋅ cos (

+ 1) θ

1 − cos θ

0

π

I n = 2 ∫ cos ( n

i.e., I n

cos (

1) θ

1 − cos θ

0

= 2∫

dθ

π

dθ + 2 ∫

dθ 1 cos (

0

)θ dθ + 2 I n+1

0

π

⎡ sin ( n + 1)θ ⎤ = 2⎢ ⎥ + 2 I n +1 n + 1 ⎦0 ⎣ = 0 + 2 n +1 . Hence In

I n = 2 I n+1 . 1

If I n

Example 5.14

∫x

n

1 − x 2 dx then prove that

0

In =

n −1 I n −2 n+2

Sol.

Here 1

In

∫x

n

1 − x 2 dx

0

1

= ∫ xn 0

1

(1

)

x 2 ⋅ x dx

1) θ

1 − cos θ

dθ

5.34

Engineering Mathematics-I

⎡ 1 − x2 ⎢ = ⎢ x n −1 −3 ⎢ ⎢⎣

(

)

3 ⎤1 2 ⎥

=

1) ∫ x n −2

⎥ −( ⎥ ⎥⎦ 0

1

In = 0 +

1) 3

−3

0

(

(n 1) n −2 x 1 x2 3 ∫0

(

3 2 2

(1 − x )

1

)

1

n −2 2 ∫ x 1 x dx −

dx

1 − x 2 dx (

0

1) 3

1

∫x

n

1 x 2 dx

0

(n 1) (n 1) I n −2 − In 3 3 ( 1) ⎛ n −1 ⎞ In = I n −2 i.e., 1 + ⎟ 3 ⎠ 3 ⎝ i.e., I n =

(n + )

i.e.,

3

In =

(n − 1) I n −2 3

n −1 I n −2 . n+2 Hence, the result is proved. i.e., I n =

Example 5.15

Show that

π 2

n ∫ cos x ssin nxdx = 0

1 ⎡ 2 2 23 2 + + + ⎢ 2 3 2 n +1 ⎣

+

2n 1 2n ⎤ + ⎥. (n 1) n ⎦

where m is any positive integer. Sol.

From Article 5.5, we have the reduction formula for π 2

J m, n = ∫ cosm x sin nxdx 0

as below J m, n =

m 1 + ⋅ Jm m+n m+n

Now let π 2

In

∫ cos 0

n

x sin nxdx

1, n 1

...(1)

5.35

Reduction Formula

So it is obvious that I n J n, n . and from (1), we obtain n 1 + ⋅ J n −11, n n+n n+n 1 1 i.e., I n = + ⋅ I n −1 2n 2 In

J n, n =

1

Now In = = In = =

⎞ 1 1 ⎛ 1 1 + ⋅⎜ + ⋅ I n −2 ⎟ ⎜ ⎟ 2 n 2 ⎝ 2 ( n 1) 2 ⎠ 1 1 1 + 2 + 2 ⋅ I n −2 2 n 2 (n − 1) 2 1 1 1 + + 2 n 22 (n − 1) 22

⎛ ⎞ 1 1 ⋅⎜ + ⋅ I n −3 ⎟ ⎜ 2 (n 2) 2 ⎟ ⎝ ⎠

1 1 1 1 + 2 + 3 + 3 ⋅ I n −3 2 n 2 (n 1) 2 ( n 2 ) 2

Proceeding in a similar manner, we obtain In =

1 1 1 + 2 + 3 + 2 n 2 (n 1) 2 ( n 2 )

1

+

2

n 1

2

n 2

2

+

1 2

n 1

⋅ I1

...(2)

Again π 2

1

∫ cos x sin xdx = 2 .

I1

0

Hence from (2), we have In =

1 1 1 + + + 2 n 22 (n 1) 23 ( n 2 )

+

=

1 1 1 + 2 + 3 + 2 n 2 (n 1) 2 ( n 2 )

+

=

1 ⎡ 2n 2 n −11 2n 2 + + + ⎢ 2 n +1 ⎣ n ( 1) ( n − 2 )

1 3

1 2

+

n 2

3

+ +

1 2

n 1

2

1 2

1

2

+ +

⎤ 23 2 2 + + 2⎥ 3 2 ⎦

Hence π 2

n ∫ cos x ssin nxdx = 0

1 ⎡ 2 2 23 2 + + + ⎢ 2 3 2 n +1 ⎣

+

2n 1 2n ⎤ + ⎥. (n 1) n ⎦

1 2

n −1

1 2n

⋅

1 2

5.36

Engineering Mathematics-I

EXERCISES

Short and Long Answer Type Questions 1. Obtain the reduction formula for π 2

∫ cos

n

xdx,

n (> 1)) is a positive integer

0

π 2

and evaluate

∫ cos

5

d .

[WBUT 2007, 2008]

0

8 ⎤ ⎡ ⎢ Ans : 2nd Part. 15 .⎥ ⎣ ⎦

2. Evaluate: π 2

(i)

∫ cos

6

π 2

xdx

(ii)

0

∫ cos

7

xdx

0

5π 16 ⎤ ⎡ ⎢ Ans : (i) 32 , (ii) 35 ⎥ ⎣ ⎦

3. Obtain a reduction formula for π 2

∫ sin

n

xdx,

n (> 1)) is a positive integer

0

π 2

and evaluate

∫ sin

5

d .

[WBUT 2006, 2009]

0

8 ⎤ ⎡ ⎢ Ans : 2nd Part. 15 .⎥ ⎣ ⎦

4. Evaluate: π 2

(i)

∫ sin

8

π 2

xdx

(ii)

0

π 2

5. If I n =

∫ 0

In

∫ sin

9

xdx

0

sin (2 n 1) x dx, show that sin x

In =

1 sin 2 nx. n

35π 428 ⎤ ⎡ ⎢ Ans : (i) 256 , (ii) 315 ⎥ ⎣ ⎦

5.37

Reduction Formula π 2

∫ sin

6. If I n

2n

xdx xd , show that

0

In =

2n 1 I n −1 . 2n π 2

7. If I n = ∫ sin 2 n 1θ dθ , where n is a positive integer, show that 0

In =

2n I n −1 . 2n 1 π 2

∫ sin θ dθ .

Use this to evaluate

7

[WBUT 2001]

0

16 ⎤ ⎡ ⎢ Ans : 2nd Part. 35 .⎥ ⎣ ⎦ π 2

∫ x cos

8. If I n

n

xdx d , where n is a positive integer, show that

0

In =

n −1 1 I n −2 − 2 . n n

9. Show that 1

(a)

(b)

∫

x6

0

1− x

1

x5

∫

1− x

0

∞

(c)

∫ 0

2

2

dx =

5 π 32

dx =

8 15

x4

(1 + x ) 2

2π

10. If I n =

4

cos nx

I n = 2π .

[Hint : put x = sin θ ]

π 32

∫ 1 − cos x dx, 0

In

dx =

[WBUT 2003] [Hint : put x = sin θ ]

[Hint : put x = tan θ ]

then prove that

5.38

Engineering Mathematics-I

11. Obtain the reduction formula for π 2

∫ sin

m

cosn x dx, where m (> 1)) a

n (> 1) are positive integers.

0

π 2

Hence evaluate

∫ sin

4

cos8 x dx

[WBUT 2008]

0

35π ⎤ ⎡ ⎢ Ans : 2nd Part. 1280 .⎥ ⎣ ⎦ 12. Evaluate: π 2

(i)

∫ sin

5

cos6 x dx

π 2

∫ sin

(ii)

0

6

cos8 x dx

0

8 5π ⎤ ⎡ ⎢ Ans : (i) 693 (ii) 4096 ⎥ ⎣ ⎦ 1

13. If I n

∫ ( cos 0

1

)

n

x dx then prove that

⎛π ⎞ n(n 1) I n −2 = n ⎜ ⎟ ⎝2⎠

In

π 4

14. If un

∫

n −1

.

n n dθ , prove that

0

n(un

un 1 ) = 1.

15. If I n = ∫ that Jn

sin nθ dθ , show that sin θ

1)( I n − I n −2 ) = 2

17. If I n = ∫ (

sin 2 nx sin (2 n 1) x dx, where n is any integer then prove dx and J n = sin x sin 2 x

J n = I n+1 .

16. If I n = ∫ (

[WBUT 2003]

(n − 1)θ .

cos nθ dθ , show that cos θ

1)( I n + I n −2 ) = 2

(n − 1)θ

[WBUT 2004]

5.39

Reduction Formula

∫ ( 4 cos θ 2

Hence evaluate π 2

∫x

18. If I n

n

)

3 dθ

[WBUT 2005]

sin xdx d , (n > 1) then prove that

0

⎛π ⎞ n(n 1) I n −2 = n ⎜ ⎟ ⎝2⎠

In

n −1

.

[WBUT 2008] π 2

19. Obtain the reduction formula for I n

∫ cos

2n

xdx and hence show that

0

π 2

∫ cos

2n

xdx =

( 2 )!

( 2 !) n

0

2

∫x

π . 2 2n 1 ⎡ ⎤ ⎢ Ans : First part I n = 2 n I n −1 .⎥ ⎣ ⎦

1

20. Prove that if un

⋅

n

tan −1 xdx then

0

(

1)

(n 1)un −2 =

n

π 1 − . 2 n

[WBUT 2002]

π 2

21. If I n = ∫ x n (sin x + cos x) dx, show that 0

π ⎞⎛ π ⎞ ⎛ n(n 1) I n −2 = ⎜ n + ⎟ ⎜ ⎟ 2 ⎠⎝ 2 ⎠ ⎝

In

1

∫x

22. If I n

n

n −1

.

cot −1 xdx then prove that

0

(

1)

(n 1) I n− n 2 =

n

π 1 + . 2 n

23. Obtain the reduction formula for ∞

In = ∫ 0

dx (x

2

a2 )n

, where n (> 1)) is a positive integer

5.40

Engineering Mathematics-I

and hence show that ∞

dx

∫ ( x2

2 2

a )

0

=

π . 4a 2

1 (2 n 3) ⎡ ⎤ ⎢ Ans : First part I n = 2 ⋅ (2 n − 2) ⋅ I n −1 .⎥ a ⎣ ⎦

Multiple-Choice Questions π 2

1. The value of

∫ sin

6

xdx is

0

a)

7π 32

7π 16

b) π 2

2. The value of

∫ cos

7

c)

5π 32

d)

5π 16

c)

16π 35

d)

8π 35

xdx is

0

a)

8 35

b)

16 35 π 2

3. The reduction formula for I n

∫ cos

n

xdx is

0

a) I n =

n −1 I n −2 n

b) I n =

n −1 I n −1 n

c) I n =

n I n −2 n −1

d) I n =

n I n −1 n −1

π 2

4. The value of

∫ sin

n

xdx is same as

0

π

a)

n ∫ cos 2 xdx

π

b)

0

n

xdx

0

π

12 cosn xdx c) 2 ∫0

∫ cos

π 2

d)

∫ cos 0

n

xdx

[WBUT 2007]

5.41

Reduction Formula π 2

5. The reduction formula for I m, n

∫ si

m

x cosn xdx is

0

a) I m, n =

n −1 Im m+n

2, n

b) I m, n =

m −1 Im m+n

2, n

c) I m, n =

m −1 I m, n − 2 m+n

d) I m, n =

n −1 Im m+n

2, n 2

π 2

6. The value of

∫ sin

m

cosn x dx is same as

0

π 2

a)

∫ cos

m +1

π 2

ssin n −1 x dx

b)

0

π 2

c)

∫ cos

∫ cos

m 1

ssin n +1 x dx

0

m

sin s n x dx

d) none of these.

0

π 2

7. The reduction formula for I m, n = ∫ cosm x sin nxdx is 0

a) I m, n =

m 1 Im + m+n m+n

c) I m, n =

m Im m+n π 2

8. The value of

∫ cos

1, n 1

1 Im m+n

1, n 1

d) none of these.

1, n 1

4

b) I m, n =

sin 3 3x dx is

0

a)

13π 35

b) ∞

9. The value of

∫ 0

a)

π a2

7 16 dx

(

2

x + a2 b)

)

2

π 4a 2

c)

13 35

d)

7π 16

c)

π 2a2

d)

π 3a 2

is

5.42

Engineering Mathematics-I π 4

10. The reduction formula for un

∫ tan

n

xdx is

0

a) un =

1 − un − 2 n −1

b) un =

1 − un −1 n −1

c) un =

1 − un − 2 n−2

d) un =

1 − un −1 n−2

π 2

11. The value of

∫ sin

5

cos6 x dx is

0

a)

2 693

b) π 2

12. The value of

∫ sin

2

8 693

c)

4 693

(d)

c)

1 32

d)

4. d 10. a

5. b 11. b

8π 693

cos4 x dx is

0

a)

π 16

ANSWERS: 1. c 7. a

b)

2. b 8. c

1 16

3. a 9. b

π 32 6. c 12. d

CHAPTER

6 Calculus of Functions of Several Variables 6.1

INTRODUCTION

In the earlier two chapters (chapters 3 and 4), we have dealt with the functions of single varable only. But we also require the fuctions of two or more variables to solve various problems in different branches of science and technology. Also, the derivatives and integrations of functions of two or more variables have a wide range of applications. Basically, in this chapter, we first discuss briefly the limit and continuity of the functions of two or more variables. Then we describe the methods of differentiations and their applications towards the optimisation of the functions.

6.2

FUNCTIONS OF SEVERAL VARIABLES

A real function of a single variable f R → R is defined by y = f ( x) where x and y R

R

Example 1 y = f ( x) = x 2 . Here, y is a function of a single variable x. A real function of two variables f R 2 → R is defined as z = f ( x, y) where ( x, y) ∈ R 2 and z R. Example 2 z = f ( x, y) = x 2 + y 2 + xy. Here, z is a function of two variables, x and y. The domain D of a function f of two variables is any closed curve on the twodimensional plane, namely, rectangular, square, circular, etc. A real function of three variables f R3 → R is defined as z = f ( x1 , x2 , x3 ) where ( 1 , x2 , 3 ) R 3 and z R.

6.2

Engineering Mathematics-I

Example 3 z = f ( x1 , x2 , x3 ) 2 x1 3x 3 x22 + 4 x1 x3 . Here, z is a function of three variables, x1 x2 and x3 . A real function of n variables f R n → R is defined as z = f ( x1 , x2 , x3 , , xn ) where ( 1 , x2 , x3 , … , n ) ∈ R n and z R. Example 4 z = f ( x1 , x2 , x3 , , xn ) = x1 + x2 + x3 + variables, x1 x2 , x3 , , xn .

6.3

+ xn . Here, z is a function of n

LIMIT AND CONTINUITY

To describe the analytical concept of limit of a function, first we define two types of δ -neighbourhood (or δ -nbd) of a point (a, b) in the two-dimensional plane.

(i) Square δ -neighbourhood Any square region consisting of the points ( x, y) and satisfying x a < δ , y b < δ , for δ > 0 is called the square δ -nbd of the point (a, b).

(ii) Circular δ -neighbourhood Any circular region consisting of the points ( x, y) and satisfying 00 is called the circular δ -nbd of the point (a, b). Suppose ( x, y) be any variable point lying in any neighbourhood of a fixed point (a, b) in a two-dimensional plane. Also, let f ( x, y) be a function defined on a certain neighbourhood of the point (a, b). Now we will check whether the function f ( x, y) tends to a real value l as ( x, y) tends to (a, b).

6.3.1

Limit of a Function of Two Variables

General Definition Let f ( x, y) be a function of two independent variables, x and y. If the function f tends to a real value l as ( x, y) (a, b) then we write lim f ( x, y) = l ( x, y )

( a, b )

or, f ( x, y) l as ( x, y) → (a, b) or, f ( x , y ) = l. x →a y b

where l is called the limit.

Calculus of Functions of Several Variables

6.3

Analytical Definition: Let f ( x, y) be a function of two independent variables x and y. The function f is said to tend to limit l as ( x, y) (a, b) if for an arbitrary small positive number ε , no matter how small, there exists a positive number δ , such that f ( x, y) l < ε for every point ( x, y) which lies in any δ -nbd N of the point (a, b). N may be a square δ -nbd of the point (a, b), i.e., x a < δ , y b < δ , for δ > 0 or, N may be a circular δ -nbd of the point (a, b), i.e., 00 or N may be any other δ -nbd. Symbolically, lim f ( x, y) = l. ( x, y )

( a, b )

Example 5 Find Sol.

(

lim

( x, y )

(

lim

( x, y )

2

y2

y).

(2, 3) 2

y2

y)

(2, 3)

( x2 )

= ( x, y )

( y2 ) ( x, y )

(2, 3)

(2, 3)

( xy) ( x, y )

(2, 3)

= 4 + 9 6 = 19. Example 6 Justify the following by analytical definition x 2 y2

lim

=0 x 2 + y2 To prove the existance of the limit, for a given ε > 0 we are to find δ > 0 such that in any δ -nbd N of (0, 0), ( x, y )

Sol.

(0, 0)

f ( x, y) l < ε or,

x2 y 2 2

x + y2 or,

0 0

and this extreme value is i) a maxima according as fxx a, b f yy ( a, b) < 0 ii) a minima according as fxx a, b f yy ( a, b) > 0 Note: Let f ( x, y) be a continuous function having second-order partial derivatives such that f x (a, b

and f y (a, b) = 0, provided they exist

Now if (i) 2

f xx (a, b) f yy (a, b) − ⎡⎣ f xy (a, b) ⎤⎦ < 0, Then f ( x, y) has no extreme value at (a, b), i.e., (a, b) is a saddle point. and if (ii) 2

f xx (a, b) f yy (a, b) − ⎡⎣ f xy (a, b) ⎤⎦ = 0 Then f ( x, y) may or may not have extreme value at (a, b), i.e., the case is undecided and further investigation is required. Alternate Conditions of Maxima and Minima Let z f ( x, y) be a continuous function having second-order partial derivatives. If df = 0 at ( a, b) then i) ( a, b) is a point of maximum if d 2 f < 0, and ii) ( a, b) is a point of minimum if d 2 f > 0.

Calculus of Functions of Several Variables

6.27

Example 25 Find the maxima and minima of the function x3 + y 3 − 3 x 12 y + 20. Find also the saddle points. [WBUT 2001, 2005] Sol.

Here, f ( x, y) = x3 + y 3 − 3 x 12 y + 20. Then f x ( x, y) f xx x, y)

3 x 2 3, f y ( x, y) 6 x, f yy x, y)

6y

Solving, f x ( x, y)

3 x2 3 = 0

and f y ( x, y)

3 y 2 12 = 0

3 y 2 − 12, f xy ( x, y) = 0

we obtain, x 1 d y= 2 Therefore, the (stationary points) critical points are (1, 2), (1, 2), 2) ( 1, 2) and ( 1 1, 2). Now, f xx ( x, y) f yy ( x, y) − ⎡⎣ f xy ( x, y) ⎤⎦ At the point (1, 2)

2

36 xy x

2

f xx (1, 2) f yy (1 1, 2) − ⎡⎣ f xy (1, 2) ⎤⎦ = 72 > 0, f xx (1, 2) = 6 > 0

f yy (1, 2) = 12 > 0

Therefore, f ( x, y) has minimum at (1, 2) and the minimum value is f (1, 2) = 2. At the point ( 1, 2) 2

f xx ( 1, 2) f yy (−1 1, 2) − ⎡⎣ f xy ( 1, 2) ⎤⎦ = −72 < 0 Therefore, f ( x, y) has neither maximum nor minimum at ( 1, 2) . At the point (1, 2) 2

f xx (1, 2) f yy (1 1, − 2) − ⎡⎣ f xy (1 (1, 2) ⎤⎦ = −72 < 0 Therefore, f ( x, y) has neither maximum nor minimum at (1, 2) . At the point ( 11, 2) 2

f xx ( 1, 2) f yy (−1 1, − 2) − ⎡⎣ f xy ( 1, 2) ⎤⎦ = 72 > 0 f xx ( 1 1, 2) = −6 < 0

f yy (−1, − 2) = 12 < 0

Therefore, f ( x, y) has maximum at ( 1 1, 2) and the minimum value is f ( 11, 2) = 38.

6.28

Engineering Mathematics-I

We have, from the above, that at the stationary points ( 1, 2) and (1, 2), the function does not have any extreme values. So the saddle points are ( 1, 2) and (1, 2). The following topic is included for further reading by interested students.

6.10.2 Maxima and Minima of Implicit Functions: (Lagrange’s Multiplier Method) For Functions of Two Variables: Let f ( x, y) be a function of two variables x and y, subject to the constraint conditions φ ( x, y) = 0. Let L( x, y f ( x, y)+λφ φ ( x, y), where λ is called the Lagrangian multiplier. The critical points can be found by solving ∂L ∂L φ ( x, y) = 0, = 0 and =0 ∂x ∂y i) the critical point is a point of maxima according as d 2 f < 0 where d 2 f is determined considering y is dependent on x or, the critical point is a point of maxima according as d 2 L < 0 ii) the critical point is a point of minima according as d 2 f > 0 where d 2 f is determined considering y is dependent on x or, the critical point is a point of minima according as d 2 L > 0 For Functions of Three Variables: Let f ( x, y, z ) be a function of two variables x y and z, subject to the constraint conditions φ ( x, y, ) = 0. Let L( x, y, z ) = f ( x, y z ) +λφ φ ( x, y, ), where λ is called the Lagrangian multiplier. The critical points can be found by solving ∂L ∂L ∂L φ ( x, y, ) = 0, = 0, = 0 and = 0. ∂x ∂y ∂z i) the critical point is a point of maxima according as d 2 f < 0 where d 2 f is determined considering z is dependent on x and y or, the critical point is a point of maxima according as d 2 L < 0 ii) the critical point is a point of minima according as d 2 f > 0 where d 2 f is determined considering z is dependent on x and y or, the critical point is a point of minima according as d 2 L > 0. Example 26 Find the optimum value of f ( x, y) Lagrangian multiplier method. Sol.

Let φ ( x, y) = x + y − 1 = 0. Now L( x, y

x 2 y 2 , subject to the condition x + y = 1 using

...(1)

f ( x, y) λ .φ ( x, y), where λ is the Lagrangian multiplier.

Calculus of Functions of Several Variables

So, L( x, y

6.29

f ( x, y) λ .φ ( x, y)

y λ ( x + y − 1) Then ∂L = 2 xy 2 + λ ∂x and ∂L = 2 x2 y + λ ∂y =

2 2

∂L ∂L = 0, = 0 and (1) we have ∂x ∂y

Now solving

1 1 and y = . 2 2

x

⎛1 1⎞ So the critical point is ⎜ , ⎟ . ⎝2 2⎠ Now 2

⎛ ∂ ⎞ ∂ d 2 f = ⎜ dx + dy ⎟ f ∂y ⎠ ⎝ ∂x =

∂2 f ∂x

2

( )2 +

∂2 f ∂y

2

( )2 + 2

∂2 f dx ⋅ dy ∂x∂y

...(2)

Again ∂2 f

= 2 y2,

∂x 2

∂2 f ∂y 2

= 2 2,

∂2 f = 4 xy ∂x∂y

and x + y 1 ⇒ dx = −dy Putting the values in (2), we have d2 f =

∂2 f ∂x 2

( dx )2 +

∂2 f ∂y 2

( dy )2 + 2

= 2 y 2 ( dx d ) + 2 x 2 ( dx ) 2

= (2 y 2

2

2xx 2 − 8 y) ( dx ) 2

∂2 f dx ⋅ dy ∂x∂y

2 ⋅ 4 y d (−dx)

2

Since

(d f ) 2

⎛1 1⎞ ⎜ 2, 2 ⎟ ⎝ ⎠

1 1 1⎞ 2 ⎛ 1 = 2 ⋅ + 2 ⋅ − 8 ⋅ ⋅ ⎟ ( dx ) 4 2 2⎠ ⎝ 4 =

( )2 < 0

⎛1 1⎞ x 2 y 2 attains maximum value at the point ⎜ , ⎟ . ⎝2 2⎠ 1 The maximum value of f ( x, y) is . 16 the function f ( x, y)

6.30

Engineering Mathematics-I

WORKED-OUT EXAMPLES Example 6.1

Show that the function

f ( x, y)

xy x

x2

y2

x2 + y 2

( x, y) = (0, 0)

= 0, is continuous at (0, 0). Sol.

, ( x, y) ≠ (0, 0)

To prove the continuity at (0, 0), we are to show f ( x, y) = f (0, 0)

lim

i.e.,

( x, y )

(0, 0)

lim

xy

( x, y )

(0, 0)

x2

y2

x2 + y 2

=0

Now to prove the existence of the above limit, for a given ε > 0 we are to find δ > 0 such that in any δ -nbd N of (0, 0), xy

x2

y2

x 2 + y2

−0 < ε

x2 or, x y

y2

x2 + y 2

0 2

and f xx ( , 0) = 6 > 0, f yy (2, 0) = 6 > 0 Therefore, f ( x, y) is minimum at (2, 0). At the point (1, 1) f xx (1, 1) f yy (1, 1) − { f xy (1, 1)} = −36 < 0 2

Therefore, f ( x, y) has no extreme value at (1, 1) At the point (0, 0) f xx (0, 0) f yy (0, 0) − { f xy (0, 0)} = 36 > 0 2

and f xx ( , 0) = 6 < 0, f yy (0, 0) = 6 < 0 Therefore, f ( x, y) is maximum at (0, 0). Example 6.23

Find the maximum and minimum of the function

f ( x, y) Sol.

x3 + y 3 − 3axy

[WBUT-2002, 2008]

Here, f ( x, y)

x3 + y 3 − 3axy

Then fx

x 2 − 3ayy, f y

f xx

x, f yy

y 2 3ax

y, f xy

a

Now, fx

x 2 3ay = 0 ⇒ x 2 − ay = 0

...(1)

and f y

y 2 3ax = 0 ⇒ y 2 − ax = 0

...(2)

Solving equations (1) and (2), the critical points are (0, 0) and ( , a) Now, f xx f yy − ( f xy )2

x 9a 2 xy

Calculus of Functions of Several Variables

a)

At the point a

f

6.51

a a

a a

2

a2 > 0

and a

a

a

6a a

0

a>0

0

a0

a

a 0. At the point (0, 0) f (0, 0)

f (0, 0)

a) if

= 9a < 0

x y) is neither maximum nor minimum at (0, 0).

Hence Example 6.24

Find the maxima and minima of the function +

x y x Find also the saddle points. Sol.

2

a) if a < 0 and minimum at

y.

Here, x y Then

+

x

x

x

x

x

x . y = 3 y2

12 y f ( x

y

f

x,

y = 12

Now to find the critical points, we solve x and

x

x

12

x,

1 4y

12x

1 4x = 0

Subtracting (1) from (2) we obtain, )=0 this implies or, ( 4) = 0 So from above we have two pairs of equations ( 2

)=0 0

and

(

4) = 0 0

2

Solving the above, we obtain the (stationary points) critical points as ( 7, 7), (3, 3), (5, 1) and ( 1, 5). Now, x

−

x

2

y − 144

...(1) ...(2)

6.52

Engineering Mathematics-I

At the point ( 7 7, 7) f xx

2

) f yy (−7, − 7) − ⎡⎣ f xy ( 77, 7) ⎤⎦ = 1620 > 0, f xx ( 77, 7) = −42 < 0 and f yy ( 7, 7) = −42 < 0

Therefore, f ( x, y) has maximum at ( 7 7, 7) and the maximum value is f (1, 2) = 2. At the point (3, 3) 2

f xx (3, 3) f yy (3 3, 3) − ⎡⎣ f xy (3, 3) ⎤⎦ = 180 > 0 f xx (3, 3) = 18 > 0 and f yy (3, 3) = 18 > 0 Therefore, f ( x, y) has minimum at (3, 3) and the minimum value is f (3, 3) = −216. At both the the points (5, 1) and ( 1, 5) 2

f xx (5, 1) f yy (5 5, − 1) − ⎡⎣ f xy (5 (5, 1) ⎤⎦ = −324 < 0 2

and f xx ( 1, 5) f yy (−1, 5) − ⎡⎣ f xy ( 1, 5) ⎤⎦ = −324 < 0 Therefore, f ( x, y) has neither maximum nor minimum at both the the points (5, 1) and ( 1, 5). So, these are the saddle points. Example 6.25

Find the point in the plane x + 2 y + 3 z = 13 nearest to the point

(1, 1, 1) using Lagrange’s multiplier method. Sol.

[WBUT-2001,2002]

Let P( x, y, z ) be any point on the plane x + 2 y + 3 z = 13. The distance between the point P( x, y, z ) and A(1, 1, 1) is D

( x − 1)2 + ( y 1)2

( z − 1)2

Let us consider f ( x, y, z ) = D 2 ( x, y z ) = ( x 1)2

( y − 1)2 + ( z 11))2

Here, we have to find the point P( x, y, z ) such that f ( x, y, z ) or D 2 ( x, y, z ) is minimum subject to

φ ( x, y, ) = x 2 y + 3 z 13 Let L( x, y, z ) = f ( x, y z ) λφ ( x y, z ) = ( x 1)2 + ( y 1)2

( z 1)2 + λ (

Now, ∂L ∂f ∂φ +λ = =0 ∂x ∂x ∂x

2 y + 3 z − 13)

Calculus of Functions of Several Variables

or, 2( x − 1) + λ = 0

6.53

...(1) ∂L ∂f ∂φ +λ = =0 ∂y ∂y ∂y

or, 2( y − 1) + 2λ = 0

or, 2( z − 1) + 3λ = 0

...(2) ∂L ∂f ∂φ +λ = =0 ∂z ∂z ∂z ...(3)

The critical points are found by solving, ∂L ∂L ∂L = 0, = 0, = 0 and φ ( x, y, z ) = 0 ∂x ∂y ∂z Putting the values of x y, z from (1), (2) and (3) in φ ( x, y, ) = 0, we get ⎛ −λ ⎞ ⎛ −3λ ⎞ ⎜ 2 + 1 ⎟ + 2(−λ + 1) + 3 ⎜ 2 + 1 ⎟ − 13 = 0 ⎝ ⎠ ⎝ ⎠ or, λ = 1 Putting the value of λ in (1), (2) and (3) we have 3 5 x y = 2, z = 2 2 ⎛3 Therefore the required point on the plane is ⎜ , 2, ⎝2 Now, 2

⎛ ∂ ∂ ∂ ⎞ d f = ⎜ dx + dy + dz ⎟ f ∂y ∂z ⎠ ⎝ ∂x 2

=

∂2 f ∂x 2

(dx)2

∂2 f ∂y 2

∂2 f

(dy)2

∂z 2

(dz )2

∂2 f ∂2 f ∂2 f dxdy + 2 dydz + 2 dzdx ∂x∂y ∂y∂z ∂ ∂x

+2

Again, ∂f ∂f ∂f = 2( x − 1), = 2( y − 1), = 2( z − 1) ∂x ∂y ∂z ∂2 f ∂x 2 and

= 2,

∂2 f ∂y 2

= 2,

∂2 f ∂z 2

=2

∂2 f ∂2 f ∂2 f = 0, = 0, =0 ∂x∂y ∂y∂z ∂ ∂x Therefore,

{

}

d 2 f = 2 (dx d )2 + (dy d )2 + (dz d )2 > 0

5⎞ . 2 ⎟⎠

6.54

Engineering Mathematics-I

So,

( d f )( 2

)

3 , 2, 5 . 2 2

{

}

= 2 (dx d )2 + (dy dy)2 + (dz d )2 > 0

5⎞ ⎛3 Therefore, f ( x, y, z ) or D 2 ( x, y, z ) is minimum at ⎜ , 2, ⎟ 2⎠ ⎝2 5⎞ ⎛3 Hence ⎜ , 2, ⎟ is the point in the plane x + 2 y + 3 z = 13 nearest to the point 2⎠ ⎝2 (1, 1, 1). Example 6.26

If xyz = a 3, find the critical points of xy + yz y + zx using Lagrange’s

multiplier method. Sol.

Here f ( x, y, z ) = xy + yz + zx subject to

φ ( x, y, ) = xyz − a 3 Let,

(

L( x, y, z ) = f ( x, y z ) λφ ( x y, z ) = xy x + yz + zx zx + λ xyz − a 3

)

Now,

or, y

or,

λ ( yz ) = 0

λ ( xz ) = 0

∂L ∂f ∂φ +λ = =0 ∂x ∂x ∂x ...(1) ∂L ∂f ∂φ +λ = =0 ∂y ∂y ∂y ...(2) ∂L ∂f ∂φ +λ = =0 ∂z ∂z ∂z

or, y λ ( xy) = 0 The critical points are found by solving,

...(3)

∂L ∂L ∂L = 0, = 0, = 0 and φ ( x, y, z ) = 0 ∂x ∂y ∂z From (1), (2) and (3), we have 1 1 1 1 1 1 + = + = + = −λ y z z x x y ⎛1 1 1⎞ or, 2 ⎜ + + ⎟ = 3λ ⎝x y z⎠ From (4) and (5), we get −2 −2 −2 x y= ,z= λ λ λ

...(4) ...(5)

6.55

Calculus of Functions of Several Variables

Putting the values of x y, z in φ ( x, y, ) = 0 we get x y=z a Therefore, the critical point is ( , a, ). Prove that the volume of the greatest rectangular parallelepiped Example 6.27 that can be inscribed in the ellipsoid x2

+

a2 is

8abc 3 3

Sol.

y2 b2

+

z2 c2

=1

.

[WBUT 2002]

The volume of the rectangular parallelepiped is 2 x ⋅ 2 y 2 z = 8 xyz

V

The problem is to find the maximum volume subject to the condition x2

+

y2

a 2 b2 Here,

+

z2 c2

=1

f (xy xxy z ) = V = 8 xyz and φ ( x, y, z ) = Let,

x2 a2

+

y2 b2

+

z2 c2

−1

⎛ x2 y 2 z 2 ⎞ L( x, y, z ) = f ( xy x , z ) + λφ ( x, y, z ) = 8 xyz + λ ⎜ 2 + 2 + 2 − 1 ⎟ ⎜a ⎟ b c ⎝ ⎠ Now, ∂L ∂f ∂φ +λ = =0 ∂x ∂x ∂x 2λ x or, 8 yz + 2 = 0 a ∂L ∂f ∂φ +λ = =0 ∂y ∂y ∂y 2λ y or, 8 zx + 2 = 0 b ∂L ∂f ∂φ +λ = =0 ∂z ∂z ∂z 2λ z or, 8 xy + 2 = 0 c Multiplying (1), (2), (3) by x y, z respectively and adding we have ⎛ x2 y 2 z 2 ⎞ 24 xyz + 2λ ⎜ 2 + 2 + 2 ⎟ = 0 ⎜a b c ⎟⎠ ⎝

...(1)

...(2)

...(3)

6.56

Engineering Mathematics-I

or, 24 xyz + 2λ = 0 or, λ = 12xyz From (1), we have 2λ x 8 yz + 2 = 0 a 2(−12 xyz ) x or, 8 yz + =0 a2 a2 or, x 2 = 3 a or, x = ± 3 Since x > 0, we consider a x= 3 Similarly from (2) and (3), −4abc b c y= ,z= and λ = 3 3 3 ⎛ a b c ⎞ Therefore, the critical point is ⎜ , , ⎟ ⎝ 3 3 3⎠ Again, ⎡ (dx d )2 (dy dy)2 (dz d )2 ⎤ d 2 L = 2λ ⎢ 2 + 2 + 2 ⎥ + 16( ydzdx d + zdxdy + xdydz d ) b c ⎦ ⎣ a Also from, x2

+

y2

+

...(5)

z2

−1 = 0 a 2 b2 c 2 we get xdx yydyy zdz + 2 + 2 =0 a2 b c 1 ⎡ dx dyy dz ⎤ or, ⎢ + b + c ⎥=0 3⎣a ⎦ dx dyy dz or, + + =0 a b c From (6) 2

⎛ dx dy ⎞ ⎛ dz ⎞ ⎜ a + b ⎟ = ⎜− c ⎟ ⎝ ⎠ ⎝ ⎠ 2

or, 2

...(4)

...(6)

2

2

dxdy ⎛ dz ⎞ ⎛ dx ⎞ ⎛ dy ⎞ = ⎜ ⎟ −⎜ ⎟ −⎜ ⎟ ab ⎝ c ⎠ ⎝ a ⎠ ⎝ b ⎠

2

Calculus of Functions of Several Variables

6.57

Similarly from (6), we obtain 2

2

2

2

2

2

2

dydz ⎛ dx ⎞ ⎛ dy ⎞ ⎛ dz ⎞ = ⎜ ⎟ −⎜ ⎟ −⎜ ⎟ bc ⎝ a ⎠ ⎝ b ⎠ ⎝ c ⎠

dxdz ⎛ dy ⎞ ⎛ dx ⎞ ⎛ dz ⎞ = ⎜ ⎟ −⎜ ⎟ −⎜ ⎟ ac ⎝ b ⎠ ⎝ a ⎠ ⎝ c ⎠ Therefore using the above results in (5) and 2

d2L = −

16 abc ⎡ (dx d )2 (dy)2 (dz d )2 ⎤ ⎢ 2 + 2 + 2 ⎥ 0) Sol. Here x + y + z = a (a > 0) or, z = a − x

y

Therefore, the lower and upper limits of z are z = 0 and z Putting z = 0 in x + y + z = a (a > 0), we have x+ y = a or, y = a − x

a−x

y

7.15

Line Integral, Double Integral and Triple Integral y

x +y =a

x x =a

y =0

O

Figure 7.7 a−x

Therefore, the lower and upper limits of y are y = 0 and y Putting z = 0; y = 0 in x + y + z = a (a > 0), we have

x a Therefore, the lower and upper limits of x are x = 0 and x = a. Therefore,

∫∫∫ (

a

2

y

∫ a

∫

⎡ = ∫ ⎢ x =0 ⎣

∫

x =0

dx

∫

y =0

⎡ 2 ∫ ⎢ x (a − x)) y =0 ⎣

2

2

( x 2 + y 2 + z 2 ) dz

z =0

y (a ( a − x) y 2 − y 3 +

x2 y 2 ( ( a − x) y − + 2

) y3 3

⎡ x2 ( )2 ( )4 ⎤ + ⎢ ⎥ dx 2 6 ⎦ ⎣ a

a5 20

∫

⎡ z dy ⎢ x 2 z + y 2 z + ⎥ 3⎦ ⎣ y =0 0

⎡1 x3 ax 4 x 6 ( )6 ⎤ = ⎢ a2 − + − ⎥ 3 4 10 30 ⎦ 0 ⎣2 =

dy

3 ⎤a x− y

∫

a

a

∫

a x− y

a x

a x

dx

x =0

=

z ) dxdyddz =

a x

dx

x =0

=

2

x =0

a

=

2

y4 ( − − 4

(

y )3 ⎤ ⎥ dy 3 ⎦ a x

y) 4 ⎤ ⎥ 12 ⎦0

dx

7.16

Engineering Mathematics-I

7.3.2

Transformation of Triple Integrals

Let us consider the integral

∫∫∫ f ( x, y, z) dxdydz R

where f ( x, y, z ) is defined over the region R. Let us take the transformation x = φ (u, v, w) y = ψ (u, v, w) and z = ρ (u, v, w) Then the Jacobian of the transformation is defined by ∂x ∂u ∂( x, y, z ) ∂y J= = ∂(u, v, w) ∂u ∂z ∂u

∂x ∂v ∂y ∂v ∂z ∂v

∂x ∂w ∂y ∂w ∂z ∂w

The transformation is invertible if J ≠ 0. If x = φ (u; v; w), y ψ (u; v; w) and d z = ρ (u; v; w) is an invertible transformation then

ÚÚÚ f ( x, y, z) dxdydz = ÚÚÚ f (f (u, v, w), Y (u, v, w), r(u, v, w)) Jdudvdw R

R1

where R1 is the region in new coordinate system Example 11

Evaluate

∫∫∫ (

2

y2

z 2 ) dxdyddz taken over the volume enclosed

by the sphere x 2 + y 2 + z 2 = 1. Sol.

Let us transform the given integral into spherical polar coordinates by putting x = r si θ φ ; y = r si θ in i φ ; z r cos θ Then ∂x ∂u ∂( x, y, z ) ∂y J= = ∂(u, v, w) ∂u ∂z ∂u

∂x ∂v ∂y ∂v ∂z ∂v

∂x ∂w ∂y s nθ ≠ 0 = r 2 si ∂w ∂z ∂w

Under this transformation, the R1 = {(r , θ , φ ) : 0 ≤ r ≤ 1;0 θ ;0 φ Therefore,

∫∫∫ (

2

y2

z 2 ) dxdyd dz

domain 2π }

of

integration

is

7.17

Line Integral, Double Integral and Triple Integral π 2π 1

=∫

∫∫

2

(r 2 sin θ ) dθ dφ d dr

0 0 0

2π

=

∫ 0

π

1

dφ ∫

θ dθ ∫ r 4 dr

0

0

1

2π

= [φ ]0 =

5 π ⎡r ⎤ θ ]0 × ⎢ ⎥ ⎣ 5 ⎦0

[

2p 5

WORKED-OUT EXAMPLES π 2π

Example 7.1

Evaluate

∫ ∫ sin ( x + y) dxdy

[WBUT-2001, 2009]

0 0

Sol. π 2π

∫ ∫ sin ( x + y) dxdy 0 0

π 2 ⎡π

= ∫ ⎢∫ 0⎢ ⎣0 π 2

=

∫ [

⎤ ( x + y) dx ⎥ dy ⎥⎦ π

( + ) ]0 dy

0

π 2

=

∫[

(π + y) − cos y ] dy

0

π 2

= 2 ∫ cos ydy 0

π

= 2 [sin y ]02 = 2 Example 7.2 y

x

Evaluate

∫∫ xy ( x

y) dxdy over the area bounded by y

x 2 and

[WBUT-2001]

7.18

Engineering Mathematics-I

Sol. y

(1, 1) y =x R

x x =1

O y =x 2

Figure 7.8 The region R is shown in Fig. 7.8 by the shaded portion. Therefore,

∫∫ xy ( x

y) dxdy

R

1

=

x

∫ ∫

x2

1

y ( x + y) dxdy −

x =0 y =0

∫ ∫

xy ( x + y) dxdy

x =0 y =0 x2

x

1 ⎡ x 2 y 2 xy 3 ⎤ ⎡ x 2 y 2 xy 3 ⎤ + = ∫ ⎢ + ⎥ dx ⎥ dx − ∫ ⎢ 2 3 ⎦ 2 3 ⎦ x =0 ⎣ x =0 ⎣ 0 0 1

=

1 1 ⎡ x6 x 7 ⎤ 5 4 x dx − ∫ ⎢ 2 + 3 ⎥ dx 6 x ∫=0 ⎦ x =0 ⎣

=

1 ⎛ 1 1 ⎞ − + ⎟ 6 ⎝ 14 24 ⎠

=

3 56 2 2 a a y

Example 7.3

Evaluate

∫ ∫ 0

Sol.

Let x = r cos θ , y coordinates.

0

r sin i θ

(

2

y 2 ) dydx by changing to polar coordinates. [WBUT-2002]

be the transform from cartesian to polar

Line Integral, Double Integral and Triple Integral

7.19

The Jacobian of the transformation is ∂x ∂( x, y) ∂r J= = ∂(r , θ ) ∂y ∂r

∂x ∂θ = cos θ sin θ ∂y ∂θ

−r i θ =r≠0 θ r

The domain of integration R = {( x, y) : 0

x

a; 0 ≤ y ≤ a 2

x 2 } i.e., the

first quardant of the circle x 2 + y 2 = a 2 . ⎧ Under the transformation the domain of integration is R1 = ⎨(r , ) : 0 ≤ ⎩ π⎫ r ≤ a;0 ≤ θ ≤ ⎬ 2⎭ Therefore, π 2

2 2 a a y

∫ ∫ 0

(

2

y 2 ) dydx =

a

∫ ∫

r 2 rdrdθ

θ =0 r =0

0

π 2

a

⎡ r4 ⎤ = ∫ ⎢ ⎥ dθ 4⎦ θ =0 ⎣ 0 π 2

4

=

=

a 4

∫

dθ

θ =0

a4 π π a4 × = 4 2 8 ax y

Example 7.4

Evaluate

∫∫∫x 000

Sol. ax y

∫∫∫x

3 2

y zdzdydx

000

a ⎡x ⎧y ⎫⎪ ⎤ ⎪ = ∫ ⎢ ∫ ⎨ ∫ x3 y 2 zdz ⎬ dy ⎥ dx 0⎢ ⎭⎪ ⎥⎦ ⎣ 0 ⎩⎪ 0 a ⎡x

y ⎤ ⎡ x3 y 2 z 2 ⎤ ⎢ = ∫ ∫⎢ ⎥ dy ⎥ dx ⎢ ⎥ 2 ⎦ 0 ⎣0 ⎣ 0 ⎦ a ⎡x 3 4 ⎤ x y dy ⎥ dx = ∫ ⎢∫ ⎥⎦ 0⎢ ⎣0 2

3 2

y zdzdydx d

[WBUT-2003]

7.20

Engineering Mathematics-I x

a ⎡ x3 y 5 ⎤ = ∫⎢ ⎥ dx 10 ⎦ 0⎣ 0 a

=

1 a9 x8 dx = ∫ 10 0 90

Example 7.5

Evaluate

∫∫∫ (

y z 1) 4 dxdyd dz over the region defined by

x ≥ 0 y ≥ 0, z ≥ 0, x + y + z ≤ 1

[WBUT-2003]

Sol. z C

O

x A

B y

Figure 7.9 The plane x + y + z = 1 cuts X , Y and Z axes at A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1). Therefore, the upper limit and lower limit of z are 0 and 1

y.

Putting z = 0 in x + y + z = 1 we have the upper limit and lower limit of y are 0 and 1 . Putting z = 0 and y = 0, we have the upper limit and lower limit of x are 0 and 1. Therefore,

∫∫∫ (

y z 1) 4 dxdyddz

1 1− x 1− x y

=∫

∫ ∫

0 0

0

( x + y + z 1) 4 dxdyddz

7.21

Line Integral, Double Integral and Triple Integral 1 1− x

=∫

∫

0 0

1 1− x

=∫

∫

0 0

⎡( ⎢ ⎣

1− x y

y 5

1⎡ 32 − ( 5⎣

1)5 ⎤ ⎥ ⎦0

dxdy

y 1)5 ⎤⎦ dxdy 1− x

1 1 ⎡ ( = ∫ ⎢32 y − 50⎣

y 1)6 ⎤ ⎥ dx 6 ⎦0

=

1 1 ⎡ ⎢32(1 5 ∫0 ⎣

=

1⎡ x 2 32 ( 1) 7 ⎤ ⎢32 x − 32 − x + ⎥ 5⎣ 2 3 42 ⎦ 0

=

1⎡ 32 64 ⎤ 117 32 − 16 − + ⎥ = 5 ⎢⎣ 3 21 ⎦ 70

)−

32 ( 1)6 ⎤ + ⎥ dx 3 6 ⎦ 1

Example 7.6

∫∫

Evaluate

R

1 2

x + y2

dxdy where R = { x ≤ 1; y ≤ 1} [WBUT-2004]

Sol.

Here R = { x ≤ 1; y ≤ 1} {( x, y) : 1

x 1; 1 ≤ y ≤ 1}

Therefore,

∫∫ R

1

1 2

x +y

2

Since

dxdy =

00

1

∫ ∫

x= 1 y = 1

1 2

x + y2

11

= 4∫ ∫

1

2

x + y2

dxdy

is an even function

1 2

x + y2

dxdy

1

1

= 4 ∫ ⎡ log ( y + x 2 + y 2 ⎤ dx ⎢⎣ ⎦0 0 1

= 4 ∫ ⎡ log (1 + 1 ⎣⎢ 0

⎧ = 4 ⎨ ⎡ x log (1 + 1 ⎢ ⎩⎣

2

) − log x ⎤ dx ⎦⎥ 2

1

1

)⎤ − ∫ ⎥⎦ 0 0 (1 + 1

x2 2

dx − [ x log x − x ]0 1

) 1+ x

2

7.22

Engineering Mathematics-I 1⎡ ⎧ x2 1 + x2 ⎪ = 4 ⎨log (1 + 2) − ∫ ⎢ − 2 1 + x2 ⎪⎩ 0⎢ ⎣ x

⎧⎪ = 4 ⎨log (1 + 2) + 1 ⎪⎩

1⎛

∫ ⎜⎜1 − 0⎝

⎫ ⎤ ⎪ ⎥ dx + 1⎬ ⎥⎦ ⎪⎭

⎞ ⎫⎪ ⎟ dx ⎬ ⎟ 1 + x 2 ⎠ ⎪⎭ 1

1

= 4{

(1

2) 1 − ⎡ x − log ( ⎣⎢

1 + x2 ) ⎤ ⎦⎥ 0

= 4{

(1

2) 1 − (1 − log (1 + 2)}

= 8 log (1 + 2)

Example 7.7 2

Evaluate

∫∫

2

circle x + y = 1 Sol.

1 − x2 − y 2 1 + x2 + y 2

dxdy over the positive quadrant of the [WBUT-2006]

Let x = r cos θ , y r sin i θ be the transform from Cartesian to polar coordinates. The Jacobian of the transformation is ∂x ∂( x, y) ∂r j= = ∂(r , θ ) ∂y ∂r

∂x ∂θ = cos θ sin θ ∂y ∂θ

−r i θ =r≠0 θ r

The domain of integration R = {( x, y) : 0

x

a; 0 ≤ y ≤ a 2

first quardant of the circle x 2 + y 2 = a 2 . Under the transformation, the domain of integration is

π⎫ ⎧ R1 = ⎨(r , ) : 0 ≤ r ≤ a;0 ≤ θ ≤ ⎬ 2⎭ ⎩ Therefore,

∫∫ R

1 − x2 − y 2 1 + x2 + y 2

dxdy = ∫∫ R1

1

=

1 − r2 1 + r2 π 2

∫ ∫

r =0 θ =0

π 2

=

∫

θ =0

1 − r2 1 + r2 1

dθ

rdrdθ

∫

r =0

rdrdθ

1 − r2 1 + r2

rdr

x 2 }, i.e., the

7.23

Line Integral, Double Integral and Triple Integral 1

=

p 2

(1 - r 2 )

Ú r =0

1- r4

1

rdr =

p 2

Ú r =0

rdr 1 - r4

1

-

p 2

Ú r =0

r 3 dr 1 - r4

p È -1 2 ˘ 1 - p p 2 p = sin r ˚ 0 3 2 Î 4 3

=

Example 7.8

d ∫ (3 xydx

Evaluate

y 2 dy) where C is the arc of the parabola

C

y

2 x 2 from (0, 0) to (1, 2) 2 x 2 ⇒ dy = 4 xdx

Here, y

Sol.

[WBUT-2007]

Therefore, d ∫ (3 xydx

y 2 dy)

C

1

=

∫ {3

2 x 2 dx − (2 x 2 )2 4 xdx}

x =0 1

=

∫

(6

3

16

5

) dx

x =0 1

1

⎡ x6 ⎤ ⎡ x4 ⎤ = 6 ⎢ ⎥ − 16 ⎢ ⎥ ⎣ 6 ⎦0 ⎣ 4 ⎦0 6 16 −7 = − = 4 6 6

EXERCISES

Short and Long Answer Type Questions 1 Evaluate ∫ [( x 2 + xy x ) dx ( x 2 + y 2 ) dy ] where C is the square formed by the lines xy) C

x

1 y = ±1.

[Ans : 0]

2. Evaluate ∫ [(cos x sin y − xy) dx + sin x cos y dy ] where C is the circle x 2 + y 2 = 1 C

3. Evaluate ∫ {( x 2 + y 2 ) d

−14 ⎤ ⎡ ⎢ Ans : 3 ⎥ ⎣ ⎦ 2 xydy d }, where C is the rectangle in the xy plane

C

bounded by x = 0, x = a; y = 0, y

a. [Ans : − 2 b2 ]

7.24

Engineering Mathematics-I

4. Evaluate the line integral

∫ {(2

y x 2 ) dx ( x

y 2 ) dy}, where C is the closed

C

curve of the region bounded by y

x2 , y 2

x. 1⎤ ⎡ ⎢ Ans : 30 ⎥ ⎣ ⎦

5. Evaluate the following double integrals: 41

a)

∫ ∫ xy ( x

y) dydx

[Ans : 8]

00

2 2 a a y

b)

∫ ∫ 0

a2

⎡ a 3π ⎤ ⎢ Ans : ⎥ 6 ⎦ ⎣

x 2 − y 2 dxdy

0

π a cosθ

c)

∫ ∫ 0

1

d)

⎡ a2 ⎤ ⎢ Ans : ⎥ 6⎦ ⎣

r s θ drdθ

0

x

∫∫(

2

3⎤ ⎡ ⎢ Ans : 35 ⎥ ⎣ ⎦

y 2 ) dydx

0 x

1 1+ x 2

e)

dydx

∫ ∫ 0

2

1+ x + y

0

∫∫

6. Evaluate

a2

quardant.

2

π ⎡ ⎢ Ans : 4 log ⎣

(

)

⎤ 2 1⎥ ⎦

x 2 − y 2 dxdy over the semicircle x 2 + y 2 = a 2 in the positive ⎡ a5 ⎤ ⎢ Ans : ⎥ 5⎦ ⎣ π a

7. Transform the integral to Cartesian form and hence evaluate

∫∫r

3

s θ cos θ drdθ

00

[Ans : 0] 8. Evaluate the following integrals: 21 1

a)

∫∫ ∫ (

2

y2

z 2 ) dxdyd dz

[Ans : 6]

0 0 −1

11

b)

c)

2

∫∫ ∫ 00

x2 + y 2

31

xy

∫∫ ∫ 11 0 x

xyzdzdydx

xyzdzdydx

3⎤ ⎡ ⎢ Ans : 8 ⎥ ⎣ ⎦ 13 1 ⎡ ⎤ ⎢ Ans : 9 − 6 log 3⎥ ⎣ ⎦

7.25

Line Integral, Double Integral and Triple Integral

9. Evaluate ∫∫∫ (

y z ) dxdyddz where R = {( x, y, z ) : 0 ≤ x ≤ 1;1

y

2;2 ≤ z ≤ 3}

R

9⎤ ⎡ ⎢ Ans : 2 ⎥ ⎣ ⎦

∫∫ (

10. Evaluate

y z ) dxdyd dz over the tetrahedron bounded by the planes

R

1⎤ ⎡ ⎢ Ans : 8 ⎥ ⎣ ⎦

x = 0, y = 0, z = 0 and x + y + z = 1. d ∫∫ ydxdy

11. Evaluate

where R is the region of the first quardant bounded by the

R

ellipse

x2 a2

12. Evaluate

+

y2 b2

∫∫∫ z

2

=1

[WBUT-2008]

dxdydz d over the region defined by z ≥ 0, x 2 + y 2 + z 2 ≤ a 2 [WBUT-2004]

Multiple-Choice Questions (1, 1)

∫

1. The value of the integral

[( x + y) d

( y x) dy is independent of the path.

(0, 0)

a) True

b) False

2. The value of the integral

∫∫ dxdy

where R = {( x, y) : x + y ≤ 1} is

R

a) 2

2

b)

c) 4

d) 3

1 8

d) 4

111

3.

∫ ∫ ∫ xyzdxdydz

is equal to

000

a) 1

b) 0

4. The value of the line integral a)

1

∫ 2 ( xdx

ydy) along any closed curve C is

C

1 2

5. The value of

c)

b) 1

∫ ( xdx

c) 0

d) none of these

dy), where C is the line joining (0, 1) to (1, 0) is

C

a)

3 2

b)

1 2

c) 0

d)

2 3

7.26

Engineering Mathematics-I π 2 cosθ

6.

∫ ∫ 0

a)

r 3 drdθ =

0

π 64

π 32

b)

c)

3π 64

d)

5π 64

c)

2 3

d) none of these

11

7.

∫ ∫(

2

y 2 ) dxdy =

00

a)

1 3

b) 0

8. The value of the integral

∫ xdy where C is the arc cut off from the parabola y

C

from the point (0, 0) to (1, −1) is a)

−1 3

b)

1 3

c) 0

d) none of these

Answers: 1. (b)

2. (a)

3. (c)

4. (c)

5. (a)

6. (c)

7. (c)

8. (a)

2

x

CHAPTER

8 Infinite Series 8.1

INTRODUCTION

This chapter basically deals with preliminary ideas of real sequences and illustrative ideas of infinite series. The first few sections elaborate the ideas of a sequence; different types of sequences, including bounded and monotone sequences, and their convergence and divergence. Each of the items are illustrated with various kinds of examples. In the later sections, we discuss the different kinds of infinite series including alternating series and also the tests of convergence of these series. Here too, useful examples are cited to illustrate the facts. Further, solutions of some important problems given in university examinations are provided in the last section.

8.2

PRELIMINARY IDEAS OF SEQUENCES

A sequence in R or a real sequence is a mapping f N → R where N is the set of natural numbers and R is the set of real numbers. So for each n N, there exists f (n) and the sequence is denoted by { f (n)} . We often denote a sequence by {an } or { xn } , etc. A sequence is also denoted by a { 1 a2 , a3 , } . Example 1 3

Let f N → R is defined by f n) = n3 , n ∈ N ; then f

{ } or {1 , 2 , 3 ,

The sequence is denoted by n3

3

3

3

}.

f (2) = 23 , f (3) = 33 ,…

8.2

Engineering Mathematics-I

Note: For our convenience, we replace { f (n)} by {an } where an is the n-th term of the sequence.

8.3

DIFFERENT TYPES OF SEQUENCES:

1) Finite Sequence: A sequence {an } having a finite number of terms is called a finite sequence, for example, {2, 5, 6, 9} is a finite sequence of four terms. 2) Infinite Sequence: A sequence {an } having infinite number of terms is called an infinite sequence, for example, {n} = { } is an infinite sequence. 1 3) Harmonic Sequence: A sequence {an } where an = , n N is a well-known n harmonic sequence. 4) Constant Sequence: A sequence {an } where an = k , n ∈ N for any real number k is called a constant sequence.

8.4

BOUNDED SEQUENCE

The real sequence {an } is said to be a bounded sequence if there exits real numbers m and M such that m ≤ an ≤ M , for example, ⎧ 1 1 ⎨1, , , ⎩ 2 3

1 1 ⎫ , , …⎬ is a bounded sequence since 0 ≤ ≤ 1. n n ⎭

(a) A sequence {an } is bounded above if there exists a real number M such that an M for all n N. M is called the upper bound of the sequence. Example 2 The sequence

n −1 1 n −1⎫ for all n < ⎬ is bounded above since 2n 2 ⎩ 2n ⎭

{an } = ⎧⎨

N.

1 is the upper bound. 2 (b) A sequence {an } is bounded below if there exists a real number m such that an ≥ m for all n N, and m is called the lower bound of the sequence.

Therefore M =

Example 3

{ }

The sequence {an } n is bounded below since n2 ≥ 1 for all n m = 1 is the lower bound.

N. Therefore,

Infinite Series

8.5

8.3

MONOTONE SEQUENCE

(a) A sequence { an } is said to be monotonic increasing if and only if an for all n N.

≥ an

Example 4 ⎧ n ⎫ The sequence {an } = ⎨ ⎬ is a monotonic increasing sequence since ⎩ n + 1⎭ ( 1) 1 n = > 0, for all n N . − ( 2) (n +1) 1)) ( 1)(n + 2) (b) A sequence { an } is said to be monotonic decreasing if and only if an for all n N.

≤ an

Example 5 ⎧1 ⎫ The sequence {an } = ⎨ ⎬ is a monotonic decreasing sequence since ⎩n⎭ 1 1 −1 − = < 0, for all n N . n + 1 n n(( 1) Observations 1) A sequence { an } is said to be strictly monotonic increasing if and only if an > an for all n N. Example 6 ⎧ n ⎫ The sequence {an } = ⎨ ⎬ is a strictly monotonic increasing sequence. ⎩ n + 1⎭ 2) A sequence { an } is said to be strictly monotonic decreasing if and only if an < an for all n N. Example 7 ⎧1 ⎫ The sequence {an } = ⎨ ⎬ is a strictly monotonic decreasing sequence. ⎩n⎭ 3) A sequence { an } is said to be monotone if { an } is either monotonic increasing or monotonic decreasing. 4) If a sequence { an } is monotonic increasing then {− } is monotonic decreasing. 5) A sequence { an } need not always be monotone.

8.4

Engineering Mathematics-I

Example 8

{2, 0, 2, 0, 2, 0…} is neither monotonic increasing nor monotonic decreasing. 8.6

LIMIT OF A SEQUENCE

A real number l is said to be a limit of a sequence {an } if for any pre-assigned positive ε , however small, there exists a natural number n0 depending on ε such that an l < ε for all n ≥ n0 . We write lim an = l . n →∞

Observation To establish the limit l of a sequence { an } , we take ε as an arbitrary positive number and then find some positive integer n0 such that the numerical magnitude of the difference ( n l ) is less than ε for every an where n > n0 . Example 9 2n + 1 = 2, let us consider a pre-assigned positive number ε , hown +1 ever small, such that To establish lim

n →∞

2n + 1 −2 < ε n +1 ⇒

1

⇒ >

1 ε

1 − 1. ε

⎡1 ⎤ Taking n0 = ⎢ − 1 , i.e., the integral part of ⎣ε ⎦

⎛1 ⎞ ⎜ ε − 1 ⎟ we have ⎝ ⎠

2n + 1 − 2 < ε for all n ≥ n0 . n +1

8.7

CONVERGENT SEQUENCE

A sequence {an } is called a convergent sequence if it has a finite real number l as its limit. We say the sequence {an } converges to l.

8.5

Infinite Series

Example 10 ⎧ n2 + 1 ⎫ The sequence ⎨ 2 ⎬ is a convergent sequence and converges to the limit 1. ⎩ n ⎭ Observations 1) A convergent sequence has at most one limit. 2) A convergent sequence is bounded. 3) Every bounded sequence is not convergent.

8.8

DIVERGENT SEQUENCE

If for any pre-asssigned positive number K, however large, there exists a natural number n0 such that an K for all n ≥ n0 , then {an } is said to diverge to ∞. We write lim an = ∞.

n →∞

If for any pre-asssigned positive number K, however large, there exists a natural K for all n ≥ n0 then {an } is said to diverge to − ∞. number n0 such that an We write lim an = −∞.

n →∞

A sequence {an } is said to be a divergent sequence if {an } either diverges to ∞ or diverges to − ∞. Example 11

{ }

The sequence n2 is a divergent sequence and diverges to the limit

∞, since lim n2 = ∞. n →∞

Note: There are also some sequences which are neither convergent nor divergent, known as oscillatory sequences. Example 12

{(

2)

}

{−22, 22, − 2, 2, …}

is a common example of an oscillatory

sequence.

8.9

INFINITE SERIES

Consider an

be a sequence of real numbers. Then a1 + a2 + a3 +

+ an +

∞ is

said to be the infinite series generated by the sequence {an } . The infinite series is ∞

denoted by

∑ an . n =1

8.6

Engineering Mathematics-I ∞

∞

n 1

n= n 1

{an } = ⎧⎨

1⎫ ⎧ 1 1 ⎬ = ⎨1, , , ⎩n⎭ ⎩ 2 3

8.10 Let

1

1

1

∑ an ∑ n = 1 + 2 + 3 +

Example 13

is a series generated by the sequence

⎫ ⎬. ⎭

CONVERGENCE AND DIVERGENCE OF INFINITE SERIES

n

+ a2 + a3 +

1

sums of the series

+ an . The sequence { n } is called the sequence of partial

∞

∞

n =1

n=1

∑ an . The infinite series ∑ an

is convergent or divergent accord-

ing to whether { n } is convergent or divergent. ∞

If lim Sn

S then S is the sum of the series

n →∞

∑ an

and if lim Sn = ∞ ( n →∞

n=1

then the infinite series is said to diverge to ∞ (or, −∞). Example 14 ⎧ 1 ⎫ Let { n } = ⎨ ⎬ be a sequence of real numbers, n )⎭ ⎩ n((

N.

So, ∞

1

1

1

∑ an = 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + n=1

The sequence of partial sums of the series is { n } where Sn =

1 1 1 + + + 1 2 2 3 3 4

⎛ 1⎞ ⎛1 1⎞ = 1− ⎟ + − ⎟+ ⎝ 2⎠ ⎝2 3⎠ = 1−

+

1 n(n 1)

1 ⎞ ⎛1 +⎜ − n n − 1 ⎟⎠ ⎝

1 n −1

From the above, we have lim Sn = 1

n →∞

∞

Therefore, the infinite series

∑ an n=1

is convergent and converges to 1.

)

8.7

Infinite Series

Example 15

{ } be a sequence of real numbers, n

Let {an } = n3

N.

So, ∞

∑ an = 13 + 23 + 33 + n=1

The sequence of partial sums of the series is { n } where Sn = 13 + 23 33 + ⎡ n(( 1) ⎤ =⎢ ⎥ ⎣ 2 ⎦

+ n3

2

From the above, we have lim Sn = ∞

n →∞

∞

Therefore, the infinite series

∑ an

is divergent and diverges to ∞.

n=1

8.10.1

The p -Series ∞

The infinite series divergent if p ≤ 1. ∞

(1)

1

∑ an =

1

1

n =1

1

1

1

1

1 1

1

p

+

∑ n2 = 12 + 22 + 32 +

1 2

p

+

1 3

P

+

∞

∞=∑

1

n =1 n

p

is convergent if p > 1 and

is convergent, since it is a p-series where p = 2.

n =1 ∞

(2)

∑n =1+ 2 +3+

is divergent, since it is a p-series where p = 1.

n =1

8.10.2

Geometric Series

A series of the form 1 + + r 2 + 3 + + n + common ratio r. The above series is (i) convergent for r < 1, i.e., −1 < r < 1 (ii) divergent for r ≥ 1 (iii) oscillatory for r ≤ −1

is called a geometric series with

8.8

Engineering Mathematics-I

8.11

PROPERTIES OF CONVERGENCE OF INFINITE SERIES ∞

1) If

∞

∑ an

and

n=1

∑ bn

are two convergent infinite series and they converge

n=1

∞

to p and q respectively then the series

∑ (c

an + d bn ) is also conver-

n =1

gent and converges to c p + d q. ∞

2) If an infinite series ∑ an is convergent then lim an = 0 n→∞ ∞

n=1

Example 16 ∞

1

∑ n2

is convergent, so we have lim

n →∞

n=1

1 n2

= 0.

The converse is not true. Example 17 1 = 0, but n →∞ n lim

∞

1

∑n

is a divergent series.

n=1

∞

3) If lim an ≠ 0 for an infinite series ∑ an then the series cannot be convergent. n→ ∞

n=1

4) If the sequence of partial sum {Sn } is not bounded then {Sn } being a monotone increasing sequence, diverges to ∞ . ∞

In this case, the series

∑ an

diverges to ∞ .

n=1

5) Addition or removal of a finite number of terms does not effect the convergence of an infinite series.

8.11.1

Series of Positive Terms ∞

An infinite series

∑ an

is called a series of positive terms if an > 0 for all n

N.

n=1

Theorem 8.1: The necessary and sufficient condition for an infinite series of posi∞

tive terms

∑ an

to be convergent is that the sequence of partial sums {Sn } is

n=1

bounded. Proof: Beyond the scope of the book.

8.9

Infinite Series

Theorem 8.2: An infinite series of positive terms either converges or diverges to ∞ . This kind of series cannot diverge to −∞ ∞ and cannot be oscillatory. Proof: Beyond the scope of the book.

8.12

DIFFERENT TESTS OF CONVERGENCE OF INFINITE SERIES

8.12.1

Comparison Test ∞

Consider

∑ an

∞

∑ bn

and

n=1

If for all n ≥ m,

are two infinite series of positive terms.

n=1

an ≤ k , k being a fixed positive number then bn

∞

i)

∑ an

∞

is convergent if

n=1

∑ bn

∞

is divergent if

n=1

8.12.2

is convergent

n=1

∞

ii)

∑ bn

∑ an

is divergent

n=1

Limit Form of Comparison Test ∞

Consider

∑ an

∞

and

n=1

∑ bn

be two infinite series of positive terms and lim

n →∞

n=1

an = l, bn

where l is a nonzero finite number. Then

∞

∞

n=1

n=1

∑ an and ∑ bn

converge and diverge together.

Example 18 Consider the following two series: ∞

∑ an = n=1

1+ 2 1+ 2 + 3 1+ 2 + 3+ 4 + + + 23 33 43

∞

and ∞

1

1

1

∑ bn = 1 + 2 + 3 + 4 +

∞

n=1

Here an =

(n 1)(n + 2) 2(n + 1)

3

=

( n 2) 2(n + 1)

2

and bn =

1 n

[WBUT-2008]

8.10

Engineering Mathematics-I

Now, lim n

an n(n + 2) 1 = lim = , a nonzer n o finite value. bn n 2(n + 1)2 2 ∞

Since

∞

1

∑ bn ∑ n n 1

∞

is a divergent series, by comparison test the series

n= n 1

∑ an

is

n=1

also divergent. Example 19 ∞

Test the convergence of the infinite series

∑ an

where an = (n

3

1 + 1) 3

− n.

n=1

[WBUT 2003, 2007] Sol.

Here, we have 1

an = (n3 + 1) 3 − n 1

⎧ ⎛ 1 ⎞ ⎫3 = ⎨ n3 1 + 3 ⎟ ⎬ − n ⎩ ⎝ n ⎠⎭ 1⎫ ⎧ 1 ⎞3 ⎪ ⎪⎛ = n ⎨ 1+ 3 ⎟ ⎬ − n ⎪⎝ n ⎠ ⎪ ⎩ ⎭

⎧ ⎪ ⎪ 1 1 i.e., an = n ⎨1 + + 3 ⎪ 3n ⎪ ⎩ =

1⎛1 ⎞ −1 3 ⎜⎝ 3 ⎟⎠ ⎛ 1 ⎞2 ⎜ 3⎟ + 2! ⎝n ⎠

1 ⎧1 1 1 + ⎨ − n 2 ⎩ 3 9 n3

⎫ ∞⎬ ⎭ ∞

Let us consider the series

⎫ ⎪ ⎪ ∞⎬ − n ⎪ ⎪ ⎭

∑ bn n=1

where bn =

1 n2

Now lim

n →∞

an ⎧1 1 1 = lim ⎨ − + bn n →∞ ⎩ 3 9 n3

⎫ 1 ∞ ⎬ = , a nonzer n o finite value. ⎭ 3

∞

Since,

∑ bn n=1

convergent.

∞

is a convergent series, by comparison test,

∑ an n=1

is also

8.11

Infinite Series

Example 20 Test the convergence of the infinite series 1+

1 22

+

22 33

+

33 44

+

44 55

+

∞

[WBUT 2005].

We know that addition and removal of finite number of terms does not affect the convergence of an infinite series. So, removing the first term from the given series, the resulting series is

Sol.

∞

22

1

33

44

∑ an = 22 + 33 + 44 + 55 + n=1

where an =

nn (n 1) n +1

. ∞

Let us consider the series ∑ bn , where bn = n=1

1 . Now n

an n n +1 = lim = 1, a nonzer n o finite value. n →∞ bn n →∞ ( n + 1) n +1 lim

∞

Since,

∑ bn is a divergent series, by comparison test, n=1

Therefore, correspondingly the given series 1+

1 22

+

22 33

+

33 44

+

44 55

+

is also divergent.

8.12.3

D’Alembert’s Ratio Test

∞

Let

∑ an

be an infinite series of positive terms and

n=1

an +1 = l , any real value. n →∞ a n lim

∞

Then, the series

∑ an n=1

i) converges if l < 1 ii) diverges if l > 1 iii) the test fails if l = 1

∞

∑ an n=1

is divergent.

8.12

Engineering Mathematics-I

Example 21 Examine the convergence of the infinite series 12 22 22 32 32 42 + + + 1! 2! 3!

∞ ∞

If we write the given series in the form of

Sol.

∑ an

then

n=1

an =

n2 (n 1)2 (n 1)2 (n + 2)2 and an +1 = . n! (n 1)!

Now, an +1 (n + 2)2 (n!) = lim 2 n →∞ an n →∞ n ( n + 1)! lim

= lim

2)2

(

1)n2

n →∞ (

= 0 < 1. Therefore, by D’ Alembert’s ratio test, the series is convergent. Example 22 Examine the convergence of the infinite series 2

2

2

⎛ 1 ⎞ ⎛ 1 2 ⎞ ⎛ 1 2 ⋅3 ⎞ ⎜ 3 ⎟ + ⎜ 3 5 ⎟ + ⎜ 3 5⋅ 7 ⎟ + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

[WBUT 2002, 2007] ∞

Sol.

If we write the given series in the form of

∑ an

then

n=1

⎛ 1 2 ⋅ 3… n ⎞ an = ⎜ ⎟ ⎝ 3 5 ⋅ 7 (2n 1) ⎠

2

1 ⋅ 2 ⋅ 3 n (n + 1) ⎛ ⎞ and so an+1 = ⎜ ⎟ ⋅ ⋅ + + 3 5 7 (2 n 1) ( 2 n 3) ⎝ ⎠

Now, lim

n →∞

an +1 = lim n →∞ an

2

1 ⋅ 2 ⋅ 3 n (n + 1) ⎛ ⎞ ⎜ 3 ⋅ 5 ⋅ 7 (2n + 1) (2n + 3) ⎟ ⎝ ⎠ ⎛ 1 ⋅ 2 ⋅ 3…n ⎞ ⎜ 3 ⋅ 5 ⋅ 7 (2 + 1) ⎟ ⎝ ⎠

2

2

8.13

Infinite Series

⎛ n +1 ⎞ = lim ⎜ 3 ⎟⎠ n →∞ ⎝ 2

2

2

1 ⎛1⎞ = ⎜ ⎟ = < 1. 4 ⎝2⎠ Therefore, by D’ Alembert’s ratio test, the given series is convergent.

8.12.4

Cauchy’s Root Test

∞

Let

∑ an

be an infinite series of positive terms and lim (an n →∞

n=1

1 )n

= l.

∞

Then the series

∑ an n=1

i) converges if l < 1 ii) diverges if l > 1 iii) the test fails if l = 1

[WBUT 2004]

Example 23 Examine the convergence of the infinite series ⎛ 22 2 ⎞ ⎜⎜ 2 − ⎟⎟ 1⎠ ⎝1 Sol.

−1

⎛ 33 3 ⎞ +⎜ 3 − ⎟ ⎜2 2 ⎟⎠ ⎝

−2

⎛ 44 4 ⎞ +⎜ 4 − ⎟ ⎜3 3 ⎟⎠ ⎝

−3

+

[WBUT 2001]

Let us consider ⎛ 22 2 ⎞ ∑ an = ⎜⎜ 12 − 1 ⎟⎟ n=1 ⎝ ⎠ ∞

−1

⎛ 33 3 ⎞ +⎜ 3 − ⎟ ⎜2 2 ⎟⎠ ⎝

−2

⎛ 44 4 ⎞ +⎜ 4 − ⎟ ⎜3 3 ⎠⎟ ⎝

−3

Then n +1 ⎪⎧⎛ n + 1 ⎞ ⎛ n 1 ⎞ ⎫⎪ −⎜ an = ⎨⎜ ⎟ ⎟⎬ ⎝ n ⎠ ⎭⎪ ⎩⎪⎝ n ⎠

−n

Now,

lim (an

n →∞

1 )n

⎡⎧ n +1 ⎪⎛ n + 1 ⎞ ⎛ n +1 ⎞ = lim ⎢ ⎨⎜ ⎟ −⎜ n ⎟ n →∞ ⎢ ⎝ n ⎠ ⎝ ⎠ ⎪ ⎣⎩

1 −n ⎤ n

⎫⎪ ⎬ ⎥⎥ ⎪⎭ ⎦

+

8.14

Engineering Mathematics-I n +1 ⎪⎧⎛ n + 1 ⎞ ⎛ n 1 ⎞ ⎫⎪ −⎜ = lim ⎨⎜ ⎟ ⎟⎬ n →∞ ⎝ n ⎠ ⎝ n ⎠ ⎭⎪ ⎩⎪

−1

⎡⎛ 1 ⎞ ⎪⎧⎛ 1 ⎞n ⎪⎫ = lim ⎢ 1 + ⎟ ⎨⎜ 1 + ⎟ − 1⎬ n →∞ ⎢⎝ n ⎠ ⎩⎪⎝ n ⎠ ⎭⎪ ⎣

⎤ ⎥ ⎥⎦

−1

= (e − 1) −1 < 1. ∞

Therefore, by Cauchy’s root test,

∑ an

is convergent.

n=1

Example 24 3

⎛ 1 ⎞ Examine the convergence of the infinite series ⎜ 1 + ⎟ n⎠ ⎝ Sol.

−n 2

Let 3 ∞

∞

⎛

∑ an ∑ ⎜1 + n =1

n =1 ⎝

1 ⎞ ⎟ n⎠

−n 2

then 3

⎛ 1 ⎞ an = ⎜ 1 + ⎟ n⎠ ⎝

−n 2

.

Now, 1

3 ⎤n ⎡ −n 2 ⎥ 1 ⎢ ⎛ 1 ⎞ ⎥ lim (an ) n = lim li ⎢⎜ 1 + ⎟ ⎥ n →∞ n →∞ ⎢⎝ n⎠ ⎢ ⎥ ⎣ ⎦ − ⎡ ⎢⎛⎜ 1 + 1 ⎞⎟ n →∞ ⎢⎝ n⎠ ⎣

= li

=

n

⎤ ⎥ ⎥ ⎦

1 < 1. e ∞

Therefore, by Cauchy’s root test,

∑ an n=1

is convergent.

[WBUT 2004]

8.15

Infinite Series

8.12.5 ∞

Let

Raabe’s Test

∑ an n=1

⎛ a ⎞ be an infinite series of positive terms and lim n ⎜ n − 1 ⎟ = l . n →∞ ⎝ an +1 ⎠ ∞

Then the series

∑ an n=1

i) converges if l > 1 ii) diverges if l < 1 iii) the test fails if l = 1 Example 25 Examine the convergence of the infinite series 1 1 1 3 1 1 3⋅5 1 1+ ⋅ + ⋅ + ⋅ + 2 3 2 4 5 2 4⋅6 7 Sol.

Since addition or removal of a finite number of terms does not affect the convergence of an infinite series, by removing the first term of the given series, let the resulting infinite series be

Âa n=1

n

=

1 1 1◊ 3 1 1◊ 3 ◊ 5 1 ◊ + + + 2 3 2 ◊4 5 2◊ 4◊6 7

Then an =

1 3 ⋅ 5 (2 n 3) 1 ⋅ 2 4 ⋅ 6 (2 n 2) (2 n − 1)

and so an+1 =

1 ⋅ 3 ⋅ 5 (2 n 3)(2 n − 1) 1 ⋅ 2 ⋅ 4 ⋅ 6 (2 n 2)2 n (2 n + 1)

Now an +1 (2 n − 1)2 = lim =1 n →∞ an n →∞ 2 n(2 n + 1) lim

Therefore, D’ Alembert’s ratio test fails. But ⎡ 2 n(2 n + 1) ⎤ ⎛ a ⎞ lim n ⎜ n − 1 ⎟ = lim n ⎢ − 1⎥ 2 →∞ n ⎝ an +1 ⎠ ⎣ (2 n − 1) ⎦

n →∞

= lim

6n2

n

n →∞ (2 n − 1)

2

=

3 >1 2 ∞

Therefore, by Raabe’s test, the series

∑ an n=1

is convergent.

8.16

Engineering Mathematics-I

8.13

ALTERNATING SERIES ∞

The infinite series of the form

∑(

) n −1 an , where an > 0 for all n

N is called an

n =1

alternating series. Example 26 Let us consider the series ∞

∑(

1) n −1

n =1

1 n2

= 1−

Here an > 0 for all n

8.13.1

+

1 32

−

1 42

+

N. So this is an alternating series.

Test of Convergence of Alternating Series (Leibnitz’s Test)

∞

∑(

Let

1 22

) n −1 an be an alterenating series with an > 0 for all n

N. Then the series

n =1

converges if i) an

< an , i.e., { n } is a monotonic decreasing sequence

ii) lim an = 0 n →∞

[WBUT-2009] Example 27 Examine the convergence of the alternating series 3 4 5 2− + − + 2 3 4 ∞

Sol.

If we write the series in the form of

∑(

1) n −1 an then

n =1

an =

(n 1) n+2 and an +1 = . n n +1

Now an

an +1 = =

(n + 1) n + 2 − n n +1 1 n((

1)

> 0 for all n

So an > an +1 , i.e., { n } is a monotonic decreasing sequence.

8.17

Infinite Series

But lim an = lim

(n + 1) =1 n

i.e., li

0.

n →∞

n →∞ n

n →∞

Therefore, by Leibnitz’s test, the alternating series is not convergent. Example 28

∞

Examine the convergence of the alternating series

cos nπ

∑ n2 + 1 .

[WBUT 2002]

n =1

Since cos nπ = (−1) n , the alternating series can be written as

Sol.

∞

∞

cos nπ

∞

( 1) n

1

∑ n2 + 1 = ∑ n2 + 1 = ∑ (−1)n n2 + 1 . n =1

n =1

n =1

Then, 1

an =

2

n +1

and so an +1 =

1 (n 1)2 1

Now an

an +1 = =

1

1

−

2

n + 1 (n + 1)2 + 1 (2 n + 1) (

2

n + 1)2 + 1) 1)((n 1)((

>0

The above implies that { n } is a monotonic decreasing sequence. Also lim an = lim

n →∞

n →∞

∞

Therefore,

1 2

n +1

cos nπ

∑ n2 + 1

=0

is convergent by Leibnitz’s test.

n =1

8.14

ABSOLUTE CONVERGENCE

∞

Let

∑ an

∞

be an infinite series. The series is said to be absolutely convergent if

n=1

is convergent.

∑ an n=1

8.18

Engineering Mathematics-I ∞

Suppose

∑(

1) n −1 an be an alternating series. The alternating series is said to be

n =1

absolutely convergent if ∞

∑(

1)

1

n =1

∞

= ∑ an n =1

is convergent.

[WBUT-2009]

Note: An absolutely convergent series is convergent, but the converse is not always true. Example 29

∞

cos nπ

n =1

n2

Examine the absolute convergence of the alternating series ∑

.

Since cos nπ = (−1) n , the alternating series can be written as

Sol.

∞

cos nπ

n =1

n2

∑

∞

( 1) n

n =1

n2

=∑

Here, an =

1 n2

∞

∞

n 1

n =1

∞

= ∑ (−1) n n =1

1 n2

.

. Then

1

∑ an ∑ n2 . ∞

But the series

1

∑ n2

is a p-series with p = 2

( > 1) .

So the series is

n=1

convergent. ∞

Correspondingly,

∑ an

is also convergent.

n=1

Hence the given series

8.15

∞

cos nπ

n =1

n2

∑

is absolutely convergent.

CONDITIONAL CONVERGENCE

∞

Let

∑ an

be an infinite series. The series is said to be conditionally convergent if

n=1

∞

∑ an is convergent but not absolutely convergent, i.e., n=1

∞

∑ an n=1

is not convergent.

8.19

Infinite Series ∞

∑(

An alternating series

1) n −1 an is said to be conditionally convergent if

n =1

∞

∑(

1) n −1 an is convergent but

n =1

∞

∑ an

is not convergent.

[WBUT-2009]

n=1

Example 30 Examine the conditional convergence of the alternating series 1 1 1 1 1− + − + − 2 3 4 5

∞ ∞

Sol.

This is an alternating series of the form

∑ ( 1) n =1

1

∞ 1 = ∑ ( 1) n −1 . n n =1

Here an =

1 1 and so an +1 = n n +1

Let us apply Leibnitz’s test for checking convergence. Now an

1 1 1 − = >0 n n + 1 n(n + 1)

an +1 =

This proves that { n } is a monotonic decreasing sequence. 1 = 0. n

Again, lim an = lim n →∞

n →∞

∞

Therefore,

∑(

∞

= ∑ ( 1)n −1

1

1)

n =1

n =1

1 is a convergent series. n

∞

Now we consider

∑ an

which is

n=1

1+

1 1 1 1 + + + + 2 3 4 5

∞ 1 =∑ . n =1 n

This is a p-series with p = 1, and so the series is divergent. ∞

Since, ∑ ( 1) n −1 an is convergent and n =1

∞

∑( n =1

1)

1

∞

= ∑ ( 1) n −1 n =1

∞

∑ an

is divergent,

n=1

1 is conditionally convergent. n

8.20

Engineering Mathematics-I

WORKED-OUT EXAMPLES Example 8.1

∑{

n 4 +1 1

Sol.

Let

∞

Examine the convergence of the infinite series

}

n4 1

n =1

∑ an ∑ { ∞

∞

n 1

n =1

[WBUT 2008]

}

n4 + 1 − n4 − 1

This is an infinite series of positive terms. Here

{ n 1 − n − 1} { n +11 n 1}{ n +11 = { n +11 n 1}

an =

=

4

4

4

4

4

4

4

{ n +11

2

4

}

n4 1

}

n4 1

Let us consider the series

∞

∞

n 1

n =1

1

∑ bn ∑ n2 ,

which is convergent since it is a

p-series for p = 2. Now, an = lim n →∞ bn n →∞ lim

2n2

{ n +1 + 4

}

n4 − 1 .

= 1, a nonzero finite value. Since

∞

∞

n 1

n =1

1

∑ bn ∑ n2

is a convergent series, by comparison test, we can

conclude that

∑ an ∑ { ∞

∞

n 1

n =1

}

n4 + 1 − n4 − 1

is also convergent.

8.21

Infinite Series

Example 8.2

Examine the convergence of the infinite series

x 2 x3 x 4 + + + 2 3 4

x+ Sol.

Let us consider ∞

∞

xn n 1 n n=

∑ an ∑ n =1

where an =

xn x n +1 and an +1 = . n n +1

Now, an +1 x n +1 n = lim n = x. n →∞ an n →∞ x ( n + 1) lim

Then by D’Alembert’s ratio test, we have i) If x < 1, the infinite series is convergent ii) If x > 1, the infinite series is divergent iii) If x = 1, the test fails For x = 1, the series becomes 1+

1 1 1 + + + 2 3 4 ∞

Since

1

∑n

∞ 1 =∑ . n =1 n ∞

is a divergent series, the infinite series

n=1

xn diverges for x ≥ 1 n=1 n

∑

and converges for x < 1. Example 8.3 1+

Examine the convergence of the infinite series

x x 2 x3 + + + 2 5 10

Sol.

[WBUT-2009]

We know that addition or removal of a finite number of terms does not alter the convergence of an infinite series. So, removing the first term, we have the series of the form x x 2 x3 + + + 2 5 10

8.22

Engineering Mathematics-I

Suppose we write the series in the form ∞

∞

n =1

n =1

xn

∑ an ∑ n2 + 1 Then an =

xn n2 + 1

and so an +1 =

x n +1 (n 1)2 1

.

Here, an +1 x n +1 = lim n →∞ an n →∞ ( n + 1) 2 + 1

xn n2 + 1

lim

n2 + 1

= x ⋅ lim

1)2 1

n →∞ (

=x

So by D’Alembert’s ratio test, we can conclude that i) If x < 1, the infinite series is convergent ii) If x > 1, the infinite series is divergent iii) If x = 1, the test fails For x = 1, the series becomes ∞

1 1 1 + + + 2 5 10 Here, bn =

n =1 n

1 2

n +1

∞

1

=∑

2

+1

= ∑ bn n =1

.

Consider the series

∞

∞

n 1

n =1

1

∑ cn ∑ n2 ,

which is convergent. Now

bn n2 = lim 2 = 1, a nonzero finite number. n →∞ cn n →∞ n + 1 lim

∞

Therefore, by comparison test,

1

∑ n2 + 1

is a convergent series.

n =1

∞

Hence, ∑ an n =1

∞

xn

∑ n2 + 1 n =1

is convergent for x ≤ 1 and divergent for x > 1.

8.23

Infinite Series

Example 8.4

Examine the convergence of the infinite series

∞

n! 2 n

n =1

nn

∑

. [WBUT 2003]

Let us consider

Sol.

∞

∞

∑ an

∑

n =1

n! ⋅ 2 n nn

nn=1

Then an =

n!2 n n

n

and so an +1 =

(n 1)!2 n +1 (n 1) n +1

Here a (n + 1)!2 n +1 lim n +1 = lim n →∞ ( n + 1) n +1 an = lim

n →∞

2 ⎛ 1⎞ ⎜1 + n ⎟ ⎝ ⎠

n

=

n!2 n nn 2 an for all n N. Example 6 ⎧ n ⎫ The sequence {an } = ⎨ ⎬ is a strictly monotonic increasing sequence. ⎩ n + 1⎭ 2) A sequence { an } is said to be strictly monotonic decreasing if and only if an < an for all n N. Example 7 ⎧1 ⎫ The sequence {an } = ⎨ ⎬ is a strictly monotonic decreasing sequence. ⎩n⎭ 3) A sequence { an } is said to be monotone if { an } is either monotonic increasing or monotonic decreasing. 4) If a sequence { an } is monotonic increasing then {− } is monotonic decreasing. 5) A sequence { an } need not always be monotone.

8.4

Engineering Mathematics-I

Example 8

{2, 0, 2, 0, 2, 0…} is neither monotonic increasing nor monotonic decreasing. 8.6

LIMIT OF A SEQUENCE

A real number l is said to be a limit of a sequence {an } if for any pre-assigned positive ε , however small, there exists a natural number n0 depending on ε such that an l < ε for all n ≥ n0 . We write lim an = l . n →∞

Observation To establish the limit l of a sequence { an } , we take ε as an arbitrary positive number and then find some positive integer n0 such that the numerical magnitude of the difference ( n l ) is less than ε for every an where n > n0 . Example 9 2n + 1 = 2, let us consider a pre-assigned positive number ε , hown +1 ever small, such that To establish lim

n →∞

2n + 1 −2 < ε n +1 ⇒

1

⇒ >

1 ε

1 − 1. ε

⎡1 ⎤ Taking n0 = ⎢ − 1 , i.e., the integral part of ⎣ε ⎦

⎛1 ⎞ ⎜ ε − 1 ⎟ we have ⎝ ⎠

2n + 1 − 2 < ε for all n ≥ n0 . n +1

8.7

CONVERGENT SEQUENCE

A sequence {an } is called a convergent sequence if it has a finite real number l as its limit. We say the sequence {an } converges to l.

8.5

Infinite Series

Example 10 ⎧ n2 + 1 ⎫ The sequence ⎨ 2 ⎬ is a convergent sequence and converges to the limit 1. ⎩ n ⎭ Observations 1) A convergent sequence has at most one limit. 2) A convergent sequence is bounded. 3) Every bounded sequence is not convergent.

8.8

DIVERGENT SEQUENCE

If for any pre-asssigned positive number K, however large, there exists a natural number n0 such that an K for all n ≥ n0 , then {an } is said to diverge to ∞. We write lim an = ∞.

n →∞

If for any pre-asssigned positive number K, however large, there exists a natural K for all n ≥ n0 then {an } is said to diverge to − ∞. number n0 such that an We write lim an = −∞.

n →∞

A sequence {an } is said to be a divergent sequence if {an } either diverges to ∞ or diverges to − ∞. Example 11

{ }

The sequence n2 is a divergent sequence and diverges to the limit

∞, since lim n2 = ∞. n →∞

Note: There are also some sequences which are neither convergent nor divergent, known as oscillatory sequences. Example 12

{(

2)

}

{−22, 22, − 2, 2, …}

is a common example of an oscillatory

sequence.

8.9

INFINITE SERIES

Consider an

be a sequence of real numbers. Then a1 + a2 + a3 +

+ an +

∞ is

said to be the infinite series generated by the sequence {an } . The infinite series is ∞

denoted by

∑ an . n =1

8.6

Engineering Mathematics-I ∞

∞

n 1

n= n 1

{an } = ⎧⎨

1⎫ ⎧ 1 1 ⎬ = ⎨1, , , ⎩n⎭ ⎩ 2 3

8.10 Let

1

1

1

∑ an ∑ n = 1 + 2 + 3 +

Example 13

is a series generated by the sequence

⎫ ⎬. ⎭

CONVERGENCE AND DIVERGENCE OF INFINITE SERIES

n

+ a2 + a3 +

1

sums of the series

+ an . The sequence { n } is called the sequence of partial

∞

∞

n =1

n=1

∑ an . The infinite series ∑ an

is convergent or divergent accord-

ing to whether { n } is convergent or divergent. ∞

If lim Sn

S then S is the sum of the series

n →∞

∑ an

and if lim Sn = ∞ ( n →∞

n=1

then the infinite series is said to diverge to ∞ (or, −∞). Example 14 ⎧ 1 ⎫ Let { n } = ⎨ ⎬ be a sequence of real numbers, n )⎭ ⎩ n((

N.

So, ∞

1

1

1

∑ an = 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + n=1

The sequence of partial sums of the series is { n } where Sn =

1 1 1 + + + 1 2 2 3 3 4

⎛ 1⎞ ⎛1 1⎞ = 1− ⎟ + − ⎟+ ⎝ 2⎠ ⎝2 3⎠ = 1−

+

1 n(n 1)

1 ⎞ ⎛1 +⎜ − n n − 1 ⎟⎠ ⎝

1 n −1

From the above, we have lim Sn = 1

n →∞

∞

Therefore, the infinite series

∑ an n=1

is convergent and converges to 1.

)

8.7

Infinite Series

Example 15

{ } be a sequence of real numbers, n

Let {an } = n3

N.

So, ∞

∑ an = 13 + 23 + 33 + n=1

The sequence of partial sums of the series is { n } where Sn = 13 + 23 33 + ⎡ n(( 1) ⎤ =⎢ ⎥ ⎣ 2 ⎦

+ n3

2

From the above, we have lim Sn = ∞

n →∞

∞

Therefore, the infinite series

∑ an

is divergent and diverges to ∞.

n=1

8.10.1

The p -Series ∞

The infinite series divergent if p ≤ 1. ∞

(1)

1

∑ an =

1

1

n =1

1

1

1

1

1 1

1

p

+

∑ n2 = 12 + 22 + 32 +

1 2

p

+

1 3

P

+

∞

∞=∑

1

n =1 n

p

is convergent if p > 1 and

is convergent, since it is a p-series where p = 2.

n =1 ∞

(2)

∑n =1+ 2 +3+

is divergent, since it is a p-series where p = 1.

n =1

8.10.2

Geometric Series

A series of the form 1 + + r 2 + 3 + + n + common ratio r. The above series is (i) convergent for r < 1, i.e., −1 < r < 1 (ii) divergent for r ≥ 1 (iii) oscillatory for r ≤ −1

is called a geometric series with

8.8

Engineering Mathematics-I

8.11

PROPERTIES OF CONVERGENCE OF INFINITE SERIES ∞

1) If

∞

∑ an

and

n=1

∑ bn

are two convergent infinite series and they converge

n=1

∞

to p and q respectively then the series

∑ (c

an + d bn ) is also conver-

n =1

gent and converges to c p + d q. ∞

2) If an infinite series ∑ an is convergent then lim an = 0 n→∞ ∞

n=1

Example 16 ∞

1

∑ n2

is convergent, so we have lim

n →∞

n=1

1 n2

= 0.

The converse is not true. Example 17 1 = 0, but n →∞ n lim

∞

1

∑n

is a divergent series.

n=1

∞

3) If lim an ≠ 0 for an infinite series ∑ an then the series cannot be convergent. n→ ∞

n=1

4) If the sequence of partial sum {Sn } is not bounded then {Sn } being a monotone increasing sequence, diverges to ∞ . ∞

In this case, the series

∑ an

diverges to ∞ .

n=1

5) Addition or removal of a finite number of terms does not effect the convergence of an infinite series.

8.11.1

Series of Positive Terms ∞

An infinite series

∑ an

is called a series of positive terms if an > 0 for all n

N.

n=1

Theorem 8.1: The necessary and sufficient condition for an infinite series of posi∞

tive terms

∑ an

to be convergent is that the sequence of partial sums {Sn } is

n=1

bounded. Proof: Beyond the scope of the book.

8.9

Infinite Series

Theorem 8.2: An infinite series of positive terms either converges or diverges to ∞ . This kind of series cannot diverge to −∞ ∞ and cannot be oscillatory. Proof: Beyond the scope of the book.

8.12

DIFFERENT TESTS OF CONVERGENCE OF INFINITE SERIES

8.12.1

Comparison Test ∞

Consider

∑ an

∞

∑ bn

and

n=1

If for all n ≥ m,

are two infinite series of positive terms.

n=1

an ≤ k , k being a fixed positive number then bn

∞

i)

∑ an

∞

is convergent if

n=1

∑ bn

∞

is divergent if

n=1

8.12.2

is convergent

n=1

∞

ii)

∑ bn

∑ an

is divergent

n=1

Limit Form of Comparison Test ∞

Consider

∑ an

∞

and

n=1

∑ bn

be two infinite series of positive terms and lim

n →∞

n=1

an = l, bn

where l is a nonzero finite number. Then

∞

∞

n=1

n=1

∑ an and ∑ bn

converge and diverge together.

Example 18 Consider the following two series: ∞

∑ an = n=1

1+ 2 1+ 2 + 3 1+ 2 + 3+ 4 + + + 23 33 43

∞

and ∞

1

1

1

∑ bn = 1 + 2 + 3 + 4 +

∞

n=1

Here an =

(n 1)(n + 2) 2(n + 1)

3

=

( n 2) 2(n + 1)

2

and bn =

1 n

[WBUT-2008]

8.10

Engineering Mathematics-I

Now, lim n

an n(n + 2) 1 = lim = , a nonzer n o finite value. bn n 2(n + 1)2 2 ∞

Since

∞

1

∑ bn ∑ n n 1

∞

is a divergent series, by comparison test the series

n= n 1

∑ an

is

n=1

also divergent. Example 19 ∞

Test the convergence of the infinite series

∑ an

where an = (n

3

1 + 1) 3

− n.

n=1

[WBUT 2003, 2007] Sol.

Here, we have 1

an = (n3 + 1) 3 − n 1

⎧ ⎛ 1 ⎞ ⎫3 = ⎨ n3 1 + 3 ⎟ ⎬ − n ⎩ ⎝ n ⎠⎭ 1⎫ ⎧ 1 ⎞3 ⎪ ⎪⎛ = n ⎨ 1+ 3 ⎟ ⎬ − n ⎪⎝ n ⎠ ⎪ ⎩ ⎭

⎧ ⎪ ⎪ 1 1 i.e., an = n ⎨1 + + 3 ⎪ 3n ⎪ ⎩ =

1⎛1 ⎞ −1 3 ⎜⎝ 3 ⎟⎠ ⎛ 1 ⎞2 ⎜ 3⎟ + 2! ⎝n ⎠

1 ⎧1 1 1 + ⎨ − n 2 ⎩ 3 9 n3

⎫ ∞⎬ ⎭ ∞

Let us consider the series

⎫ ⎪ ⎪ ∞⎬ − n ⎪ ⎪ ⎭

∑ bn n=1

where bn =

1 n2

Now lim

n →∞

an ⎧1 1 1 = lim ⎨ − + bn n →∞ ⎩ 3 9 n3

⎫ 1 ∞ ⎬ = , a nonzer n o finite value. ⎭ 3

∞

Since,

∑ bn n=1

convergent.

∞

is a convergent series, by comparison test,

∑ an n=1

is also

8.11

Infinite Series

Example 20 Test the convergence of the infinite series 1+

1 22

+

22 33

+

33 44

+

44 55

+

∞

[WBUT 2005].

We know that addition and removal of finite number of terms does not affect the convergence of an infinite series. So, removing the first term from the given series, the resulting series is

Sol.

∞

22

1

33

44

∑ an = 22 + 33 + 44 + 55 + n=1

where an =

nn (n 1) n +1

. ∞

Let us consider the series ∑ bn , where bn = n=1

1 . Now n

an n n +1 = lim = 1, a nonzer n o finite value. n →∞ bn n →∞ ( n + 1) n +1 lim

∞

Since,

∑ bn is a divergent series, by comparison test, n=1

Therefore, correspondingly the given series 1+

1 22

+

22 33

+

33 44

+

44 55

+

is also divergent.

8.12.3

D’Alembert’s Ratio Test

∞

Let

∑ an

be an infinite series of positive terms and

n=1

an +1 = l , any real value. n →∞ a n lim

∞

Then, the series

∑ an n=1

i) converges if l < 1 ii) diverges if l > 1 iii) the test fails if l = 1

∞

∑ an n=1

is divergent.

8.12

Engineering Mathematics-I

Example 21 Examine the convergence of the infinite series 12 22 22 32 32 42 + + + 1! 2! 3!

∞ ∞

If we write the given series in the form of

Sol.

∑ an

then

n=1

an =

n2 (n 1)2 (n 1)2 (n + 2)2 and an +1 = . n! (n 1)!

Now, an +1 (n + 2)2 (n!) = lim 2 n →∞ an n →∞ n ( n + 1)! lim

= lim

2)2

(

1)n2

n →∞ (

= 0 < 1. Therefore, by D’ Alembert’s ratio test, the series is convergent. Example 22 Examine the convergence of the infinite series 2

2

2

⎛ 1 ⎞ ⎛ 1 2 ⎞ ⎛ 1 2 ⋅3 ⎞ ⎜ 3 ⎟ + ⎜ 3 5 ⎟ + ⎜ 3 5⋅ 7 ⎟ + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

[WBUT 2002, 2007] ∞

Sol.

If we write the given series in the form of

∑ an

then

n=1

⎛ 1 2 ⋅ 3… n ⎞ an = ⎜ ⎟ ⎝ 3 5 ⋅ 7 (2n 1) ⎠

2

1 ⋅ 2 ⋅ 3 n (n + 1) ⎛ ⎞ and so an+1 = ⎜ ⎟ ⋅ ⋅ + + 3 5 7 (2 n 1) ( 2 n 3) ⎝ ⎠

Now, lim

n →∞

an +1 = lim n →∞ an

2

1 ⋅ 2 ⋅ 3 n (n + 1) ⎛ ⎞ ⎜ 3 ⋅ 5 ⋅ 7 (2n + 1) (2n + 3) ⎟ ⎝ ⎠ ⎛ 1 ⋅ 2 ⋅ 3…n ⎞ ⎜ 3 ⋅ 5 ⋅ 7 (2 + 1) ⎟ ⎝ ⎠

2

2

8.13

Infinite Series

⎛ n +1 ⎞ = lim ⎜ 3 ⎟⎠ n →∞ ⎝ 2

2

2

1 ⎛1⎞ = ⎜ ⎟ = < 1. 4 ⎝2⎠ Therefore, by D’ Alembert’s ratio test, the given series is convergent.

8.12.4

Cauchy’s Root Test

∞

Let

∑ an

be an infinite series of positive terms and lim (an n →∞

n=1

1 )n

= l.

∞

Then the series

∑ an n=1

i) converges if l < 1 ii) diverges if l > 1 iii) the test fails if l = 1

[WBUT 2004]

Example 23 Examine the convergence of the infinite series ⎛ 22 2 ⎞ ⎜⎜ 2 − ⎟⎟ 1⎠ ⎝1 Sol.

−1

⎛ 33 3 ⎞ +⎜ 3 − ⎟ ⎜2 2 ⎟⎠ ⎝

−2

⎛ 44 4 ⎞ +⎜ 4 − ⎟ ⎜3 3 ⎟⎠ ⎝

−3

+

[WBUT 2001]

Let us consider ⎛ 22 2 ⎞ ∑ an = ⎜⎜ 12 − 1 ⎟⎟ n=1 ⎝ ⎠ ∞

−1

⎛ 33 3 ⎞ +⎜ 3 − ⎟ ⎜2 2 ⎟⎠ ⎝

−2

⎛ 44 4 ⎞ +⎜ 4 − ⎟ ⎜3 3 ⎠⎟ ⎝

−3

Then n +1 ⎪⎧⎛ n + 1 ⎞ ⎛ n 1 ⎞ ⎫⎪ −⎜ an = ⎨⎜ ⎟ ⎟⎬ ⎝ n ⎠ ⎭⎪ ⎩⎪⎝ n ⎠

−n

Now,

lim (an

n →∞

1 )n

⎡⎧ n +1 ⎪⎛ n + 1 ⎞ ⎛ n +1 ⎞ = lim ⎢ ⎨⎜ ⎟ −⎜ n ⎟ n →∞ ⎢ ⎝ n ⎠ ⎝ ⎠ ⎪ ⎣⎩

1 −n ⎤ n

⎫⎪ ⎬ ⎥⎥ ⎪⎭ ⎦

+

8.14

Engineering Mathematics-I n +1 ⎪⎧⎛ n + 1 ⎞ ⎛ n 1 ⎞ ⎫⎪ −⎜ = lim ⎨⎜ ⎟ ⎟⎬ n →∞ ⎝ n ⎠ ⎝ n ⎠ ⎭⎪ ⎩⎪

−1

⎡⎛ 1 ⎞ ⎪⎧⎛ 1 ⎞n ⎪⎫ = lim ⎢ 1 + ⎟ ⎨⎜ 1 + ⎟ − 1⎬ n →∞ ⎢⎝ n ⎠ ⎩⎪⎝ n ⎠ ⎭⎪ ⎣

⎤ ⎥ ⎥⎦

−1

= (e − 1) −1 < 1. ∞

Therefore, by Cauchy’s root test,

∑ an

is convergent.

n=1

Example 24 3

⎛ 1 ⎞ Examine the convergence of the infinite series ⎜ 1 + ⎟ n⎠ ⎝ Sol.

−n 2

Let 3 ∞

∞

⎛

∑ an ∑ ⎜1 + n =1

n =1 ⎝

1 ⎞ ⎟ n⎠

−n 2

then 3

⎛ 1 ⎞ an = ⎜ 1 + ⎟ n⎠ ⎝

−n 2

.

Now, 1

3 ⎤n ⎡ −n 2 ⎥ 1 ⎢ ⎛ 1 ⎞ ⎥ lim (an ) n = lim li ⎢⎜ 1 + ⎟ ⎥ n →∞ n →∞ ⎢⎝ n⎠ ⎢ ⎥ ⎣ ⎦ − ⎡ ⎢⎛⎜ 1 + 1 ⎞⎟ n →∞ ⎢⎝ n⎠ ⎣

= li

=

n

⎤ ⎥ ⎥ ⎦

1 < 1. e ∞

Therefore, by Cauchy’s root test,

∑ an n=1

is convergent.

[WBUT 2004]

8.15

Infinite Series

8.12.5 ∞

Let

Raabe’s Test

∑ an n=1

⎛ a ⎞ be an infinite series of positive terms and lim n ⎜ n − 1 ⎟ = l . n →∞ ⎝ an +1 ⎠ ∞

Then the series

∑ an n=1

i) converges if l > 1 ii) diverges if l < 1 iii) the test fails if l = 1 Example 25 Examine the convergence of the infinite series 1 1 1 3 1 1 3⋅5 1 1+ ⋅ + ⋅ + ⋅ + 2 3 2 4 5 2 4⋅6 7 Sol.

Since addition or removal of a finite number of terms does not affect the convergence of an infinite series, by removing the first term of the given series, let the resulting infinite series be

Âa n=1

n

=

1 1 1◊ 3 1 1◊ 3 ◊ 5 1 ◊ + + + 2 3 2 ◊4 5 2◊ 4◊6 7

Then an =

1 3 ⋅ 5 (2 n 3) 1 ⋅ 2 4 ⋅ 6 (2 n 2) (2 n − 1)

and so an+1 =

1 ⋅ 3 ⋅ 5 (2 n 3)(2 n − 1) 1 ⋅ 2 ⋅ 4 ⋅ 6 (2 n 2)2 n (2 n + 1)

Now an +1 (2 n − 1)2 = lim =1 n →∞ an n →∞ 2 n(2 n + 1) lim

Therefore, D’ Alembert’s ratio test fails. But ⎡ 2 n(2 n + 1) ⎤ ⎛ a ⎞ lim n ⎜ n − 1 ⎟ = lim n ⎢ − 1⎥ 2 →∞ n ⎝ an +1 ⎠ ⎣ (2 n − 1) ⎦

n →∞

= lim

6n2

n

n →∞ (2 n − 1)

2

=

3 >1 2 ∞

Therefore, by Raabe’s test, the series

∑ an n=1

is convergent.

8.16

Engineering Mathematics-I

8.13

ALTERNATING SERIES ∞

The infinite series of the form

∑(

) n −1 an , where an > 0 for all n

N is called an

n =1

alternating series. Example 26 Let us consider the series ∞

∑(

1) n −1

n =1

1 n2

= 1−

Here an > 0 for all n

8.13.1

+

1 32

−

1 42

+

N. So this is an alternating series.

Test of Convergence of Alternating Series (Leibnitz’s Test)

∞

∑(

Let

1 22

) n −1 an be an alterenating series with an > 0 for all n

N. Then the series

n =1

converges if i) an

< an , i.e., { n } is a monotonic decreasing sequence

ii) lim an = 0 n →∞

[WBUT-2009] Example 27 Examine the convergence of the alternating series 3 4 5 2− + − + 2 3 4 ∞

Sol.

If we write the series in the form of

∑(

1) n −1 an then

n =1

an =

(n 1) n+2 and an +1 = . n n +1

Now an

an +1 = =

(n + 1) n + 2 − n n +1 1 n((

1)

> 0 for all n

So an > an +1 , i.e., { n } is a monotonic decreasing sequence.

8.17

Infinite Series

But lim an = lim

(n + 1) =1 n

i.e., li

0.

n →∞

n →∞ n

n →∞

Therefore, by Leibnitz’s test, the alternating series is not convergent. Example 28

∞

Examine the convergence of the alternating series

cos nπ

∑ n2 + 1 .

[WBUT 2002]

n =1

Since cos nπ = (−1) n , the alternating series can be written as

Sol.

∞

∞

cos nπ

∞

( 1) n

1

∑ n2 + 1 = ∑ n2 + 1 = ∑ (−1)n n2 + 1 . n =1

n =1

n =1

Then, 1

an =

2

n +1

and so an +1 =

1 (n 1)2 1

Now an

an +1 = =

1

1

−

2

n + 1 (n + 1)2 + 1 (2 n + 1) (

2

n + 1)2 + 1) 1)((n 1)((

>0

The above implies that { n } is a monotonic decreasing sequence. Also lim an = lim

n →∞

n →∞

∞

Therefore,

1 2

n +1

cos nπ

∑ n2 + 1

=0

is convergent by Leibnitz’s test.

n =1

8.14

ABSOLUTE CONVERGENCE

∞

Let

∑ an

∞

be an infinite series. The series is said to be absolutely convergent if

n=1

is convergent.

∑ an n=1

8.18

Engineering Mathematics-I ∞

Suppose

∑(

1) n −1 an be an alternating series. The alternating series is said to be

n =1

absolutely convergent if ∞

∑(

1)

1

n =1

∞

= ∑ an n =1

is convergent.

[WBUT-2009]

Note: An absolutely convergent series is convergent, but the converse is not always true. Example 29

∞

cos nπ

n =1

n2

Examine the absolute convergence of the alternating series ∑

.

Since cos nπ = (−1) n , the alternating series can be written as

Sol.

∞

cos nπ

n =1

n2

∑

∞

( 1) n

n =1

n2

=∑

Here, an =

1 n2

∞

∞

n 1

n =1

∞

= ∑ (−1) n n =1

1 n2

.

. Then

1

∑ an ∑ n2 . ∞

But the series

1

∑ n2

is a p-series with p = 2

( > 1) .

So the series is

n=1

convergent. ∞

Correspondingly,

∑ an

is also convergent.

n=1

Hence the given series

8.15

∞

cos nπ

n =1

n2

∑

is absolutely convergent.

CONDITIONAL CONVERGENCE

∞

Let

∑ an

be an infinite series. The series is said to be conditionally convergent if

n=1

∞

∑ an is convergent but not absolutely convergent, i.e., n=1

∞

∑ an n=1

is not convergent.

8.19

Infinite Series ∞

∑(

An alternating series

1) n −1 an is said to be conditionally convergent if

n =1

∞

∑(

1) n −1 an is convergent but

n =1

∞

∑ an

is not convergent.

[WBUT-2009]

n=1

Example 30 Examine the conditional convergence of the alternating series 1 1 1 1 1− + − + − 2 3 4 5

∞ ∞

Sol.

This is an alternating series of the form

∑ ( 1) n =1

1

∞ 1 = ∑ ( 1) n −1 . n n =1

Here an =

1 1 and so an +1 = n n +1

Let us apply Leibnitz’s test for checking convergence. Now an

1 1 1 − = >0 n n + 1 n(n + 1)

an +1 =

This proves that { n } is a monotonic decreasing sequence. 1 = 0. n

Again, lim an = lim n →∞

n →∞

∞

Therefore,

∑(

∞

= ∑ ( 1)n −1

1

1)

n =1

n =1

1 is a convergent series. n

∞

Now we consider

∑ an

which is

n=1

1+

1 1 1 1 + + + + 2 3 4 5

∞ 1 =∑ . n =1 n

This is a p-series with p = 1, and so the series is divergent. ∞

Since, ∑ ( 1) n −1 an is convergent and n =1

∞

∑( n =1

1)

1

∞

= ∑ ( 1) n −1 n =1

∞

∑ an

is divergent,

n=1

1 is conditionally convergent. n

8.20

Engineering Mathematics-I

WORKED-OUT EXAMPLES Example 8.1

∑{

n 4 +1 1

Sol.

Let

∞

Examine the convergence of the infinite series

}

n4 1

n =1

∑ an ∑ { ∞

∞

n 1

n =1

[WBUT 2008]

}

n4 + 1 − n4 − 1

This is an infinite series of positive terms. Here

{ n 1 − n − 1} { n +11 n 1}{ n +11 = { n +11 n 1}

an =

=

4

4

4

4

4

4

4

{ n +11

2

4

}

n4 1

}

n4 1

Let us consider the series

∞

∞

n 1

n =1

1

∑ bn ∑ n2 ,

which is convergent since it is a

p-series for p = 2. Now, an = lim n →∞ bn n →∞ lim

2n2

{ n +1 + 4

}

n4 − 1 .

= 1, a nonzero finite value. Since

∞

∞

n 1

n =1

1

∑ bn ∑ n2

is a convergent series, by comparison test, we can

conclude that

∑ an ∑ { ∞

∞

n 1

n =1

}

n4 + 1 − n4 − 1

is also convergent.

8.21

Infinite Series

Example 8.2

Examine the convergence of the infinite series

x 2 x3 x 4 + + + 2 3 4

x+ Sol.

Let us consider ∞

∞

xn n 1 n n=

∑ an ∑ n =1

where an =

xn x n +1 and an +1 = . n n +1

Now, an +1 x n +1 n = lim n = x. n →∞ an n →∞ x ( n + 1) lim

Then by D’Alembert’s ratio test, we have i) If x < 1, the infinite series is convergent ii) If x > 1, the infinite series is divergent iii) If x = 1, the test fails For x = 1, the series becomes 1+

1 1 1 + + + 2 3 4 ∞

Since

1

∑n

∞ 1 =∑ . n =1 n ∞

is a divergent series, the infinite series

n=1

xn diverges for x ≥ 1 n=1 n

∑

and converges for x < 1. Example 8.3 1+

Examine the convergence of the infinite series

x x 2 x3 + + + 2 5 10

Sol.

[WBUT-2009]

We know that addition or removal of a finite number of terms does not alter the convergence of an infinite series. So, removing the first term, we have the series of the form x x 2 x3 + + + 2 5 10

8.22

Engineering Mathematics-I

Suppose we write the series in the form ∞

∞

n =1

n =1

xn

∑ an ∑ n2 + 1 Then an =

xn n2 + 1

and so an +1 =

x n +1 (n 1)2 1

.

Here, an +1 x n +1 = lim n →∞ an n →∞ ( n + 1) 2 + 1

xn n2 + 1

lim

n2 + 1

= x ⋅ lim

1)2 1

n →∞ (

=x

So by D’Alembert’s ratio test, we can conclude that i) If x < 1, the infinite series is convergent ii) If x > 1, the infinite series is divergent iii) If x = 1, the test fails For x = 1, the series becomes ∞

1 1 1 + + + 2 5 10 Here, bn =

n =1 n

1 2

n +1

∞

1

=∑

2

+1

= ∑ bn n =1

.

Consider the series

∞

∞

n 1

n =1

1

∑ cn ∑ n2 ,

which is convergent. Now

bn n2 = lim 2 = 1, a nonzero finite number. n →∞ cn n →∞ n + 1 lim

∞

Therefore, by comparison test,

1

∑ n2 + 1

is a convergent series.

n =1

∞

Hence, ∑ an n =1

∞

xn

∑ n2 + 1 n =1

is convergent for x ≤ 1 and divergent for x > 1.

8.23

Infinite Series

Example 8.4

Examine the convergence of the infinite series

∞

n! 2 n

n =1

nn

∑

. [WBUT 2003]

Let us consider

Sol.

∞

∞

∑ an

∑

n =1

n! ⋅ 2 n nn

nn=1

Then an =

n!2 n n

n

and so an +1 =

(n 1)!2 n +1 (n 1) n +1

Here a (n + 1)!2 n +1 lim n +1 = lim n →∞ ( n + 1) n +1 an = lim

n →∞

2 ⎛ 1⎞ ⎜1 + n ⎟ ⎝ ⎠

n

=

n!2 n nn 2 1. Examine the convergence of the series

Example 8.8 1+ Sol.

x x 2 x3 + + + 2 5 10

∞

[WBUT-2004]

Since addition or removal of a finite number of terms does not affect the convergence of an infnite series, by removing the first term of the given series, Let the resulting infinite series be ∞

∞

∑ an

xn

∑ n2 + 1

n =1

n =1

Then xn

an =

n2 + 1

and so an +1 =

x n +1 (n 1)2 1

Now,

lim

n →∞

an +1 an

= lim

n →∞

x n +1 n2 + 1 (n + 1)2 + 1 lim = lim = x n →∞ n →∞ ( n + 1)2 + 1 xn n2 + 1 n2 + 1

n2 + 2n 2

1+ x = lim

n →∞

1 n2

2 2 1+ + 2 n n

x=x

8.27

Infinite Series

Therefore, by D’Alembert’s ratio test, i) If x < 1, the infinite series is convergent ii) If x > 1, the infinite series is divergent iii) If x = 1, the test fails For x = 1, the series becomes ∞

∞

n 1

n =1

1

∑ an ∑ n2 + 1 Consider the series ∞

∞

n 1

n =1

1

∑ bn ∑ n2 Then a n2 lim n = lim 2 = 1 (a nonzero finite number). n →∞ bn n →∞ n + 1 Since ∞

∞

n 1

n =1

1

∑ bn ∑ n2 is a convergent series, by comparison test ∞

∞

n 1

n =1

1

∑ an ∑ n2 + 1 is also a convergent series. Therefore, ∞

∞

n =1

n =1

xn

∑ an ∑ n2 + 1 is convergent for x ≤ 1 and divergent for x > 1. Test the convergence of the series

Example 8.9 1 a Sol.

3 12

2

+ b

a

3 22

3

+ b

a

Let the infinite series be ∞

n

∑

3 n =1 a ⋅ n2

+b

where an =

n 3 a ⋅ n2

+b

3 32

+ b

∞

(a > 0)

[WBUT-2004]

8.28

Engineering Mathematics-I

Consider the divergent series ∞

∞

n 1

n= n 1

1

∑ bn ∑ n . Now, n a lim n = lim n →∞ bn n →∞

3 a ⋅ n2

1 n

+ b = lim

n →∞

1

= lim

n →∞

3

=

b

a+

n2 3

a ⋅ n2 + b

1 (a nonzer n o finite number) a

3 2 n

Therefore, by comparison test ∞

n

∑

3 n =1 a ⋅ n2

+b 1

i.e, a

2

+

3 12

b

a

3

+

3 22

b

a

3 32

+

∞

b

is a divergent series. Test the convergence of the series

Example 8.10 1+ Sol.

22 32

x+

22 42

x2 +

32 52

22 42 ⋅ 62 32 52 ⋅ 72

x3 +

∞,

1

[WBUT-2004, 2009]

Since addition or removal of finite number of terms does not effect the convergence of an infinite series, by removing the first term of the given series, Let the resulting infinite series be ∞

∞

22 ⋅ 42 ⋅ 62

∑ an ∑ 32 ⋅ 52 ⋅ 72 n =1

n =1

(2 n)2 (2 n + 1)

2

xn .

Then an =

22 42 ⋅ 62 2

2

3 5 ⋅7

2

(2 n)2 (2n 1) 2

and so an +1 =

2 ⋅ 42 ⋅ 62 32 ⋅ 52 ⋅ 72

2

xn (2 n)2 (2 n + 2)2 (2n + 1)2 (2n + 3)3

x n +1

8.29

Infinite Series

Now,

lim

n →∞

an +1 an

22 ⋅ 42 ⋅ 62 (2 n)2 (2 n + 2)2 n +1 x 32 ⋅ 52 ⋅ 72 (2n + 1)2 (2n n + 3)3 = lim n →∞ 22 ⋅ 42 ⋅ 62 (2 n)2 n x 2 3 ⋅ 52 ⋅ 72 (2 n + 1)2 = lim

(2 n + 2)2

n →∞ (2 n + 3)2

x=x

Therefore, by D’Alembert’s ratio test, i) If x < 1, the infinite series is convergent ii) If x > 1, the infinite series is divergent Test the convergence of the series

Example 8.11 ⎛ 1 sin ⎜⎜ 3 ⎜ 2 ⎝1 Sol.

⎞ ⎛ ⎟ + sin ⎜ 1 ⎟ ⎜ 3 ⎟ ⎜ 2 ⎠ ⎝2

⎛ ⎞ ⎟ + sin ⎜ 1 ⎜ 3 ⎟ ⎟ ⎜ 2 ⎠ ⎝3

⎞ ⎛ ⎟ + sin ⎜ 1 ⎟ ⎜ 3 ⎜ 2 ⎟ ⎝4 ⎠

⎞ ⎟+ ⎟ ⎟ ⎠

∞

Let the infinite series be of the form ∞

∑ an n 1

⎛ 1 sin ∑ ⎜⎜ 3 n =1 ⎜ 2 ⎝n ∞

⎞ ⎟. ⎟ ⎟ ⎠

Then ⎛ 1 an = sin ⎜⎜ 3 ⎜ 2 ⎝n

⎞ ⎟ ⎟ ⎟ ⎠

Consider the p-series for p = ∞

∞

n 1

n =1

∑ bn ∑ where, 1 bn = 3

1

3 as 2

3

n2

n2 Now, ⎛ ⎞ 1 sin ⎜⎜ 3 ⎟⎟ ⎜ 2⎟ a ⎝ n ⎠ = 1, a nonzero lim n = lim o finite value. n →∞ bn n →∞ ⎛ ⎞ ⎜ 1 ⎟ ⎜ 3⎟ ⎜ 2⎟ ⎝n ⎠

[WBUT-2005]

8.30

Engineering Mathematics-I

Since ∞

∞

n 1

n =1

1

∑ bn ∑

3 2 n

is a convergent series, by comparison test, ⎛ 1 sin ∑ ⎜⎜ 3 n 1 n =1 ⎜ 2 ⎝n is also convergent. ∞

∞

∑ an

⎞ ⎟ ⎟ ⎟ ⎠

∞

Test the convergence of the series

Example 8.12

∑ n4 e− n

2

[WBUT-2005]

n =1

Sol.

Let the infinite series be of the form ∞

∞

n =1

n =1

∑ an ∑ n4 e−n

2

.

Then an

2

n 4 e− n and so an +1 = (n 1) 4 e

( n +1)2

Now, 2

a (n + 1) 4 e −( n +1) 1 ⎛ n +1 ⎞ lim n +1 = lim = lim ⎜ ⎟ ( n 1)2 n →∞ an n →∞ n →∞ ⎝ n ⎠ 4 − n2 e n e 4

n2

4

⎛ 1⎞ 1 1 + ⎟ 2n n →∞ ⎝ n⎠ e

= li

1

= 0 0

n =1

[WBUT-2006] Sol.

Let the infinite series be of the form ∞

∑ an n =1

∞

n2 − 1

∑ n2 + 1 x n . n =1

Then an =

n2 − 1 n2 + 1

x n and an +1 =

(n 1)2 1 (n 1)2 1

x n +1

8.31

Infinite Series

Now,

lim

n →∞

an +1 an

(n + 1)2 − 1 n +1 x (n2 + 2 )( 2 1) (n + 1)2 + 1 lim x = lim = n →∞ n →∞ ( 2 2 n + 2)( 2 1) n2 − 1 n x n2 + 1 1 ⎞ ⎛ 2 ⎞⎛ ⎜1 + n ⎟ ⎜1 + 2 ⎟ ⎝ ⎠⎝ n ⎠ = lim n →∞ ⎛ 2 2 ⎞⎛ 1 ⎜1 + n + 2 ⎟ ⎜1 − 2 n ⎠⎝ n ⎝

⎞ ⎟ ⎠

x=x

Therefore, by D’Alembert’s ratio test, the series is (i) convergent if x < 1 (ii) divergent if x > 1 and (iii) the test fails for x = 1 For x = 1, the series becomes ∞

∞

n 1

n =1

n2 − 1

∑ an ∑ n2 + 1 Now, lim an = lim

n →∞

n →∞

n2 − 1 n2 + 1

=1≠ 0

So the series is divergent for x = 1. Hence, the series is convergent for x < 1 and divergent for x ≥ 1. For what values of x is the following series convergent?

Example 8.14

x x2 x3 + + + 1 3 3 5 5 7 Sol.

∞

Let the infinite series be of the form ∞

∞

n =1

nn=1

xn

∑ an ∑ (2n − 1)(2n + 1) . Then an =

xn x n +1 and so an +1 = (2 n 1)(2 n + 1) (2 n +1)(2 1)(2 n + 3)

[WBUT-2006]

8.32

Engineering Mathematics-I

Now, an +1 an

lim

n →∞

x n +1 (2 n + 1)(2 n + 3) = lim n →∞ xn (2 n − 1)(2 n + 1) = lim

(2 n − 1)

n →∞ (2 n + 3)

x=x

Therefore, by D’Alembert’s ratio test, the series is (i) (ii) (iii) For

convergent if x < 1 divergent if x > 1 and the test fails for x = 1 x = 1, the series becomes

∞

∞

∑ an

1

∑ (2n − 1)(2n + 1)

n 1

nn=1

Consider another series ∞

∞

n 1

n =1

1

∑ bn ∑ n2 where, bn =

1 n2

Now, 1 an n2 (2n 1)(2 n + 1) lim = lim = lim 1 n →∞ bn n →∞ n →∞ (2 n − 1)(2 n + 1) 2 n = lim

1 1 = (a nonzero finite value) 1 ⎞⎛ 1⎞ 4 ⎜2 − n ⎟⎜2 + n ⎟ ⎝ ⎠⎝ ⎠

n →∞ ⎛

Since ∞

∞

n 1

n =1

1

∑ bn ∑ n2 being a p-series for p = 2, is a convergent series,

8.33

Infinite Series

By comparison test ∞

∞

n 1

nn=1

1

∑ an ∑ (2n − 1)(2n + 1) is also convergent. Therefore, the given series is convergent for x ≤ 1. ∞

Sol.

n

⎛ nx ⎞ ∑ ⎜⎝ n + 1 ⎟⎠ n =0 Since addition or removal of a finite number of terms does not affect the convergence of an infnite series, by removing the first term of the given series, Examine the convergence of the series

Example 8.15

Let the resulting infinite series be ∞

∞

n =1

nn=1

⎛ nx ⎞

∑ an ∑ ⎜⎝ n + 1 ⎟⎠

n

Then ⎛ nx ⎞ an = ⎜ ⎟ ⎝ n +1 ⎠ Now,

n

1

lim (an

n →∞

⎧⎪⎛ nx ⎞n ⎫⎪ n ⎛ nx ⎞ = lim ⎨⎜ ⎬ = lim ⎜ ⎟ ⎟ n →∞ ⎝ n + 1 ⎠ n →∞ ⎝ n + 1 ⎠ ⎩⎪ ⎭⎪

1 )n

⎛ ⎞ ⎜ 1 ⎟ = lim ⎜ ⎟x = x n →∞ ⎜⎜ 1 + 1 ⎟⎟ n⎠ ⎝ Therefore, by Cauchy’s root test, the series is (i) (ii) (iii) For

convergent for x < 1 divergent for x > 1 test fails for x = 1 x = 1, the series becomes

∞

∑ an n =1

∞

⎛ n ⎞ ∑ ⎜⎝ n + 1 ⎟⎠ n 1 n=

n

Now, n

1 1 ⎛ n ⎞ lim an = lim ⎜ = lim = ≠0 ⎟ n n →∞ n →∞ ⎝ n + 1 ⎠ n →∞ e ⎛ 1⎞ ⎜1 + n ⎟ ⎝ ⎠

8.34

Engineering Mathematics-I

Therefore, the series is divergent in this case. Hence the given series is convergent for x < 1 and divergent for x ≥ 1. Examine the convergence of the series

Example 8.16 Sol.

∞

(1 + nx) n

n =1

nn

∑

Let the infinite series be of the form ∞

∞

n =1

nn=1

∑ an ∑ Then an =

(1 + nx) n nn

.

(1 + nx) n nn

Now, lim (an

n →∞

1 )n

1

⎧ (1 + nx) n ⎫ n ⎛ 1 + nx ⎞ = lim ⎨ ⎬ = lim ⎜ ⎟ n n →∞ n →∞ ⎝ n ⎠ ⎭ ⎩ n

⎛1 ⎞ = lim ⎜ + x = x n →∞ ⎝ n ⎠ Therefore, by Cauchy’s root test, the series is (i) convergent for x < 1 (ii) divergent for x > 1 (iii) test fails for x = 1 For x = 1, the series becomes ∞

∞

⎛1+ n ⎞ n ⎟⎠ n 1 n=

∑ an ∑ ⎜⎝ n =1

n

Now, n

n

⎛1+ n ⎞ ⎛ 1⎞ lim an = lim ⎜ li ⎜ 1 + ⎟ = e ≠ 0 ⎟ = nlim n →∞ n →∞ ⎝ n ⎠ →∞ ⎝ n⎠ Therefore, the series is divergent in this case. Hence, the series is convergent for x < 1 and divergent for x ≥ 1. Example 8.17

Examine the convergence of the series 2

3

1 2 ⎛3⎞ ⎛4⎞ + x + ⎜ ⎟ x 2 + ⎜ ⎟ x3 + 2 3 ⎝4⎠ ⎝5⎠ Sol.

∞

Since addition or removal of a finite number of terms does not affect the convergence of an infnite series, by removing the first term of the given series,

8.35

Infinite Series

Let the resulting series be ∞

n

∞

⎛ n +1 ⎞ ∑ ⎜⎝ n + 2 ⎟⎠ x n . n 1 n=

∑ an n =1

Then n

⎛ n +1 ⎞ n an = ⎜ ⎟ x ⎝n+2⎠ Now, 1

⎧⎪⎛ n + 1 ⎞n n ⎫⎪ n ⎧⎛ n + 1 ⎞ ⎫ lim (an = lim ⎨⎜ ⎨⎜ ⎟ x ⎬ = nlim ⎟ x⎬ = x n →∞ n →∞ ⎝ n + 2 ⎠ →∞ ⎩⎝ n + 2 ⎠ ⎭ ⎪⎩ ⎪⎭ Therefore, by Cauchy’s root test, the series is 1 )n

(i) convergent for x < 1 (ii) divergent for x > 1 (iii) test fails for x = 1 For x = 1, the series becomes ∞

∞

n =1

nn=1

⎛ n +1 ⎞

∑ an ∑ ⎜⎝ n + 2 ⎟⎠

n

Now, n

1 ⎛ n +1 ⎞ lim an = lim ⎜ = lim ⎟ n n →∞ n →∞ ⎝ n + 2 ⎠ n →∞ 1 ⎞ ⎛ + 1 ⎜ n +1 ⎟ ⎝ ⎠

= lim

n →∞

1 ⎞ ⎛ ⎜1 + n + 1 ⎟ ⎝ ⎠ 1 ⎞ ⎛ ⎜1 + n + 1 ⎟ ⎝ ⎠

n +1

=

1 ≠0 e

Therefore, the series is divergent in this case. Hence, the series is convergent for x < 1 and divergent for x ≥ 1. Example 8.18 Sol.

Examine the convergence of the series

∞

1 + n log n

n =2

n2 + 5

∑

Since the addition or removal of finite number of terms does not effect the 1 1 ⋅ log 1 convergence of an infnite series, by adding 2 as the first term to the 1 5 given series,

8.36

Engineering Mathematics-I

Let the resulting series be ∞

∞

1 + n log n

n 1

n =1

n2 + 5

∑ an ∑

.

Then an =

1 + n log n

n2 + 5 Consider the well-known divergent series ∞

∞

∑ bn

1

∑n

n 1

n= n 1

where 1 bn = n Now, 1 + n log n 2 n2 + 5 = lim n + n log n 1 n →∞ n2 + 5 n 1 + log n = lim n =∞ 5 n →∞ 1+ 2 n Therefore, by comparison test, the series a lim n = lim n →∞ bn n →∞

∞

1 + n log n

n =1

n2 + 5

∑

is divergent. Examine the convergence of the series

Example 8.19

1 32 1 3 ⋅ 52 1 3 ⋅ 5 72 + + + 2 42 2 4 ⋅ 62 2 4 ⋅ 6 82 Sol.

Since addition or removal of a finite number of terms does not affect the convergence of an infinite series, by removing the first term of the given series, let the resulting infinite series be ∞

∑ an n 1

Then,

∞

an =

∞

∑

1⋅ 3 ⋅ 5

n =1

1 3⋅5

(2 n 1)(2 n + 1)2

2 ⋅ 4 ⋅ 6 … 2 n(2 n + 2)2 (2 n 1)(2 n + 1)2

2 4 ⋅ 6 … 2 n(2 n + 2)2

8.37

Infinite Series

and so an+1 =

1⋅ 3 ⋅ 5

(2 n + 1)(2 n + 3)2

2⋅4⋅6

(2 n + 2)(2 n + 4)2

Now, an +1 (2 n + 3)2 (2 n + 2) = lim n →∞ an n →∞ (2 n + 4)2 (2 n + 1) lim

2

= lim

n →∞

3⎞ ⎛ 2⎞ ⎛ ⎜2 + n ⎟ ⎜2 + n ⎟ ⎝ ⎠ ⎝ ⎠ 2

4⎞ ⎛ 1⎞ ⎛ ⎜2 + n ⎟ ⎜2 + n ⎟ ⎝ ⎠ ⎝ ⎠

=1

So, it is obvious from the above that D’Alembert’s ratio test fails. Therefore, we apply Raabe’s test. Now, ⎧ (2 n + 4)2 (2 n + 1) ⎫ ⎧ a ⎫ lim n ⎨ n − 1⎬ = lim n ⎨ − 1⎬ 2 n →∞ ⎩ an +1 ⎭ n →∞ ⎩ (2 n + 3) (2 n + 2) ⎭ ⎧ 4 3 6n 6n2 − 2n ⎫ = lim ⎨ ⎬ n →∞ (2 n + 3)2 (2 n + 2) ⎭ ⎩ 4+ = lim

n →∞

6 2 − n n2

2 ⎞⎛ 3⎞ ⎛ ⎜2 + n ⎟⎜2 + n ⎟ ⎝ ⎠⎝ ⎠

2

=

4 1 = 0

8.38

Engineering Mathematics-I

and an +1 =

22⋅ 42 … (2 n)2 (2 n + 2)2 n 4 x 2 n+ 3 ⋅ 4 ⋅ 5 ⋅ 6 … (2 n + 2)(2 n + 3)(2 n + 4)

Now, an +1 (2 n + 2)2 = lim x2 n →∞ an n →∞ (2 n + 3)(2 n + 4)

x2

lim

By D’Alembert’s ratio test, the series is (i) convergent if x 2 < 1, i.e., 0 < x < 1 (ii) divergent if x 2 > 1 i.e., x > 1 (iii) The test fails for x = 1. For x = 1, the series becomes ∞

∞

n 1

n =1

22⋅ 42 … (2 n)2

∑ an ∑ 3 ⋅ 4 ⋅ 5 ⋅ 6… (2n + 2) Then an =

22 42 … (2 n)2 3 4 ⋅ 5 6 … (2 n + 2)

and an+1 =

22⋅ 42 … (2 n)2 (2 n + 2)2 3 ⋅ 4 ⋅ 5 ⋅ 6 … (2 n + 2)(2 n + 3)(2 n + 4)

Now, ⎛ (2 n + 3)(2 n + 4) ⎞ ⎛ a ⎞ lim n ⎜ n − 1 ⎟ = lim n ⎜ − 1⎟ 2 n →∞ ⎝ an +1 ⎠ n →∞ ⎝ (2 n + 2) ⎠ = lim

n(6 n + 8)

n →∞ (2 n + 2)

2

=

6 3 = >1 4 2

Therefore, by Raabe’s test, the series is convergent for x = 1. ∞

Test the convergence of the series

Example 8.21

∑

n=1

Sol.

sin(2 (

1)

π 2

1)

Since

π = (−1) n −1 2 the given series can be represented as sin(2

1)

∞

∑( n =1

1)

1

∞

= ∑ ( 1) n −1 n =1

1 (

1)

So, this is an alternating series and we apply Leibnitz’s test for testing its convergence.

8.39

Infinite Series

Here 1

an =

(n 1)

1

and so an +1 =

( n 2)

Now, an

1 1 1 − = > 0 for all n n + 1 n + 2 (n + 1)(n + 2)

an +1 =

Since, an Also,

N

< an , so { n } is monotonically decreasing. 1

lim an = lim

n →∞ n + 1

n →∞

=0

Therefore, by Leibnitz’s test the alternating series is convergent. Test the convergence of the series

Example 8.22 1 1 23 Sol.

−

12 + 22 2 33

12 + 22 + 32

+

3 43

−

12 + 22

32

42

4 53

+

∞

The given series can be represented as ∞

∑ ( 1)

∞

= ∑ ( 1) n −1

1

n =1

(12 + 22

32 n((

n =1

42 +

n2 )

1)3

Then an =

n(n 1)(2 n + 1) 6 n(n + 1)

3

=

(2n + 1) 6(n + 1)2

So, this is an alternating series and we apply Leibnitz’s test for testing its convergence. Now, an +1 (2 n + 3)(n + 1)2 = an (2 n + 1)(n + 2)2 =

2

= 1−

3

2

n + 12 n + 4 9n 9 2n2

2

3

4n + 1 2

9n + 12 n + 4

1. ln 3 ∞

Therefore, the series (2) is divergent and consequently, the series is also divergent.

∑( n=1

1 3) n

Infinite Series

Example 8.24

8.41

Show that the following series is divergent:

1 2 + 2 3 + 3 4 + 4 5 +�∞ Sol.

The given series is ∞

∞

n 1

nn=1

∑ an ∑ n(n + 1) Now, lim an = lim n(n + 1)

n →∞

n →∞

∞

Therefore, the given series is divergent. Example 8.25

Test the convergence of the following series:

6 8 10 + + + 1 3⋅5 3 5⋅ 7 5 7 9 Sol.

∞

Let the given series be ∞

∞

∑ an

2n + 4

∑ (2n − 1)(2n + 1)(2n + 3) .

n 1

nn=1

Then an =

2n 4 (2 n 1)(2 n + 1)(2 n + 3)

Consider the convergent series ∞

∞

n 1

n =1

1

∑ bn ∑ n2 where bn =

1 n2

Now, 2n + 4 an (2 n 1)(2 n + 1)(2 n + 3) lim = lim 1 n →∞ bn n →∞ n2 (2 n + 4)n2 n →∞ (2 n − 1)(2n 1)(2 n + 1)(2 n + 3)

= lim

[WBUT-2008]

8.42

Engineering Mathematics-I

4⎞ ⎛ ⎜2 + n ⎟ ⎝ ⎠ 1 = lim = (a nonzeroo finite value) n →∞ ⎛ 1 ⎞⎛ 1 ⎞⎛ 3⎞ 4 ⎜2 − n ⎟⎜2 + n ⎟⎜2 + n ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ Therefore, by comparison test ∞

∞

n 1

nn=1

2n + 4

∑ an ∑ (2n − 1)(2n + 1)(2n + 3) is convergent. Example 8.26 1+

Test the convergence of the following series:

2 p 3p 4 p + + + �∞ 2! 3! 4!

[WBUT-2008]

Let the given series be

Sol.

∞

∞

np ∑ n! . n 1 n=

∑ an n =1

Then an =

np (n 1) p and so an +1 = n! (n 1)!

Now, an +1 n →∞ an lim

(n + 1) p (n + 1) p n! (n + 1)! lim = lim = n →∞ n p ( n + 1)! n →∞ np n! p

⎛ 1⎞ 1 1+ ⎟ = 0 1)

and so convergent.

Here bn =

1 n2

Therefore, for all n an

bn ⇒

an bn

N,

< 1. ∞

So, by comparison test,

∑ an

is convergent.

n=1 ∞

Hence, the given series

∑ an n 1

Example 8.29 x−

∞

∑

n =1

cos nx n2

is absolutely convergent.

Prove that the infinite series

n x 2 x3 x 4 n 1 x + − + � (−1) n+ +� 2 3 4 n

is absolutely convergent when x < 1 and conditionally convergent when x = 1. [WBUT 2001, 2007] Sol.

Let the given series be ∞

∞

n =1

n =1

∑ an ∑ (−1)n+1

xn n

…(1)

where an = (−1) n +1

xn . n

Now we consider the series ∞

∑ an

…(2)

n=1

where n

an = (−1) n +1

x xn = . n n

So, n +1

an +1 = (−1) n + 2

x x n +1 = . n +1 n +1

8.45

Infinite Series

Therefore an +1

lim

n →∞

n

= lim

x

n →∞ ( n + 1)

an

1

= lim

n →∞ ⎛

1⎞ ⎜1 + n ⎟ ⎝ ⎠

x = x.

Then by D’Alembert’s ratio test, we have i) If x < 1, the infinite series (2) is convergent ii) If x > 1, the infinite series (2) is divergent Hence, the given series (1) is absolutely convergent for x < 1. For x = 1, the series becomes 1 1 1 1 − + − +� 2 3 4 which is conditionally convergent. (For proof see the example 30 of Article 8.15). Example 8.30 1−

1 2

Sol.

p

+

1 3

p

−

Test the convergence of the series:

1

+�∞

4p

[WBUT-2003]

The given series can be written as ∞

∑ ( 1)

∞

= ∑ ( 1) n −1

1

n =1

n =1

1 np

.

So this is an alternating series and an =

1 n

p

and an +1 =

1

( n + 1) p

.

Then 1 an +1 ( n + ) np = = < 1 for p > 0 and n ∈ N . p 1 an n+ ) ( np p

Since, an

< an for p > 0, so { n } is monotonically decreasing for p > 0.

8.46

Engineering Mathematics-I

Also, for p > 0 lim an = lim

n →∞

n →∞

1 np

=0

Hence by Leibnitz’s test, the alternating series is convergent for p > 0. But for p < 0, lim an = lim

n →∞

n →∞

1 np

= ∞.

Therefore, the series can’t be convergent for p < 0.

EXERCISES

Short and Long Answer Type Questions (A) Test the convergence of the following series: 1)

1 1 1 + + + 1 3 3 5 5 7

∞ [Ans : Convergent]

2)

12 22 22 32 32 42 + + + 1! 2! 3!

∞ [Ans : Convergent]

∞

3)

n +1 1

∑

n 1 n

n =1

[Ans : Convergent] 4)

1 1 2

−11

+

1 1 2

2

+

1 1 2

−33

+

1 1 2

4

+

∞ [Ans : Divergent]

5)

1 3 7 15 + 2 + 3 + 4 + �∞ 2 2 2 2 [Ans : Divergent] ∞

6)

2

n

∑ n3 n =1

[Ans : Convergent] ∞

7)

⎛

∑ ⎜⎝1 + n =1

1 ⎞ ⎟ n⎠

3 −n2

[WBUT-2004] [Ans : Convergent]

8.47

Infinite Series ∞

8)

∑

n =1

n!2 n nn [Ans : Convergent] 2

2

2

⎛ 1 ⎞ ⎛ 1 2 ⎞ ⎛ 1 2⋅3 ⎞ 9) ⎜ ⎟ + ⎜ ⎟ +⎜ ⎟ + ⎝ 3 ⎠ ⎝ 3 5 ⎠ ⎝ 3 5⋅ 7 ⎠

∞

[WBUT-2002, 2007] [Ans : Convergent]

∞

10)

n3 − n + 1 n! n =1

∑

[Ans : Convergent] 11)

∞

n +1

n =1

n6

∑

[Ans : Convergent] ∞

12)

∑

n4

n =1 e n

2

[Ans : Convergent] (B) Examine the convergence of the following series for different values of x: ∞

13)

∑

n =1

n 2

n +1

xn [Ans : Convergent if −1 ≤ x < 1, divergent if x ≥ 1 or x < 1]

∞

14)

1

∑ xn + x

n

n =1

, >0 [Ans : Convergent if x > 1 or 0 < x < 1]

∞

15)

⎛ nx ⎞

∑ ⎜⎝ n + 1 ⎟⎠

n

n =1 ∞

16)

∑

n =1

[Ans : Convergent if x < 1, divergent if x ≥ 1] n n x n +1 [Ans : Convergent if x < 1, divergent if x ≥ 1]

17) Prove that ∞

∑

n =1

( 1) n −1 cos 2nx 3

n2

is absolutely convergent.

8.48

Engineering Mathematics-I

18) Prove that ( 1) n −1

∞

∑ n((

g )2

n =1

is absolutely convergent. (C) Test the convergence of the following series: ∞

1)

n!

∑ nn n =1

[Ans : convergent] ∞

2)

1

∑

n =2

( log n )n [Ans : convergent]

∞

3)

1

∑ sin n n=1

[Ans : divergent] 4)

12

2 4

1

x+

22

2

2

4

x2 +

32 3

2 4

x3 +

∞

[WBUT-2002]

[Ans : Convergent if x ≤ 1, divergent if x > 1] ∞

5)

⎛

1⎞

∑ ⎜⎝1 + n ⎟⎠

n

2

n =1

[Ans : divergent] 2

6) 1 +

3

x x x + + +�∞ 2 5 10

[WBUT-2004] [Ans : Convergent if x ≤ 1, divergent if x > 1]

∞

7)

n +1

∑

n

n =1

n

p

1 1⎤ ⎡ ⎢ Ans : Convergent if p > 2 , divergent if p ≤ 2 ⎥ ⎣ ⎦ 8) 1 +

22 2

3

x+

22 42 2

2

3 5

x2 +

22 42 ⋅ 62 2

2

3 5 ⋅7

2

x3 +

∞,

1

[WBUT-2004, 2009]

[Ans : Convergent if x < 1, divergent if x > 1] 9)

2p q

1

+

3p 2

q

+

4p 3

q

+

5p 4q

+ � ∞ ( p, q > 0 ) [Ans : Convergent if q > p 1, divergent if q ≤ p + 1]

Infinite Series

10)

x x2 x3 + + + 1 3 3 5 5 7

∞

8.49

[WBUT-2006] [Ans : Convergent if x ≤ 1, divergent if x > 1]

11)

5 7 9 11 − + − + 2 4 6 8

∞ [Ans : Not convergent]

∞

12)

∑

sin(2

1)

(

n=1

π 2

1) [Ans : Convergent]

13) 1 2 + 2 3 + 3 4 + 4 5 + � ∞ [Ans : Divergent] 14) x 2 +

22 4 22 42 6 22 42 ⋅ 62 x + x + x8 + ∞, x > 0 3 4 3 4⋅5 6 3 4⋅5 6⋅7 8 [Ans : Convergent if 0 < x ≤ 1, divergent if x > 1]

+ ⎧⎪⎛ n + 1 ⎞n +1 n + 1 ⎫⎪ − 15) ∑ ⎨⎜ ⎬ ⎟ n ⎪⎭ n =1 ⎪ ⎩⎝ n ⎠ ∞

∞

16)

1

∑

n

n =1

sin

−n

[Ans : Convergent]

1 n

[WBUT-2001] [Ans : convergent] 2

17)

3

1 2 ⎛3⎞ ⎛4⎞ + x + ⎜ ⎟ x 2 + ⎜ ⎟ x3 + 2 3 ⎝4⎠ ⎝5⎠

∞ [Ans : Convergent if x < 1, divergent if x ≥ 1]

(a + x) a + x (a + x) + + + 1! 2! 3! 2

18)

3

∞, x > 0

⎡ ⎢ Ans : Convergent if ⎣ 19) 1 +

1 1 1 + + + 2 12 + 1 2 22 + 1 2 32 + 1

1 1⎤ divergent if x ≥ ⎥ e e⎦

∞ [Ans : Convergent]

20)

23 1p

3p

+

33 2p

5p

+

43 3p

7p

+

∞

[Ans : Convergent if p > 4, divergent if p ≤ 4]

8.50

Engineering Mathematics-I

21) 1 +

1 2! 3! 4! x + 2 x 2 + 3 x3 + 4 x 4 + 2 3 4 5

∞, > 0

[Ans : Convergent if 0 < x < , divergent if x ≥ e ] (D) Examine the convergence of the following alternating series: 22)

1 1 1 1 − + − + �∞ log 2 log 3 log 4 log 5 [Ans : Convergent]

23) 1 −

1 2

2

+

1

−

2

3

1

+ �∞

42

[WBUT-2009] [Ans : Convergent]

24)

∞

( 1) n 1 2 n

n =1

n2

∑

[Ans : Divergent] 25)

∞

( 1) n −1 x n

n =1

n2 − n

∑

,0 < x 0 1 5

1

+

a) divergent 3. The series

1 5

2

+

1 5

3

c) p > 1 1

+

5

4

+ � ∞ is

b) convergent 1

3 5

1

a) convergent

+

1 3

2

5

+

1 3

5

+

3

b) oscillatory

d) p < 0

c) oscillatory 1 3

45

d) none of these

+ � ∞ is c) divergent

d) none of these

c) convergent

d) oscillatory infinitely

4. The sequence {(−1) n ⋅ 2 n } is a) monotone

b) bounded

⎧1 ⎫ 5. The sequence ⎨ sin n ⎬ is ⎩n ⎭ a) oscillatory c) convergent with limit 1

b) divergent to ∞ d) convergent with limit 0

⎧ 1 ⎫ 6. The sequence ⎨ ⎬ is ⎩ 4 3n ⎭ a) decreasing and unbounded c) decreasing and bounded

b) increasing and bounded d) none of these

1⎫ ⎧ 7. The sequence ⎨3 (−1) n ⎬ is n⎭ ⎩ a) oscillatory c) convergent

b) monotone d) bounded but not convergent

⎧ n3 + 1 ⎫ 8. The sequence ⎨ 2 ⎬ is ⎩ n ⎭ a) bounded c) convergent

b) divergent to ∞ d) none of these

8.52

Engineering Mathematics-I

9. Which of the following sequence is convergent? ⎧1 ⎫ a) ⎨ n + 2 n ⎬ ⎩4 ⎭

⎧ n4 + 1 ⎫ b) ⎨ 3 ⎬ ⎩ n ⎭

1⎫ ⎧1 c) ⎨ n + n ⎬ 4 ⎭ ⎩2

d) {1 + ( 1) n }

∞

10. The series

n +1 is n =1 n

∑

b) divergent to ∞ d) none of these

a) convergent c) oscillatory ∞

11. If

∑ an

is convergent then

n=1

a) { n } is monotone

b) { n } is convergent with limit 0 a d) lim n +1 < 1. n →∞ an

c) lim an = 1 n →∞

∞

12. The series

1

∑(

1) n

n=1

is

a) divergent to ∞ c) oscillatory ∞

13. The series

∑(

b) convergent d) none of these

n+

)

n is

n =1

a) convergent c) oscillatory

b) divergent d) none of these.

∞

14. The series

⎛ n +1 ⎞ is convergent if p ⎟ ⎠ n =1 n

∑ ⎜⎝

a) p > 2

b) p > 1

c) p ≤ 2

d) p > 0

15. Which of the following infinite series is convergent? ∞

a)

⎛ 1

∞

⎞

∑ ⎜⎝ n2 + 4 ⎟⎠

b)

n =1

c) 1 +

1 2

4

+

1 3

4

+

⎛8⎞

∑ ⎜⎝ 5 ⎟⎠

n

n =1

1 4

4

+ �∞

∞

d)

3

2

∑ 4 n3 n=1

1 1

8.53

Infinite Series

16. The infinite series 1 2

1

−

1+ x 2+ x is convergent

2

1

+

3+ x

2

−

1 4 + x2

+

a) only for −1 ≤ x ≤ 1 c) for no real values of x

∞

b) only for x = 0 d) for all real values of x

17. The infinite series 1−

1 2

+

1 3

−

1 4

+ �∞

is a) absolutely convergent c) conditionally convergent ∞

18. The series

⎛ n! ⎞

∑ ⎜⎝ nn ⎟⎠

b) oscillatory d) none of these

is

n =1

a) convergent b) divergent c) neither convergent nor divergent d) none of these 1

19. Let

∑ an be an infinite series of positive terms. If lim (an ) n = n →∞

a) convergent c) oscillate infinitely

4 then 3

∑ an

is

b) divergent d) none of these

20. If an > 0 for all n and a1 ≥ a2 ≥ a3 ≥ � and lim (an )an = 0 then the series n →∞

∞

∑(

1)

n −1

an

n =1

b) is divergent to −∞ d) none of these

a) oscillates infinitely c) is convergent Answers: 1. (c) 2. (a) 10. (b) 11. (b) 19. (b) 20. (c)

3. (a) 12. (b)

4. (d) 13. (b)

5. (d) 14. (a)

6. (c) 15. (c)

7. (c) 16. (d)

8. (b) 17. (c)

9. (c) 18. (a)

CHAPTER

9 Vector Analysis 9.1

INTRODUCTION

Vectors are very important part of any branch of science and technology. Vectors and their differentiation and integrations have wide range of applications for solving problems in many practical situations. Basically, we shall divide the chapter into three parts. In the first part of this chapter, we discuss Vector Algebra which includes various kinds of vectors, different terminologies, different kinds of products of vectors, equations of straight line, plane and sphere in vector form and of course their applications too. In the second part of the chapter, we deal with vector differentiations, gradient, divergence, curl, directional derivative along with their applications. In the third part of the chapter, we give theorems on vector integrations (Green’s theorem, Divergence theorem, Stoke’s theorem) and their applications to physical problems.

PART-I (VECTOR ALGEBRA) 9.2

SCALARS AND VECTORS

Any physical quantity which has magnitude only is known as a scalar. The examples of scalars are area, volume, mass, speed, etc. Any physical quantity which has magnitude as well as direction is known as a vector. The examples of vectors are displacement, velocity, force, etc. Generally, a vector is represented by a directed line segment. Any vector from point A to point B is denoted by AB. Let us consider any vector AB = a, where the length of the line segment AB is a (which is always positive). Then the magnitude or absolute value of the vector is denoted by AB , and is given by AB = a a.

9.2

Engineering Mathematics-I

9.2.1

Different Kinds of Vectors

Here we give a few definitions. (1) Like Vectors: Vectors having the same direction are known as like vectors. (2) Null Vector or Zero Vector: Any vector having the magnitude zero is known as a null vector or zero vector. It is generally denoted by 0 or θ . (3) Unit Vector: Any non-null vector of unit length (or, having the unit magnitude) is known as a unit vector. Let a be any non-null vector. Then the unit vector in the direction of a is given a a by aˆ = or . a a In the three-dimentional Cartesian coordinate system, the unit vectors along the x-axis, y-axis and z-axis are iˆ, ˆj and kˆ respectively. These are called fundamental unit vectors. (4) Equal Vectors: Two vectors a and b are called equal if they have the same magnitudes as well as the same direction. Then we write a = b . Two parallel vectors having the same magnitude are equal. (5) Negative of a Vector: Let AB = a be a vector. Then a vector having the same magnitude but opposite direction is known as the negative of AB and is given by BA = AB = −a. (6) Position Vector: The position of a point P with respect to any arbitrary point O is represented by a vector OP. Here, O is called the initial point or the vector origin and the vector OP is called the position vector w.r.t O.

9.2.2

Addition and Subtraction of Vectors

Additon of Vectors using Triangle Law

C

c b

A

a

Figure 9.1

B

Triangle Law

9.3

Vector Analysis

Let AB = a, BC = b and AC = c . Then by Triangle Law, (see Fig. 9.1) we have AB + BC = AC or, a b = c .

Additon of Vectors using Parallelogram Law a

D

c

b

a

A

Figure 9.2

C

b

B

Parallelogram Law

Let AB = a, AD = b and AC = c . Then by Parallelogram Law (see Fig. 9.2), we have AB + AD = AC or, a b = c Properties of Vector Addition (i) Addition of vectors is commutative, i.e., a b = b a. (ii) Addition of vectors is associative, i.e., (

b) + c =

(b ( b c ).

Subtraction of Vectors Let AB = a; BC = b ; then their subtraction is given by AB − BC = AB + (− BC ) i.e., a b = a ( b ).

9.2.3

Scalar Multiplication of a Vector

Multiplication of a vector a by any scalar m, +ve or −ve, is denoted by the vector ma. Now ma = m a , i.e., magnitude of the vector ma is m multiple of magnitude of the vector a.

9.4

Engineering Mathematics-I

Direction of the vector ma is same or opposite of the direction of the vector a, accordingly as m is positive or negative. In particular, if m = 0 then ma = 0 ⋅ a = 0. Properties )a=m ma a + na

(i) (

ma + mb

(ii) m( a b ) (iii) m( na )

9.2.4

mna = m( na )

Collinear Vectors

Any set of vectors having the same or different magnitudes is said to be collinear if all of them have the same directions. In particular, when two vectors a and b are collinear then we can write a λ b for any scalar λ . Here, we state a theorem on the collinearity of three points. Theorem 9.1: Any set of three distinct points A, B and C will be collinear (i.e., lie on the same line) iff there exists three scalars α , β , γ (not all zero) such that

α

β

γ c = 0 and α β γ = 0

where a, b and c are the position vectors of A, B and C respectively w.r.t a vector origin. Proof: Beyond the scope of the book.

9.2.5

Coplanar Vectors

Any set of vectors are called coplanar if all of them are parallel to the same plane. Here, we state a theorem on the coplanarity of four points. Theorem 9.2: Any set of four distinct points , B, C and D (no three of them are collinear) will be coplanar (i.e., lie on the same plane) iff there exists four scalars α , β , γ , δ (not all zero) such that

α

β

γ

δ d = 0 and α β γ

δ =0

where a, b, c and d are the position vectors of A, B, C and D respectively w.r.t a vector origin. Proof: Beyond the scope of the book.

9.5

Vector Analysis

9.2.6

Resolution of Vectors in Rectangular Cartesian coordinate System Z

P k

O i

Y j

M N X

Figure 9.3 Let the unit vectors along the three axes OX , OY and OZ be iˆ, ˆj, kˆ respectively and P( x, y, ) be any point. From Fig. 9.3, it clear that PN is perpendicular to the XY plane and MN is parallel to Y -axis. So, OM = x, MN = y, NP = z and correspondingly, OM = xiˆ, MN = yjˆ, NP = zkˆ. Now from ΔOMN M , ON = xiˆ + yjˆ. So from ΔONP N , we have OP = xiˆ + yjˆ + zkˆ. Hence we can say for any point P ( x, y, ), the position vector OP w.r.t some vector origin O is given by OP = xiˆ + yjˆ + zkˆ, where the vector components of OP along the directions of X-axis, Y-axis, and Z-axis are respectively xiˆ, yjˆ and zkˆ. Now OP = x 2 + y 2 + z 2 . So, the unit vector along the direction of OP is

OP OP

=

xiˆ + yjˆ + zkˆ x2 + y 2 + z 2

.

Observation: Let us consider the two points A and B whose coordinates are ( x1 , y1 , 1 ) and ( x2 , y2 , 2 ). So the position vectors of A and B are OA = x1 iˆ + y1 ˆj + z1 kˆ and OB = x iˆ + y ˆj + z kˆ respectively. 2

2

2

So the vector AB, joining two points A and B is given by AB = OB − OA

( x2 − x1 ) iˆ + ( y2 − y1 ) ˆj + ( z2 − z1 ) kˆ.

9.6

Engineering Mathematics-I

9.3 9.3.1

SCALAR PRODUCT OR DOT PRODUCT OF VECTORS Definition

The scalar product or dot product of two vectors a and b, where θ is the smallest angle between their directions, is defined by a b = a b cos θ where a and b are magnitudes of a and b respectively (see Fig. 9.4). B

b

θ A

O

a

N

Figure 9.4

9.3.2

Geometrical Interpretations

The scalar product of two vectors is nothing but the product of the length of one vector and the projection of the other to the former one.

9.3.3

Properties of Scalar Product

1) The scalar product of two vectors always yields a scalar quantity. 2) The scalar product of two vectors is commutative, i.e., a b = b a. 3) Two non-null vectors a and b are perpendicular if and only if a b = 0. 4) a 2

a a

a

a cos 0

a

2

2 5) Here, iˆ iˆ = ˆ 2 = ˆ = 1. Similarly, ˆj 2 = kˆ 2 = 1. Again, iˆ ˆj = ˆj kˆ =

kˆ iˆ = 0, where iˆ , ˆj , kˆ respectively are the unit vectors along the three coordinate axes. 6) Let a = a1 iˆ a2 ˆj Then, a b = ( a1 iˆ

a3 kˆ , and b = b1 iˆ a2 ˆj

a3 kˆ ) ( b1 iˆ

b2 ˆj b2 ˆj

b3 kˆ b3 kˆ ) = a1b1 + a2 b2 + a3 b3

9.7

Vector Analysis

7) The angle between the vectors a and b is given by ⎛ ⎞ a b ⎟ θ = cos −1 ⎜ ⎜⎝ a b ⎟⎠ Ê a1b1 + a2 b2 + a3 b3 = cos-1 Á ÁË a12 + a22 + a32 b12 + b22 + b32

ˆ ˜ ˜¯

8) Distributive property: a (b c)

a b a c

9) Component of a along b is given by a cos θ

a b a b = | a| | b | b

a

10) If a b = a c then a ( b c ) = 0 implies the following facts. a = 0 or b c = 0 or a is perpendicular to b c . i.e., a = 0 or b

c or a is perpendicular to b c .

Example 1 If α = 3 and β = 4, then find the values of the scalar μ for which the vectors

α + μβ and α μβ will be perpendicular to each other. Sol.

The vectors α + μ β and α (

2

or, μ 2 = or, μ = ±

9.4.1

μβ will be perpendicular to each other if

μβ ) (α μβ ) = 0

i.e., α

9.4

[WBUT 2005]

μ2 β α

2

β

2

=

2

=0

9 16

3 4

VECTOR OR CROSS PRODUCTS OF VECTORS Definition

The cross product, or vector product, of two vectors a and b, where θ is the smallest angle between their directions, is defined by a ¥ b = | a | | b | sin q n

9.8

Engineering Mathematics-I

where a and b are magnitudes of a and b respectively and nˆ is a unit vector perpendicular to both a and b and the direction of nˆ is same as the direction of the motion of a right-handed screw rotating from a to b (see Fig. 9.5). B

b nˆ

θ O

a

A

Figure 9.5

9.4.2

Geometrical Interpretations

We have a × b =

sin θ ⋅ ˆ.

So, a b = a b sin θ = OA ⋅ OB ⋅ sin θ = 2 × area of ΔOAB = area of the parallelogram with the adjacent sides OA and OB (see Fig. 9.5). Hence, a b represents the vector area of the parallelogram whose adjacent sides are the vectors a and b.

9.4.3

Properties of Cross Products of Vectors

1) The cross product of two vectors always yields a vector quantity. 2) Let a and b be any two vectors. Then, ( uct is non-commutative.

b ) = (b (b

a ), i.e., cross prod-

3) Two non-null vectors a and b are parallel or collinear if and only if a b = 0. 4) For any vector a, we have a × a = 0. 5) Here, iˆ

iˆ = ˆj

ˆj = kˆ

kˆ = 0. Again iˆ

ˆj = kˆ , ˆj

kˆ = iˆ , kˆ iˆ = ˆj , where

iˆ , ˆj , kˆ respectively are the unit vectors along the three coordinate axes.

9.9

Vector Analysis

6) Let a = a1 iˆ Then, iˆ a b = a1 b1

b2 ˆj

b3 kˆ

kˆ a3 b3

ˆj a2 b2

= (a2 b3

a3 kˆ , and b = b1 iˆ

a2 ˆj

a3 b2 ) iˆ ( a3 b1

a1 b3 ) ˆj ( a1 b2

a2 b1 ) kˆ

7) The unit vector nˆ , which is perpendicular to both a and b, is given by nˆ =

a b a b

8) The angle between the vectors a and b is given by ⎛ a b⎞ θ = sin −1 ⎜ ⎟. ⎜⎝ a b ⎟⎠ 9) Distributive property: a (b c) 10) If a b

a× b a c

a × c then a ( b c ) = 0 implies the following facts.

a = 0 or b c = 0 or a is parallel to b c i.e., a = 0 or b

c or a is parallel to b c .

Example 2 Find a unit vector perpendicular to each of the vectors 2iˆ − ˆj + 2 kˆ and 3iˆ + ˆj − kˆ and obtain the angle between them. Sol.

Let a = 2iˆ − ˆj + 2 kˆ and b = 3iˆ + ˆj − kˆ. Now, ˆj kˆ iˆ a × b = 2 −1 2 3 1 −1 = iˆ (1 − 2) − ˆj (−2 − 6) + kˆ (2 + 3) = −iˆ + 8 ˆj + 5kˆ So, a ×b

−iˆ + 8 ˆj + 5kˆ = 12 + 82 + 52 = 90

9.10

Engineering Mathematics-I

Therefore, the unit vector is given by nˆ =

a ×b

−iˆ + 8 ˆj + 5kˆ

=

a ×b

90

Now, the angle is given by a ×b

θ = sin −1

= sin −1

90

= sin −1

22 12 + 22 32 +1 12 12

a b

⎛ 10 ⎞ = sin −1 ⎜ . ⎜ 11 ⎟⎟ 3 11 ⎝ ⎠

3 10

Example 3

(

Show that a b Sol.

)

2

= a 2 b 2 − (a b)2 . 2

For any vector a, we have a 2 = a . So,

(a b )

2

= a b = a

2

= = a =

9.5 9.5.1

2

2

2 2

2

2

b ⋅ sin 2 θ 2

2

2

(1

b −a

2

)

2

b ⋅ cos2 θ

((a a b )2 .

SCALAR TRIPLE PRODUCTS Definition

Let a, b and c be three vectors. Then the scalar triple product of a, b and c is defined as a (b × c ). It is always a scalar quantity and is denoted by [ b ]. Geometrically, it represents the volume of a parallelepiped whose coterminus edges are a, b and c.

9.11

Vector Analysis

9.5.2

Properties of Scalar Triple Product

1) Let a = a1 iˆ a1 ⎡ abc ⎤ = b1 ⎣ ⎦ c1

a3 kˆ , b = b1 iˆ

a2 ˆj a2 b2 c2

b2 ˆj

b3 kˆ and c = c1 iˆ c2 ˆj c3 kˆ then

a3 b3 c3

2) [ b ] = [ bca]] = [ b ] = [[acb acb] = −[ b ] = [ cba]. 3) [ iˆ , ˆj , kˆ ] = [ ˆj , kˆ , iˆ ] = [ kˆ , iˆ , ˆj ] = 1, where iˆ , ˆj , kˆ respectively are the unit vectors along the three coordinate axes. 4) The three nonzero vectors a, b and c are coplanar if and only if ⎡ abc ⎤ = 0 i.e., a ( b × c ) = 0 ⎣ ⎦

9.6 9.6.1

VECTOR TRIPLE PRODUCTS Definition

Let a, b and c be three vectors in three dimensions. Then the vector triple product of a, b and c is defined as a ( b × c ) = ( a c ) b − ( a b) c

9.6.2

Properties of Vector Triple Products

1) Let a, b and c be three nonzero vectors. Then, a ( b × c ) ≠ ( a × b ) c . 2) If any two of the nonzero vectors a, b and c are parallel or equal then, a (b × c) = 0 3) ( b

9.7 9.7.1

)× a =

(b (b

c) = (

b) c (

c ) b.

STRAIGHT LINE Equation of a Line Passing Through a Given Point and Parallel to a Given Vector

Let A be the given point whose position vector is a w.r.t the origin O, and also suppose the line is parallel to the given vector b. Let P be any point on the line and its position vector is given by r. Then the equation of the required line in vector form is given by r = a + tb, where t is any scalar.

9.12

Engineering Mathematics-I

b

P

A

r

a O

Figure 9.6 Note: The equation of the line passing through the origin is r = tb tb , since a = 0. Example 4

Find the equation of the line through the point ( 2, 1, 0) and paral-

lel to the vector 5 ˆ − 3 ˆj + 4 ˆ. Sol.

Here, the given point is ( 2, 1, 0), whose position vector is a and the vector b = 5iˆ − 3 ˆj + 4 kˆ represents the point (5, 3, 4). Also, r is the position vector of any arbitrary point ( x, y, ). We have from the above section, the equation of the line as r scalar t. Therefore,

a + tb for any

( x, y, ) = (−2, 1, 0) + t (5, 3, 4) which implies x + 2 y −1 z − 0 = = =t 5 −3 4 Hence the required equation of the line is x + 2 y −1 z = = . 5 −3 4

9.7.2

Equation of a Line Passing Through Two Points

Let A and B be the given points whose position vectors are a and b respectively w.r.t the origin O. Let P be any point on the line and its position vector is given by r.

9.13

Vector Analysis P

B A b r

a O

Figure 9.7 Then the equation of the required line in vector form is given by r

tta a

(1 t ) b

…(1)

where t is any scalar. Note: Let the coordinate of the points A and B are ( x2 , y2 , 2 ) and ( x1 , y1 , 1 ) w.r.t the rectangular Cartesian coordinate system. Also, P( x, y, z ) be any point on the line. Then, from (1) we have ( x, y, ) = ( x2 , y2 , z2 )

(1 t ) ( x1 , y1 ,

1)

which gives x x1 y y1 z z1 = = = t. x2 x1 y2 y1 z2 z1 This is the equation of a line through two given points in three-dimensional Cartesian coordinate system. Example 5

Find the equation of the line through the points (2, 3, 4) and

(3, 4, 5). Sol.

Here, ( x1 , y1 ,

1)

≡ (2, 3, 4) and ( x2 , y2 ,

Hence the required equation of the line is x x1 y y1 z z1 = = x2 x1 y2 y1 z2 z1 or,

x 2 y −3 z −4 = = 3−2 4−3 5−4

or, x 2

y − 3 = z − 4.

2)

≡ (3, 4, 5).

9.14

9.8 9.8.1

Engineering Mathematics-I

PLANE Equation of a Plane Perpendicular to the Unit Vector nˆ and Passing Through a Point whose Position Vector is a

r −a P

A N r nˆ

a

O

Figure 9.8 Let A be the given point whose position vector is a. Suppose ON is perpendicular to the plane and ON = p, length of the perpendicular from the origin. Also, consider P to be any point on the plane whose position vector is r. Here ON = p ⋅ nˆ and OP = r . Then the required equation of the plane is given by r n = p. This is known as normal form of the equation of the plane. Note: If the plane passes through the origin then the equation becomes r n = 0.

9.8.2

Equation of a Plane Passing Through a Point whose Position Vector is a and Parallel to Two Vectors b and c

Let P be any point on the plane whose position vector is r. Then the required equation of the plane is [ b ] = [ abc ].

9.8.3

Equation of a Plane Passing Through Three Given Points

Let the position vectors of three points be a, b and c respectively. Then b a

(

)

and c a lie in the same plane. So, b a × ( c − a ) is perpendicular to the plane.

Vector Analysis

Let P be any point on the plane whose position vector is r. Then also r

(

)

9.15

a is per-

pendicular to b a × ( c − a ) . So,

(r

{(

}

)

a ) ⋅ b a × (c − a ) = 0

which implies r u = [ abc ] where u

a × b b × c + c × a.

is the required equation of the plane. Example 6

Find the equation of the plane through the points A(2, 1, 4),

B(3, 4, 7) and C( 2 2, 3 3, 1). Here AB = (1, 5, 3) and AC = (− 4 4, 4 4, − 5).

Sol.

Let P ( x, y, z ) be any point on the plane. Here

AB × AC = (−37, − 7, 24)

is perpendicular to the plane and

AP = ( x 2 y + 1, z − 4) lies on the plane. So, AP is perpendicular to AB × AC . Therefore

{

}

AP ⋅ AB × AC = 0 or, (−37 37, − 7, 24) ⋅((

2, y 1,

or, − 37( x − 2) − 7 ( +

) + 24(

4) = 0 − 4) = 0.

This is the required equation of the plane.

9.8.4

Distance of a Point from a Plane

Let the position vector of the given point be a and the equation of the plane be r n = p, where n is normal to the plane. Then the required distance is p a⋅ n n

.

9.16

9.9 9.9.1

Engineering Mathematics-I

SPHERE General Equation of a Sphere P r −c C r c

O

Figure 9.9 Let the radious of the sphere be a, and c be the position vector of its centre C. Also, let P be any point on the sphere whose position vector is r. It is clear from the figure that CP = r c . 2

Again CP = a 2 . Therefore

(r

c ) = a2 2

i.e., ( r )

2

2 r ⋅ c + ( c ) − a2 = 0 2

This is the required equation of the sphere. Note: (1) If the centre of the sphere is the origin, i.e., c = 0 then the equation of the sphere becomes

( r )2 = a 2 . (2) If the origin lies on the sphere then ( c ) = a 2 , and correspondingly the equation becomes 2

( r )2 9.9.2

2 r ⋅ c = 0.

Equation of the Sphere with Given Diameter Ends

Let a and b be the position vectors of the ends of the diameter of the sphere.

9.17

Vector Analysis

Then the equation of the sphere is given by (

)⋅(

b) = 0

where r is the position vector of any arbitrary point on the sphere.

WORKED-OUT EXAMPLES Given two vectors α = 3iˆ − ˆj

Example 9.1 the form of β

and β = 2iˆ + ˆj − 3kˆ, express β in

β 2 , where β1 is parallel to α and β 2 is perpendicular to α . [WBUT-2005]

Sol.

Since β 1 is parallel to α and β 2 is perpendicular to α then

β

α and β α = 0 where k is any scalar

Now,

β

β1 + β

β = kα

β2

Therefore,

β α = kα α + β α or, β α = k α or, k = or, k =

2

β .α α

2

(2 ˆ + ˆj − 3 ˆ)(3 ˆ − ˆj ) (3) + 1 2

2

=

5 1 = 10 2

Therefore, 1 α = (3ˆ − ˆj )

β1

2

and

β

1 1 3 β − β1 = (2iˆ + ˆj − 3kˆ) − ((3iˆ − ˆj ) = iˆ + ˆj − 3kˆ 2 2 2

Example 9.2

Let α β , γ be unit vectors satisfying α β = 0 and α γ = 0. If

the angle between β and γ is Sol.

Here α

β = γ =1

Since,

α β = 0 and α γ = 0.

π then show that α 6

β γ ).

9.18

Engineering Mathematics-I

Therefore α is perpendicular to β and γ Therefore,

α

β γ ), ) t being a scalar

Therefore,

α

2

β ×γ )

2 2

π⎫ ⎧ or, 1 = t 2 ⎨ β γ sin ⎬ 6⎭ ⎩ 2

1⎫ ⎧ or, 1 = t 2 ⎨1 ⋅ 1 ⋅ ⎬ 2⎭ ⎩ 2 2 or, t ⇒ t = ±2 Hence

α

β γ)

1 If aˆ , bˆ, cˆ are unit vectors such that aˆ × (bˆ × cˆ) = bˆ, find the angles 2 which aˆ makes with bˆ and cˆ. Example 9.3

Sol. Here, 1 aˆ × (bˆ × cˆ) = bˆ 2 Therefore, (aˆ ⋅ cˆ) ⋅ bˆ − (aˆ ⋅ bˆ) ⋅ cˆ =

1ˆ b 2

Equating coefficients of bˆ and cˆ, we get 1 2

…(1)

( ˆ ⋅ bˆ) = 0

…(2)

( ˆ ⋅ ˆ) = and

From (1), aˆ cˆ cos θ

1 where θ is the angle between aˆ and cˆ 2

or, θ = cos−11

1 π = since ˆ = ˆ = 1 2 3

9.19

Vector Analysis

From (2), a bˆ cos

φ i he angle between a and bˆ π 2

or,

Therefore, the angle between aˆ and bˆ is Example 9.4

2

and the angle between aˆ and cˆ is ×c

c, prove that a

Given three vectors

) ⋅ c. Sol.

a

3

.

b

[WBUT-2005]

Let a=a ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

3k

Now, iˆ b

b ×c

kˆ

ˆj

3 3

=ˆ

ˆ

ˆ

2 3

1

)

Therefore, ×

×

iˆ a1

=

kˆ a3

ˆj a2

) =ˆ

2

1 2

ˆ +ˆ = ˆ[(a2

)]

3

2 )] 2

+

ˆ

+

+

+

1 1

+ˆ

+

+

ˆ

+

+

ˆ

+

+

ˆ

+

+

2 2

+

ˆ+

+ +

+

ˆ+

ˆ+

ˆ

ˆ + 3 kˆ)

] 1]

1 1

3]

9.20

Engineering Mathematics-I

= {( ˆ

ˆ ˆ

−

ˆ + ˆ − 3 ˆ,

ˆ

ˆ ˆ

ˆ ˆ

ˆ

ˆ ˆ

3k )

ˆ

ˆ

3 k ).

⋅b c

=( Example 9.5

ˆ

ˆ− ˆ+ ˆ

Find the constant m such that the vectors = 3 ˆ + ˆ + 5kˆ are coplanar.

[WBUT-2004].

Sol. The three nonzero vectors a b and c are coplanar if and only if [ which implies 2 1 3

] = 0,

1 1 2 −3 5 (5 +

or, 2(10

+ 1(m 6) = 0

or 7m + 28 = 0 or,

= −4

Example 9.6 [

b

c are three vectors, show that

If

c 2.

c a = a

[WBUT 2006, 2009]

Sol. ×

={ ={

}(c a )

⋅

), where p = a × b)

c c c×

=(

)

c a)

⋅b c

=( ⋅

,

= {c = [c a = [a b

=[

c ]2

[

a] = 0

Vector Analysis

9.21

PART-II (VECTOR DIFFERENTIATION AND GRADIENT, DIVERGENCE, CURL) 9.10

VECTOR FUNCTION OF A SCALAR VAR V IABLE

9.10.1 Definition � � If for each value of a scalar variable t, there corresponds a unique vector f then f is � called a vector function of the scalar variable t and is denoted by f ( ). � If the components of a vector function f ( ) along the coordinate axes be � � � � f 1(t), f 2(t), f 3(t) then the vector function f ( ) is written as � � � �� f (t ) f 1 (t )iˆ f 2 (t ) ˆj f3 (t )kˆ. The position vector of a point P in three-dimensional space with respect to a vector origin is a function of a scalar variable t and is denoted by, � r (t ) x(t )iˆ y(t ) ˆj z(t )kˆ.

9.10.2 Limit and Continuity of a Vector Function Limit of a Vector Function � � A Vector function f ( ) is said to tend to a limit a as t tends to s, if for any preassigned positive number ε , there exists a small positive number δ , such that � � f ( ) a < , when t s < δ Limit of a vector function is denoted by � � lim f (t ) a t →s

Continuity of a Vector Function � A vector function f ( ) is said to be continuous at a point s if � � lim f (t ) = f (s ) t →s

9.11

DIFFERENTIATION OF VECTOR FUNCTIONS

� The ordinary derivative of a single-valued function f ( ) with respect to t is defined as � � � d (t ) lim f (t δ t ) − f (t ) = δ t →0 dt δt provided the limit exists.

9.22

Engineering Mathematics-I

9.11.1 General Rules for Differentiation 1) 2) 3) 4)

5)

6)

� � � d (a b ) da db = ± dt dt dt � � � � d (a b ) � db � da +b ⋅ = a⋅ dt dt dt � � � � d (a b ) � db da � + ×b = a× dt dt dt � � � ⎡ da� � � ⎤ ⎡ � db � ⎤ ⎡ � � dc� ⎤ d [ a, b , c ] = ⎢ , b , c ⎥ + ⎢ a, , c ⎥ + ⎢ a, b , ⎥ dt dt ⎦ dt ⎣d ⎦ ⎣ dt ⎦ ⎣ � � � � � d da � � � ⎛ db � ⎞ � ⎛ � dc ⎞ {a (b × c )} = (b c ) + a × ⎜ × c ⎟ + a × ⎜ b × ⎟ dt dt d dt ⎠ ⎝ ⎝ dt ⎠ � � d (φ a ) da d =φ where φ is any scalar dt dt d

9.12

SCALAR AND VECTOR POINT FUNCTION

If for every position of a point in space, a physical quantity has one or more definite values assigned to it then it is said to be a point function. If the point function has only one value at each point then the function is called a single-valued function.

9.12.1 Scalar Point Function � � f is said to be a scalar point function of r if for every value of r, there corresponds a definite scalar quantity f . The scalar point function is denoted by f r ) or f ( x, y, z ) � where r is the position vector corresponding to any point P ( x, y, ) in the space. The scalar point function will constitute a scalar field, for example, f ( x, y, z ) = x 2 + y + xyz.

9.12.2 Vector Point Function � � � F is said to be a vector point function of r if for every value of r, there � corresponds a definite vector quantity F. The vector point function is denoted by � � � F ( ) o F ( , y, z ) � where r is the position vector corresponding to any point P ( x, y, ) in the space. The vector point function will constitute a vector field, for example, � F( x, y, ) = x 2 iˆ yz ˆj + z 2 kˆ.

9.23

Vector Analysis

9.13 GRADIENT OF A SCALAR POINT FUNCTION 9.13.1 Definition Let , , ) be a scalar point function differentiable at each point in a certain region R of space. Then the gradient of , denoted by grad is defined by, grad

, z)

φ =

x

=

9.13.2

i+

y

x

y

+

z

kˆ

z

( x, , )

kˆ

roperties of Gradient of a Scalar Point Function

1) The necessary and sufficient condition for a scalar point function be a constant is � , , )=0 2) If �

, ) and ( , ,

3) If �

x

, , ) and (

4) If �

, , , ) and

, , ) to

, , ) are two scalar point functions then, � � , )} x, , ) , , ) are two scalar point functions then, � )} = ( , , , , ) ( , x, , ) are two scalar point functions then, � � z

, , , )

, z )} 2

5) If c is a constant and , , ) is a scalar point function, then � x, ) (c x z ) c Example 7 i) ∇ ii) ∇

1 r

−

If r � r

� � r where

xˆ

ˆ z ˆ prove that

r3

= nr

−2

r

[WBUT 2004]

9.24

Sol.

Engineering Mathematics-I

� Here, r = xiˆ yj yˆj zkˆ, so r

� r = x2 + y 2 + z 2 ⇒ r 2 = x2 + y 2 + z 2 .

� ⎛1⎞ ∂ ⎛1⎞ ∂ ⎛1⎞ ∂ ⎛1⎞ i) ∇ ⎜ ⎟ = iˆ ⎜ ⎟ + j ⎜ ⎟ + kˆ ⎜ ⎟ ∂ ∂ ∂ r x r y r z⎝r ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −1 ∂ ⎞ ⎛ −1 ∂r ⎞ ˆ ⎛ −1 ∂r ⎞ = iˆ ⎜ 2 ⎟+k⎜ 2 ⎟+ j⎜ ⎟ ⎝ r ∂x ⎠ ⎝ r 2 ∂y ⎠ ⎝ r ∂z ⎠ 1 x 1 y 1 z = −iˆ 2 − j 2 − kˆ 2 r r r r r� r −1 r = 3 ( ˆ yjˆ kˆ ) = − 3 r r ii) Since r 2

x 2 + y 2 + z 2 , we have

∂r ∂r x = 2x ⇒ = ∂x ∂x r ∂r y ∂r z Similarly, = and = ∂y r ∂z r � ∂ ∂ ∂ ∇( n ) = ˆ (r n ) + ˆj (r n ) + kˆ (r n ) ∂x ∂y ∂z ∂r ∂r ∂r = iˆ nr n −1 + ˆj nr n −1 + kˆ nr n −1 ∂x ∂y ∂z x y z = iˆ nr n −1 + ˆj nr n −1 + kˆ nr n −1 r r r � = n 2 ( xiˆ + yjˆ + zkˆ ) = nr n−2 r 2

9.14

LEVEL SURFACE

Let f r ) or f ( x, y, z ) be a scalar point function over a region in space. Then the points ( x, y, ) in the region satisfying the equation f ( x, y, z ) = c constitute a family of surfaces.This family of surfaces is called a level surface determined by f , for example, x3 + y 2 − 2 z = 5 is a level surface, determined by the function f ( x, y, z ) = x3 + y 2 − 2 z for c = 5.

9.15

DIRECTIONAL DERIVATIVE OF A SCALAR POINT FUNCTION

9.15.1 Definition Let φ ( x, y, ) be a scalar point function possessing first-order derivatives. Then the directional derivative of φ ( x, y, ) at P( x, y, z ) along the unit vector a� , where � a a1iˆ a2 ˆj a3 kˆ is given by

9.25

Vector Analysis

∂φ ∂φ ˆ ⎞ � ⎛ ∂φ ∇φ a = ⎜ i+ j+ k ( a1 iˆ ∂y ∂z ⎠ ⎝ ∂x = a1

a2 ˆj

a3 kˆ ).

∂φ ∂φ ∂φ + a2 + a3 ∂x ∂y ∂z

9.15.2 Observations 1) Directional derivative is the rate of change of φ at ( x, y, ) in the direction of a� . 2) The directional derivative along along any straight line can be expressed in terms of those along the coordinate axes. 3) The directional derivative at P ( x, y, z ) along the reverse direction will be dφ ∂φ ∂φ ∂φ = − a1 − a2 − a3 ds ∂x ∂y ∂z Find the directional derivative of φ ( x, y, ) = xy 2 z + x 2 z

Example 8

at (1,1, 2) in the direction (2iˆ Sol.

ˆj − 2 kˆ ).

Here, � ⎛ ∂φ ∂φ ˆ ∂φ ˆ ⎞ ∇φ ( x, y, ) = ⎜ i + j+ k⎟ x ∂ ∂y ∂z ⎠ ⎝ = ( y 2 z + 2 )i )iˆ + 2 y ˆj ( xy 2 + x 2 )kˆ � Now, ∇φ (1, 1, 2) = 6iˆ + 4 ˆj + 2 kˆ. The unit vector in the direction (2iˆ ˆj − 2 kˆ ) is aˆ =

(2iˆ 2

ˆj − 2 kˆ ) 2

2

=

2ˆ 1ˆ 2 ˆ i j− k 3 3 3

2 1 2 The directional derivative is � 1 2 ⎞ ⎛2 ∇φ (1, 1, 2) ⋅ ˆ = (6iˆ + 4 ˆj + 2 kˆ ) ⋅ iˆ + ˆj − kˆ ⎟ = 4 3 3 ⎠ ⎝3

9.15.3 Properties of Directional Derivative of a Scalar Point Function Theorem 9.3: The directional derivative of a scalar field φ at a point P ( x, y, z ) in the direction of the unit vector aˆ is given by dφ � = ∇ φ ⋅ aˆ ds

9.26

Engineering Mathematics-I

where s is the distance of the point P( x, y, z ) from some fixed point in the direction � of a. Theorem 9.4: The directional derivatives of a scalar point function φ ( x, y, ) in the directions of X , Y and Z-axes are ∂φ ∂φ ∂φ , and respectively. ∂x ∂y ∂z Theorem 9.5: Let nˆ be a unit normal vector to the level surface φ ( x, y, ) = c at a point P( x, y, z ), n being the distance of P ( x, y, z ) measured from a fixed ˆ Then, point in the direction of n. � ⎛ dφ ⎞ ∇ φ = ⎜ ⎟ nˆ ⎝ dn ⎠ � i.e., the direction of ∇φ is normal to the level surface. Theorem 9.6: The directional derivative of a scalar field function φ ( x, y, ) is maximum along the normal to the level surface φ ( x, y, ) = c, and the maximum value is � dφ ⎛ dφ ⎞ n= ∇φ = ⎝ dn ⎠ dn Example 9 2

Find the maximum value of the directional derivative of

φ ( x, y, ) = x + z − y 2 at the point (1, 3, 2). Sol.

2

Here, � ⎛ ∂φ ∂φ ˆ ∂φ ˆ ⎞ ∇φ ( x, y, ) = ⎜ i + j+ k⎟ x ∂ ∂y ∂z ⎠ ⎝ = 2 xiˆ − 2 yjˆ 2 zkˆ. � At (1, 3, 2), ∇φ (1, 3, 2) = 2iˆ − 6 ˆj + 4 kˆ is the direction in which the directional derivative is maximum. The maximum value of the directional derivative is � ∇φ (1, 3, 2) = 2iˆ − 6 ˆj + 4 kˆ 22 + 62 + 42 = 2 14

9.16

TANGENT PLANE AND NORMAL TO A LEVEL SURFACE

Let φ ( x, y, ) = c be the equation of a level surface. Then, i) the equation of tangent plane at P( x, y, z ) is ∂φ ∂φ ∂φ ( ) +( y) +( ) =0 ∂x ∂y ∂z

Vector Analysis

9.27

ii) the equation of the normal at P( x, y, z ) is ( ) ( y) ( ) = = . ∂φ ∂φ ∂φ ∂x ∂y ∂z Find the equation of the tangent plane to the surface xyz = 4 at the Example 10 point (1, 2, 2). Find also the equation of the normal line at that point. Sol.

The equation of the level surface is φ ( x, y, ) ≡ xyz − 4 = 0. Now, ∂φ ∂φ ∂φ = yz y, = zx, = xy ∂x ∂y ∂z and

∂φ ∂φ ∂φ d (1, 2, 2) = 4, (1, 2, 2) = 2 and (1, 2, 2) = 2. ∂x ∂y ∂z

The equation of the tangent plane at (1, 2, 2) is (

)

∂φ +( ∂x

y)

∂φ +( ∂y

)

∂φ =0 ∂z

or, ( x 1)4 ( y − 2)2 ( z − 2)2 = 0 or, 4 x + 2 y 2 z − 12 = 0 The equation of the normal at (1, 2, 2) is ) ( y) ( ) = = ∂φ ∂φ ∂φ ∂x ∂y ∂z ( x 1) ( y 2) ( 2) or, = = 4 2 2 (

Find a unit normal to the surface x 2 − y 2 + z = 2 at (1, 1, 2)

Example 11 Sol.

The surface is given by φ ( x, y, ) = x 2 − y 2 + z − 2. Therefore, � ⎛ ∂φ ∂φ ˆ ∂φ ˆ ⎞ ∇φ ( x, y, ) = ⎜ i + j+ k ⎟ = 2 xiˆ − 2 yˆj + kˆ ∂x ∂y ∂z ⎠ ⎝ � and ∇φ (1, − 1, 2) = 2iˆ + 2 ˆj + kˆ. The unit normal to the surface is 2iˆ 2 ˆj + kˆ 2 + 2 +1 2

2

2

=

2ˆ i 3

2ˆ 1ˆ j + k. 3 3

Another unit normal is −

⎛ 2 ˆ 2 ˆ 1 ˆ⎞ i + j + k ⎟ in the other side of the surface. 3 3 ⎠ ⎝3

9.28

9.17

Engineering Mathematics-I

DIVERGENCE AND CURL OF A VECTOR POINT FUNCTION

9.17.1 Definition � Let F F1iˆ F2 ˆj F3 kˆ be any continuously differentiable vector point function. The divergence of the vector point function is defined by � � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ div F ∇ F = ⎜ iˆ + j + k ( F1 iˆ F2 ˆj F3 kˆ ) ⎝ ∂x ∂y ∂z ⎠ =

∂F1 ∂F2 ∂F3 + + . ∂x ∂y ∂z

� div F can also be written in the summation form as � � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ � ∂F ˆ ˆ ∇⋅F = ⎜ i + j + k ⎟ ⋅ F = ∑i ⋅ , ∂y ∂z ⎠ ∂x ⎝ ∂x ˆ the summation being taken over all iˆ, ˆj and k. The curl of the vector point function is defined by � � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ curl F ∇ F iˆ + j + k ( F1 iˆ F2 ˆj F3 kˆ ) ⎝ ∂x ∂y ∂z ⎠ i ∂ = ∂x F1

j ∂ ∂y F2

k ∂ ∂z F3

� curl F can also be written in the summation form as � � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ � ∂F ˆ ˆ ∇× F = ⎜ i + j + k ⎟ × F ∑i ⋅× , ∂y ∂z ⎠ ∂x ⎝ ∂x ˆ ˆ ˆ the summation being taken over all i , j and k.

9.17.2 Properties of Divergence and Curl � � 1) If F and G be two differentiable vector point functions then, � � � � i) div( = div F div i G � � � � ii) curl ( = curl F curl G Proof: � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ � � j + k ⎟ (F ± G ) (i) div ( F G ) = ⎜ iˆ + ∂y ∂z ⎠ ⎝ ∂x � � ⎛ ∂F ∂G ⎞ ∂ � � ˆ ˆ =∑ (F G ) = ∑ i ⋅ ⎜ ± ⎟ ∂x ⎝ ∂x ∂x ⎠

9.29

Vector Analysis

� = ∑i

±

i⋅

∂ � � � � ∇⋅ ±∇⋅

�

(ii) curl (

�

ˆ

)=

∂

=

∂

= cur

. �

(

)

� � ˆi × ∂F ± ∂G ∂x ∂

(

� G

±

∂G iˆ × x �

�

∑ iˆ ×

�

±

∂ ˆ k z

ˆ

ˆ

� ∂G ∂ �

curl

2) If F be vector point function and � � � � i) div div F � ii) curl curl F

be a scalar point function. Then WBUT-2004]

Proof � (i) di φ F)

ˆ

ˆ z

∂ = ∑ iˆ ⋅ =∑ =

i �

� (ii) curl

�

∂ � iˆ F ∂

=

⋅

ˆ

� ˆi φ ∂F ⎝ ∂x ⎠ ∂F iˆ ∂

F �

� F

+φ

ˆ

=

kˆ ( F

∂ ˆ k z

ˆ ∂

� F)

∂ � ⎫ = ∑ iˆ × =∑

∂ iˆ ∂

(

×

�

iˆ ×

∂F

� ∂F iˆ × ∂x

9.30

Engineering Mathematics-I

∂ ∑ iˆ ∂x × � � = × + ⋅

� F

� ˆ × ∂F ∂

� 3) If F and G be two differentiable vector point functions, then � � � � � � � � � � i) grad G F+ curl + curl � � � � � � ii) div = curl curl � � � � � � � � � � � � iii) curl div div + ) [WBUT-2003]. Proof:

� (i) grad ( ) � � � ∂ � � ( ⋅ G ) = ∑ iˆ ( ) x � � � � ∂G ˆ ˆi F ∂G + ⋅ ∂F ∑ ∂x ∑ F x ∂ ⎠ ⎝ � � � � ∂G = F i i× × + ∂ � � � � � � � � = (F F +

� � ∂F i ⋅ ∂x � ˆ ∂ F G x �

� � ∂ � � (ii) div ( F G ) = ∑ iˆ ( ) ∂x � ⎧ ∂F � � ∂G = ∑ iˆ × × ∂x ∂F � = ∑ iˆ × x � =∑ i× ⋅G x � = ∑ i× ⋅G � =G � (iii) curl (

) �

= ∑ iˆ ×

∂x

�

� � F F � F × G)

� � ∂G × x � ˆ ⋅ ∂G × F ∂x � � ˆi × ∂G F x⎠

iˆ

� G

� ∂F iˆ × ×

×

� ∂G ∂x

� ˆ ∂F x

9.31

Vector Analysis

� =∑ i× (iˆ G )

=

=

i

x

�

�

� )}

� ˆ ∂F ∂

∂x

� ∂F � G ∂x

−∑

(i G )

= G

ˆ×(

×G

∂F iˆ x

F

−G

�

+ (G

� �

�

� � � � ˆi ⋅ ∂G F ( ˆ ⋅ ) ∂G x x⎠ � i⋅ � ˆ ⋅ ∂G ∂x � � � F G

� ∂G ⋅F ∂x ⎩ ∂ � F − F ∑ iˆ G ∂x

9.17.3 Irrotational and Solenoidal Vectors � � A vector point function F is said to be irrotational if curl F = 0 and solenoidal if � div F = 0. � If the vectors A and B are irrotational then show that the vectors Example 12 � � [WBUT 2004, 2006] A B is solenoidal. � Sol If the vector functions A and B are irrotational then �� � curl A = 0 and curl B = 0 Now, �� �� � �� �� �� � �� �� div )= A )=B 0 0=0 �� �� �� �� Since, div ( A B) = 0 therefore ( ) is solenoidal.

9.17.4 Some Results on Second-Order Differential Operators Laplacian Operator The Laplacian operator is defined as ∇2

∂2

If ∇2 = � If v x

x2

+

,

∂2 y2

+

∂2 ∂z 2

be a scalar point function then 2

∂φ

∂φ

∂φ

x2

y2

∂z 2

is a scalar quantity.

z be a vector point function then � � v v � ∂2v ∇ 2 v = 2 + 2 + 2 is a vector quantity. x y z

9.32

Engineering Mathematics-I

The equation ∇ 2ϕ = 0 is known as the Laplace Equation � i)

Let

,

�

be a scalar point function then, div grad

Proof: di

grad

= ∑i

∂ ∂x

i

∂x

∂φ

∂ϕ

x2

x2

2

=

iv i

∂2

=

∂x

2

∂2

+

x

2

+

∂f ∂x

f ∂f +k ∂y ∂z

ˆj ∂ + k ∂ ∂y ∂z ∂2

+

x2 ∂2

=∇

x2

ii) Let

, , ) be a scalar point function then, � � � curl grad = 0,

[WBUT-2005]

Proof: curl ( grad

) ˆj ∂ + kˆ ∂ ∂y ∂z

= curl iˆ = ∑i ×

∂

i

∂

=

∂

j

x

∂ + kˆ ∂y ∂z

∂ iˆ +

2 ˆj + ∂ x

∂2 x

kˆ

=0 � iii) Let F x

� z = � �

F k be a vector point function then,

div curl F

F) = 0

Proof: div curl F

div

=

∂ x

=0

� F kˆ}

∂F ∂F − − iˆ + ∂y ∂z ∂z ∂x F − ∂z y

+

∂ y

F − ∂z x

ˆj + ∂F − kˆ x x +

∂ z

F − x ∂x

Vector Analysis

9.33

� � � � iv) Let F ( x, y, z ) = F1 iˆ + F2 ˆj + F3 kˆ be a vector point function then, � � � � �� � � curl curl F = ∇ × (∇ × F ) = ∇(∇F ) − ∇ 2 F

(

)

Proof: � � � ∇ × (∇ × )

� � � ∂ ⎛ ˆ ∂F ˆ ∂F ˆ ∂F ⎞ i j × k + × + × ⎜ ⎟ ∂x ⎝ ∂x ∂y ∂z ⎠ � � � ⎛ ∂2 F ∂2 F ˆ ∂2 F ⎞ ˆ ˆ ˆ = ∑i ⎜i × 2 + j × +k× ⎟ ⎜ ∂x ∂x∂y ∂x∂z ⎟⎠ ⎝ � � � � ⎡⎛ ∂ 2 F ⎞ ⎛ ∂2 F ⎞ ⎛ ⎛ ∂2 F ⎞ ∂2 F ⎞ ˆ ˆ ˆ j i j = ∑ ⎢⎜ iˆ ⋅ 2 ⎟ ⋅ iˆ iˆ ⋅ iˆ ⎜ 2 ⎟ + ⎜ ⎜ iˆ − ⋅ ( ) ⎟ ⎜ ⎟ ⎜ ∂x ⎟ ⎜ ⎜ ∂x∂y ⎟⎟ ∂x∂y ⎟⎠ ⎠ ⎝ ⎠ ⎝⎝ ⎣⎢⎝ ∂x ⎠ = ∑ iˆ ×

� � ⎛ ⎛ ∂2 F ⎞ ∂2 F ⎞ ˆ ˆ ˆ ˆ +⎜⎜i k − (i ⋅ k ) ⎟ ⎜ ⎜ ∂x∂z ⎟⎟ ∂x∂z ⎟⎠ ⎠ ⎝⎝

⎤ ⎥ ⎥⎦

� � � � � � � � � Using the formula a (b × c) = (a ⋅ c) b − (a b) ⋅ c and since iˆ iˆ = ˆj ˆj = kˆ kˆ = 1, iˆ ˆj = ˆj kˆ = kˆ iˆ = 0, we have from above � � � ∇ × (∇ × ) � � � � ⎛ ∂2 F ⎞ ⎛ ∂2 F ⎞ ⎛ ∂2 F ⎞ ˆ ∂2 F ˆ ˆ ˆ ˆ ˆ k −∑ 2 = ∑⎜i ⋅ 2 ⎟⋅i ⎜i j i ⎜ ∂x ⎟ ⎜ ∂x∂y ⎟⎟ ⎜ ∂x∂z ⎟⎟ ∂x ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ � � � � ∂ ⎛ ∂F ˆ ∂F ˆ ∂F ⎞ 2 = ∑ iˆ ⋅ ⎜ iˆ ⋅ +j +k ⎟−∇ F ∂x ⎝ ∂x ∂y ∂z ⎠ � � � � 2 = ∇(∇ ⋅ F)) − ∇ F

WORKED-OUT EXAMPLES �� �� � If φ = xy + yz y + zx and A x 2 yiiˆ + y 2 zzjˆ z 2 xkˆ then find a) A ⋅ ∇φ Example 9.7 � � � � b) φ ⋅ (∇ ⋅ A) c) ∇φ × A at the point (3, 1, 2) Sol. Here,

φ = xy + yz y + zx

9.34

Engineering Mathematics-I

Now, � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ j + k ⎟ (xy xy + yz + zx ) ∇φ = ⎜ iˆ + ∂y ∂z ⎠ ⎝ ∂x = ( y + z ) ˆ ( x + z) z ) ˆj ( x + y)kˆ � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ 2 ˆ 2 ˆ 2 ˆ ∇ ⋅ A = ⎜ iˆ + j + k ⎟ ( x yii + y zj z + z xk ) ∂ ∂ x y ∂z ⎠ ⎝ =

∂ 2 ∂ 2 ∂ 2 ( x y) ( y z) ( z x) ∂x ∂y ∂z

= 2( y

y

)

At the point (3, 1, 2) � � ∇φ (3, − 1, 2) = ˆi + 5 ˆj + 2 kˆ and A(3, − 1, 2) = −9iˆ + 2 ˆj + Therefore, � � a) A ⋅∇φ = ( 9iˆ + 2 ˆj kˆ ) ⋅ (ˆi + 5 ˆj + 2 kˆ ) = 9 + 10 + 24 = 25 � � b) φ ⋅ (∇ ⋅ A) = 2( xy x + yz + zx)) ⋅ ( y + y +

) = 2(( y + y +

kˆ

)2

Therefore, � � φ ⋅ (∇ ⋅ A)(3, − 1, 2) = 2( −3 − 2 + 6)2 = 2 � � c) ∇φ × A =

iˆ ˆj kˆ 1 5 2 −9 2 12

= iˆ(60 4) ˆj (12 + 18) + kˆ(2 + 45) = 56 ˆ 30 ˆj 47kˆ �� If A 2 x 2 iˆ 2 yzjˆ + xz 2 kˆ and f Example 9.8 �� A grad f at (1, 1, 1) Sol. Here, ⎛ ∂ ∂ ˆ ∂ ˆ⎞ grad f = ⎜ iˆ + j + k ⎟ ( z − x 3 y) x y ∂ ∂ ∂z ⎠ ⎝ = 3x 3 2 yiˆ − x 3 ˆj + 2 kˆ Therefore,

�� z − x3 y find a) A

f b)

9.35

Vector Analysis

� a) A

f = (2 x iˆ 2 y ˆ x ˆ −3 x y

ˆ)

x

z2

6 At (1, 1, 1) � A f =6 � b) A

i x2

f 3

j yz

2

3

ˆ(

k xz 2 2 +ˆ

) ˆ x

At (1, 1, 1) � A f = 5iˆ

5

)

ˆj − 8kˆ

� � Find div F and curl F where F

Example 9.9

−6

3

x

− 3 xyz )

[WBUT-2001, 2009] Sol. Here, F

3

x ˆ + kˆ

3 xyz ) ˆj ∂ ( ∂y

+ z3 ∂ ( ∂z

3

3 xyz )

z

+ z 3 3 xyz ) ˆ

3 y)kˆ

xz ˆ

Therefore, div F

iˆ

∂

∂ ∂ + ˆj + kˆ {(3 ∂y ∂z

x

iˆ (3

2

∂ z

y

2

xz ˆ (3z 2 − 3 )kˆ} 3 y)

z and curl F

iˆ

∂

∂ ∂ + ˆj + kˆ ∂y ∂z

x

yz iˆ

−

)ˆ

−

) ˆ}

9.36

Engineering Mathematics-I

iˆ ∂ ∂x

= (3 x 2

ˆj kˆ ∂ ∂ ∂y ∂z 3 yz ) (3 y 2 − 3 ) (3z 2 3 y)

3 x )iˆ (−3 y 3 y) ˆj (−3 3z 3z )kˆ = 0

= (−3 3

Find the directional derivative of f x, y, z ) = 2 x 2 + 3 y 2 + z 2 at the kˆ point (2, 1, 3) in the direction of the vector iˆ k. [WBUT-2001] Example 9.10

Sol.

Here, f x, y, z ) = 2 x 2 + 3 y 2 + z 2 Therefore, � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ 2 ∇f = ⎜ iˆ + j + k ⎟ (2 x + 3 y 2 + z 2 ) = 4 xiˆ + 6 yjˆ + 2 zkˆ ∂y ∂z ⎠ ⎝ ∂x At the point (2, 1, 3) � ∇f ∇ 8i + 6 j + 6 k If aˆ is the unit vector in the direction of iˆ 2 kˆ, then kˆ

iˆ

1 ˆ 2 ˆ i k 5 5 5 Therefore, the required directional derivative is � ⎛ 1 ˆ 2 ˆ ⎞ −4 ∇f ⋅ aˆ = (8i + 6 j + 6 k ) ⎜ i− k⎟ = 5 ⎠ 5 ⎝ 5 � Example 9.11 Show that A = (6 xxyy z 3 )iˆ (3 x 2 − z ) ˆj (3xz 2 y kˆ is irrotational. � � Find the scalar function φ such that A = ∇φ . [WBUT-2002, 2004] aˆ =

=

Sol. Here,

� curl A =

ˆj iˆ kˆ ∂ ∂ ∂ ∂x ∂y ∂z (6 xy + z 3 ) (3 (3xx 2 z ) (3 xxz 2 y)

⎡∂ ⎤ ∂ ∂ ⎡∂ ⎤ (3 2 y) − (3 x 2 z ) ˆj (3 2 y) − (6 xy + z 3 )⎥ ∂ y ∂ z ∂ x ∂ z ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ∂ ∂ 2 3 + kˆ ⎢ (3 x − z ) − (6 xyy + z ) ⎥ ∂y ⎣ ∂x ⎦ 2 ˆ(6 6 x ) = 0 ˆ ˆ = i ( 1 1) j (3z 3z 2 ) k(6 � Therefore, A = (6 xxyy z 3 )iˆ (3 x 2 − z ) ˆj (3xz 2 y kˆ is irrotational. =ˆ

9.37

Vector Analysis

Since � � A ∇φ

∂φ ∂φ ˆ ∂φ iˆ +j +k ∂x ∂y ∂z

therefore, ∂φ ∂φ ∂φ = (6 xyy + z 3 ), = (3 x 2 − z ), = (3 xz 2 − y) ∂x ∂y ∂z Now, ∂φ ∂φ ∂φ dφ = dx + dy + dz ∂x ∂y ∂z or, dφ = (6 xy + z 3 ) d (3 2 ) dy + (3 xz 2 − y) dz or, dφ = (6 xydx d +3

2

y) + ( 3 dx 3 xz 2 dz) z ) ( xdy + ydz )

or, dφ = 3[ yd ( x 2 ) + x 2 dy] dy] [ z 3 dx + xd ( 3 )] ( xdy + ydz ) or, dφ = 3d ( x 2 y) d(x ( xz 3 ) d ( yz ) Integrating, we get

φ ( x, y, ) = 3 x 2 y xxz 3 Example 9.12

yz + c where c is arbitrary constant.

Show that curl grad f = 0 where f x, y, z ) = x 2 y 2 xy + z 2 . [WBUT-2003]

Sol.

Here, f x, y, z ) = x 2 y 2 xy + z 2 Now, ⎛ ∂ ∂ ˆ ∂ ˆ⎞ 2 grad f = ⎜ iˆ + j + k ⎟ ( x y + 2xy 2 xxy z 2 ) x y z ∂ ∂ ∂ ⎝ ⎠ 2 = (2 y 2) ˆ ( x + 2 x ) ˆj + 2 zkˆ Therefore, ˆj iˆ kˆ ∂ ∂ ∂ curl grad f = ∂x ∂y ∂z 2 (2 xy + 2) ((xx 2 x ) 2 z = iˆ(0 0) ˆj (0 0) kˆ(2 ( x 2−2

Example 9.13

In what direction from the point (1, 1,

tive of φ ( x, y, ) = x tional derivative.

2

2

2

2) = 0 ) is the directional deriva-

2 y + 4 z a maximum? Obtain the magnitude of the direc2y [WBUT-2003, 2006, 2007]

9.38

Sol.

Engineering Mathematics-I

� Directional derivative of φ is maximum in the direction of ∇φ . Here,

φ ( x, y, ) = x 2 22yy 2 + 4 z 2 Therefore, � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ 2 ∇φ = ⎜ iˆ + j + k ⎟ ( x − 2 y2 + 4 z2 ) x y ∂ ∂ ∂z ⎠ ⎝ = 2 xiˆ − 4 yjˆ 8 zkˆ At (1, 1, 1) � ∇φ ⋅ (1, 1, − 1) = 2iˆ − 4 ˆj − 8kˆ Therefore, the directional derivative is maximum along 2iˆ 4 ˆj − 8kˆ. The magnitude of the directional derivative is � 2 ∇φ + (−4)2 + (−8)2 = 2 21 Example 9.14

Sol.

In what direction from the point (1, 2, ) is the directional deriva-

2 z 2 a maximum? Also find the value of this maximum directional [WBUT-2004] � Directional derivative of f is maximum in the direction of ∇f ∇ . Here,

tive of f x derivative.

2

y

2

f x, y, z ) = x 2

y2

2z2

Therefore, � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ 2 ∇f = ⎜ iˆ + j + k ⎟ ( x − y2 + 2 z2 ) ∂ ∂ x y ∂z ⎠ ⎝ = 2 xiˆ − 2 yjˆ 4 zkˆ At (1, 2, ) � ∇φ ⋅ (1, 2, 3) = 2i − 4 j + 12 k Therefore, the directional derivative is maximum along 2iˆ 4 ˆj + 12 kˆ. The magnitude of the directional derivative is � 2 ∇f + (−4)2 + (12)2 = 162 Example 9.15

� If r

xiˆ yyjˆ zkˆ and r

θ is the null vector. Sol.

Here, r

� r = x 2 + y2 + z2

� � r , show that grad f r ) × r = θ where [WBUT-2005]

Vector Analysis

9.39

Now, grad f r ) =

∂f ˆ ∂f ˆ ∂f ˆ i+ j+ k ∂x ∂y ∂z

Since f is a function of r Therefore, ∂f df ∂r = ∂x dr ∂x = f ′(r )

1 2

1 2

x + y2 + z2

⋅ 2x

xf ′( ) r Similarly, ∂f df ∂r yf ′(r ) = = ∂y dr ∂y r =

and ∂f df ∂r zf ′(r ) = = ∂z dr ∂z r Therefore, ∂f ∂f ˆ ∂f ˆ grad f r ) = iˆ + j+ k ∂x ∂y ∂z xf ′( ) ˆ yf ′( ) ˆ zf ′( ) ˆ = i+ j+ k r r r f ′(r ) = ( yj k ) r f ′(r ) � = r r Now, � ⎛ f ′ r) � ⎞ � grad f r ) × r = ⎜ r ×r ⎝ r ⎠ f ′(r ) � � = ( ) =θ r Example 9.16 Sol.

� � � � 1 � � If v = w × r , prove that w = curl v where w is a constant vector. 2

Let, � r xiˆ yyjˆ zkˆ Therefore, � � � curl v curll w × r

�

w1iˆ w2 ˆj w3 kˆ

9.40

Engineering Mathematics-I

� i = ∇× x � = (

j

k w3 z

y

ˆ (w

iˆ ∂ x

=

) ( ∂ ( ∂y

) ( )

∂ ( x

+ kˆ ˆ(w

Example 9.17

)

⎤ ˆ ∂ j ( x

∂ ( y

)

∂ ( ∂z

)

⎤

) ˆ(

ˆw

= 2[ ˆ Therefore, 1 w= v 2

)

∂ ( z )−

)kˆ

1

kˆ ∂ ∂z

ˆj ∂ ∂y

( =ˆ

ˆ

3

3)

ˆ w kˆ ] = 2 w�

Find a unit normal to the surface x 2 y + 2

= 4 at the point

(2, 2, 3). Sol.

Let xz − 4

, = Now, � ∇ , , ) =(

∂ ˆ ˆj + ∂ kˆ i+ y ∂x ∂z

x2

ˆ

xkˆ

ˆ

− )

At the point (2, 2, 3), the normal to the given surface is − ˆ + 4 ˆ + 4 kˆ The unit normal to the surface is ˆ + 4 ˆ + 4 kˆ 1 ± − + + 2 ˆ) 3 � Determine the constant a so that the vector v + az ) kˆ is solenoidal.

Example 9.18 y Sol.

ˆ

� � Since the vector v is solenoidal div v = 0

x+

ˆ+

9.41

Vector Analysis

⎛ ∂ ∂ ˆ ∂ ˆ⎞ or, ⎜ iˆ + j + k ⎟ x + y)iˆ + ( y − 2 z ) ˆj + ( x + az )kˆ} = 0 ∂ x ∂ y ∂z ⎠ ⎝ ∂ ∂ ∂ or, ( x + 3 y) + ( y − 2 z ) + ( x + az ) = 0 ∂x ∂y ∂z or, 1 +1 1 a=0 or, a = 2 Example 9.19

A particle moves on a curve x

2t 2 , y

t 2 − 4t and z

3t − 5

where t denotes time. Find the components of velocity and acceleration at time t = 1 in the direction iˆ 3 ˆj + 2 kˆ. [WBUT-2002, 2009] � Sol. Let r be the position vector of any point on the given curve. Then � r xiˆ yyjˆj zzkkˆ 2t 2iˆ (t 2 − 4t ) ˆj + (3t 5)kˆ Therefore, velocity of the particle is � � dr v= = 4tiˆ (2t (2t 4) ˆj 3kˆ dt At t = 1, the velocity is [ v ] = 4iˆ 2 ˆj 3kˆ t =11

The acceleration of the particle is � � d 2r a = 2 = 4iˆ 2 ˆj dt At t = 1, the acceleration is � a ] = 4iˆ 2 ˆj t 1

Let � α = iˆ 3 ˆj + 2 kˆ � Therefore, the component of velocity along α is � � v (4iˆ − 2 ˆj kˆ )(iˆ 3 ˆj + 2 kˆ ) 8 14 = � = α 7 14 � Therefore, the component of acceleration along α is � � a α (4iˆ + 2 ˆj )(iˆ − 3 ˆj kˆ ) − 14 = � = α 7 14 Example 9.20

� � � Evaluate [ , r ′, ′′] where r = a cos uiˆ a sin ujˆ bukˆ. [WBUT-2008]

Sol. Here r = a cos uiˆ a sin ujˆ bukˆ.

9.42

Engineering Mathematics-I

dr = du

=

cos uˆ bkˆ

ˆ

d r

ˆ

sin ujˆ

du Therefore, u u

r, ] −

a a −a

bu b 0

sin u bua cosu 2

−ba cosu

+ ba u sin

Example 9.21

2

2

a 2

u

u+

cos

sin u}

u 2

u

2

= a u

Find the angle between the surfaces

z3 3

z 2 x − 5 xyz = 8 at the point (1,0,1).

x y

[WBUT-2008]

Sol. Let x, , Now � f=

+ z 3 3 xyz 5

z

2

x

x − 5 xyz − 8

f ˆ ˆj + f kˆ i+ x y ∂z ˆ + 3( 2 x ˆ (

�

)ˆ

(

iˆ 3 ˆj + kˆ ∂

� ∇

∂y

x =(

ˆj + ∂ kˆ ∂z z) ˆ x

x ˆ

2

� ∇ , 0, 1) = ˆ 4 ˆ + kˆ Let be the angle between the surfaces. Then � � ∇ 1 cos � � = cos−1 = cos

1

(3ˆ

ˆ

4

2 ) 2

21 7 = cos−1 3 27 21

= 5 and

y)kˆ

9.43

Vector Analysis

Find the directional derivative of f x, y, z ) = x 2 yz 4 xz 2 at the point (1, 2, 1) in the direction of the vector 2iˆ ˆj 2 kˆ [WBUT-2008] Example 9.22

Sol. Here, f x, y, z ) = x 2 yz 4 xz 2 � ⎛ ∂f ∂f ˆ ∂f ˆ ⎞ ∂f ∇f = ⎜ iˆ + j + k⎟ ∂y ∂z ⎠ ⎝ ∂x 2 ˆ 2 ˆ = (2 xyz 44zz ) j ( x 2 y + 8 x z )kˆ Therefore, � ∇f ( , − 1) = − ˆj − 6 kˆ Therefore, the directional derivative in the direction 2iˆ � 2i − j − 2 k 1 13 ∇f ( , − ) = (− j − k ) ( i − j − k ) = 3 3 2i − j − 2 k Example 9.23 2 z Sol.

2

ˆj 2 kˆ is

Find the equation of the tangent plane and normal line to the surface

33xy xy − 4 x = 7 at the point (1, 1, ). Here, φ ( x, y, ) = 2 x z 2 3 x y 4 x − 7 Therefore, ∂φ ∂φ ∂φ = 2 z 2 − 3 y − 4, = −3 x, = 4 xz ∂x ∂y ∂z At the point (1, − 1, 2). ∂φ ∂φ ∂φ (1, − 1, 2) = 7, (1 − 1, 2) = −3, (1, − 1, 2) = 8 ∂x ∂y ∂z The equation of the tangent plane at the point (1, − 1, 2) is given by ∂φ ∂φ ∂φ ( x − 1) + ( y + 1) + ( z − 2) = 0 ∂x ∂y ∂z or, 7( 1) 3( y 1) 8( 2) = 0 or, 7 x 3 y 8 z = 26 The equation of the normal to the surface at (1, − 1, 2) is given by ( 1) ( y 1) ( 2) = = ∂φ ∂φ ∂φ ∂z ∂y ∂x or,

( x 1) ( y 1) ( 2) = = 7 8 −3

9.44

Engineering Mathematics-I

Example 9.24 Sol.

Prove

� � � � d ⎡ � dr d 2r ⎤ ⎡ � dr d 3r ⎤ , 2 ⎥ = ⎢r , , 3⎥ ⎢r , dt ⎣ d dt dt ⎦ ⎣ dt dt ⎦

Now, � � d ⎡ � dr d 2r ⎤ , 2⎥ ⎢r , dt ⎣ d dt dt ⎦ � � � � � � ⎡ dr� d dr d 2r ⎤ ⎡ � d 2r d 2r ⎤ ⎡ � dr d 3r ⎤ =⎢ , , 2 ⎥ + ⎢r , 2 , 2 ⎥ + ⎢r , , 3⎥ ⎣ dt dt dt ⎦ ⎣ dt dt ⎦ ⎣ dt dt ⎦ ⎡ � dr� d 3r� ⎤ ⎡ dr� dr� d 2r� ⎤ ⎡ � d 2r� d 2r� ⎤ = ⎢r , , 3 ⎥ since ⎢ , , 2 ⎥ = 0, 0 ⎢r , 2 , 2 ⎥ = 0 ⎣ dt dt ⎦ ⎣ dt dt dt ⎦ ⎣ dt dt ⎦

Example 9.25

Find the constants a and b so that the surface a x 2 − byz = (a + 2) x

will be orthogonal to the surface 4 x 2 y z 3 = 4 at the point (1, 1, 2). Sol.

Since, (1, 1, 2) lies on the surface a x 2 − b yz = (a + 2) x. Therefore, a b a ⇒ b =1 Let the given surfaces are, f x, y, z ) = a x 2 byz by (a + 2) x and

φ ( x, y, ) = 4 x 2 y z 3 − 4 Therefore, � ⎛ ∂ ∂ ∂ ⎞ ∇f x, y, z ) = ⎜ iˆ + ˆj + kˆ ⎟ {a x 2 − byz − (a + 2) x} ∂y ∂z ⎠ ⎝ ∂x = {2 (a 2)}iˆ − zjˆ − ykˆ (a � ⎛ ∂ ∂ ∂ ⎞ ∇φ ( x, y, ) = ⎜ iˆ + ˆj + kˆ ⎟ (4 x y z⎠ ∂ ∂ ∂ ⎝ 2 2 = 8 x yiˆ 4 ˆj 3z kˆ

2

y+

3

− 4)

Now, � � ∇f (1, − 1, 2) = (a − 2)iˆ − 2 ˆj + ˆ ∇φ (1, − 1, 2) = −8iˆ + 4 ˆj + Since the surfaces are orthogonal � � ∇f (1, − 1, 2)∇φ ( , − 1, 2) = 0 or, {( 2) ˆ 2 ˆj kˆ}{−8iˆ 4 ˆj + 12 kˆ} = 0 or, 8(

2) 8 + 12 = 0

kˆ

9.45

Vector Analysis

or, a =

5 2

Therefore, a =

5 and b = 1. 2

� Show that F = ( y sin z sin x )iˆ ( x sin z 2 yz) ˆj ( xy cos z y 2 )kˆ � � is irrotational. Find a scalar function φ such that F = ∇φ . Example 9.26

Sol. Here, � F = ( y sin z sin x )iˆ ( x sin z 2 yz) ˆj ( xy cos z y 2 )kˆ Now, ˆj iˆ kˆ � ∂ ∂ ∂ curl F = ∂x ∂y ∂z ( y sin z sin x ) ( x i z 2 yz ) ( xxy cos z + y 2 ) = ( cos 2 y cos 2 y) ˆ ( y cos y cos ) ˆj (sin sin ) ˆ = 0 � Therefore, F is irrotational. Since, � � F = ∇φ . ∂φ ∂φ ˆ ∂φ ˆ or, (y sin z sin x )iˆ ( x sin z 2 yz ) ˆj ( xy cos z y 2 )kˆ = ˆi + j+ k ∂x ∂y ∂z Therefore, ∂φ ∂φ ∂φ = ( y sin z − sin x ), = ( x sin z + 2 yz y ), = ( xyy z + y2 ) ∂x ∂y ∂z Now, dφ =

∂φ ∂φ ∂φ dz dx + dy + ∂x ∂y ∂z

= ( y sin z − sin si x) x ) dx ( x sin z + 2 yz ) dy + ( xy = (6 y

3 x 2 dy) ( z 3 dx 3 x 2 d ) ( dy ydz )

= 3d ( x 2 y) d (

3

) − d( y )

Integrating, we get,

φ ( x, y, ) = 3 x 2 y xxz 3

yz + c

where c is an arbitrary constant.

y 2 ) dz

9.46

Engineering Mathematics-I

PART-III (VECTOR INTEGRATION) 9.18 GREEN’S THEOREM IN A PLANE 9.18.1 Cartesian Form Let us consider two continuous functions M ( x, y) and N ( x, y) of x and y possess∂M ∂N and in a region R on the two-dimensional ∂y ∂x xy plane bounded by a closed curve C.

ing continuous partial derivatives

Then, Green’s theorem states that ⎛ ∂N ∂M ⎞ ( x, y) dx N ( x, y) dy} = ∫∫ ⎜ − dxdy ⎝ ∂x ∂y ⎟⎠ R

�∫ { C

where the line integral along the curve C is taken in the anticlockwise direction.

9.18.2

Vector Form

� � Let F = M ( x, y) iˆ N ( x, y) ˆj and r

x iˆ y ˆj where M and N have continuous

partial derivatives in a region R on the xy plane bounded by a closed curve C . Then,Green’s theorem states that � � � � �∫ Fd r = ∫∫ (∇ × F ) ⋅ kˆ dxdy C

R

Note: Using Green’s theorem, we are able to transform a double integral over a closed region into a line integral along the boundary of the region and vice-versa. Example 13

Verify Green’s theorem in the plane for

�∫ {( xy + y

2

) dx x 2 dy}

C

where C is the closed curve of the region bounded by y = x and y = x2. [WBUT 2001, 2003] Sol.

We have Green’s theorem as

�∫ { C

⎛ ∂N ∂M ⎞ ( x, y) dx N ( x, y) dy} = ∫∫ ⎜ − ⎟ dxdy. ∂x ∂y ⎠ R ⎝

So, for the given problem M ( x, y) ( xxy y 2 ) and N ( x, y) Then ∂M ∂N = x + 2 y and = 2 x. ∂y ∂x

x2 .

9.47

Vector Analysis Y A (1, 1) C1: y =x 2

C2: y =x R

X

O

Figure 9.10 From Fig. 9.10, it is clear that

�∫ (

dy) = ∫ ( Mdx Md Ndy Ndy)

d

C

C1

( Mdx ∫ (Mdx

Ndy)

C 2

where C1 y = x 2 and C2 y = x. So,

�∫ (

dy) =

d

C

∫ {(

y y 2 )dx x 2 dy}

C 1

∫ {(

y y 2 )dx x 2 dy}

C 2

Now, on the curve C1 y = x 2 , dy = 2 x dx and on the curve C2 y = x, dy = dx. Then from above

�∫ (

dy)

d

C

1

0

Ú

2

= {(

4

) dx

0 1

2

x ◊ 2 xdx}}

Ú {( x ◊ x + x

2

)

1 0

Ú

3

= (3

4

Ú

) dx + 3 x 2 dx

0

1

1

⎡ 3x x ⎤ + ⎥ +[ =⎢ 5⎦ ⎣ 4 0 4

5

3 0 ]1

=−

1 20

Now, we find the double integral 1 y x

⎛ ∂N ∂M ⎞ ∫∫ ⎜⎝ ∂x − ∂y ⎟⎠ dxdy = ∫ 0 R

∫

y x2

{2 x − ( x + 2yy

dxdy

2

d}

9.48

Engineering Mathematics-I 1 y= x

=∫

∫

0 y= x2

1

( x 2 y) dydx = ∫ [ xy − y 2 ]y = x 2 dx 0

1

= ∫ (x4

x3 ) d =

0

y= x

−1 20

In both the cases, we have the same value for the integrals. Hence the Green’s theorem is verified. Example 14

Evaluate using Green’s theorem

�∫ {(cos x sin y − xy) dx + sin x cos ydy} C

where C is the circle x 2 + y 2 = 1. Sol.

[WBUT 2004]

Let M = ( x sin y xy x ) and N i x cos yy. Now by Green’s theorem, we have ⎛ ∂N ∂M ⎞ ( x, y) dx N ( x, y) dy} = ∫∫ ⎜ − ⎟ dxdy ∂x ∂y ⎠ R ⎝

�∫ { C

Now, ∂M ∂N = cos x cos y − x and = cos x cos y ∂y ∂x So,

�∫ {(cos x sin y − xy) dx + sin x cos ydy} C

= ∫∫ (cos x cos y − cos x cos y + x ) dxd dy R

= ∫∫ xdxdy

…(1)

R

Let x = r cosθ and y r sin θ . Then (1) becomes

�∫ {(cos x sin y − xy) dx + sin x cos ydy} C

= ∫∫ xdxdy = ∫ R

=

2π

∫

1

r cosθ ⋅ rdθ dr

θ =00 r =00

(∫

2π

θ =00

θ θ

1 = 0× = 0 3

)( ∫ ) 1

=0

2

9.49

Vector Analysis

9.19 GAUSS’ DIVERGENCE THEOREM 9.19.1 Cartesian Form � Let F

ˆ Then Gauss’ divergence theorem can be written as F1iˆ F2 ˆj F3 k. ⎛ ∂F1

∫∫∫ ⎜⎝ ∂x V

+

∂F2 ∂F3 ⎞ dxdydz = ∫∫ ( F1 dydz F2 dzdx + F3 dxdy) + ∂y ∂z ⎟⎠ C

where V is the volume enclosed by the closed surface S.

9.19.2 Vector Form

� Let F be a vector point function possessing continuous first-order partial derivatives in the volume V bounded by a closed surface S. Then Gauss’ divergence theorem states that � � � —◊ F ◊ nˆ ◊ dS

ÚÚÚ

ÚÚÚÚ

V

S

where nˆ is the outward drawn unit normal vector to the surface S. � �

Example 15

Evaluate the volume integral

� where F = ( x 2 − z 2 )iˆ 2 xyjˆ + (y ( y2

∫∫∫ ∇ ⋅ FdV V

z ) kˆ bounded by the planes x

and x = y = z = 1. Sol.

Here, � F = ( x 2 − z 2 )iˆ 2 xyjˆ + (y ( y2

z )kˆ

and � � ∂ ∂ ∂ ∇ ⋅ F = ( x 2 − z 2 ) + (2 xyy) + ( y 2 + z ) ∂x ∂y ∂z = 2x + 2x 1 = 4x +1 Therefore, 1 1 1

ÚÚÚ V

— ◊ FdV =

Ú Ú Ú (4 x + 1) dxdydz 0 0 0

11

= ∫ ∫ [2

2

]10 dydz

00

11

= ∫ ∫ 3dydz = 3 00

y=z=0

9.50

Engineering Mathematics-I

Verify the divergence theorem for the vector function Example 16 � F 4 xxziˆ − y 2 ˆj + yzkˆ taken over a cube bounded by x = 0, x = 1; y = 0, y = 1; z = 0, z = 1.

[WBUT2002]

Sol. Z C

D

F E B O Y A G X

Figure 9.13 Here in Fig. 9.13, x = 0 and x = 1 are the equations of the planes OBDC and AGEF. y = 0 and y = 1 are the equations of the planes OAFC and BGED. z = 0 and z = 1 are the equations of the planes OAGB and CFED. From the divergence theorem, � � � — ◊ FdV = F ◊ nˆ ◊ dS

ÚÚÚ

ÚÚ

V

S

Here, � F 4 xxziˆ − y 2 ˆj + yzkˆ and � � ∂ ∂ ∂ ∇ ⋅ F = (4 xz ) − ( y 2 ) + ( yz y ) ∂x ∂y ∂z = 4z 2 y y = 4z − y Then � � ∫∫∫ ∇ ⋅ FdV V

111

∫ ∫ ∫ (4 z − y 000

dxdydz

9.51

Vector Analysis 11

= ∫ ∫ [2 00 11

11

yz ]10 dxdy = ∫ ∫ (2 y) dxdy

2

00

1

= ∫∫d

∫ (2

00

0

3 y) dy = . 2

Again, let S1 be side FEGA G , S2 be the side BDCO, S3 be side BDEG, S4 be the side OAFC, S5 be side DCFE, S6 be the side BGAO. Then � � � � � � � ˆ = ∫∫ F ⋅ ndS ˆ dS + ∫∫ F ⋅ ndS ˆ dS + ∫∫ F ⋅ ndS ˆ dS + ∫∫ F ⋅ ndS ˆ + ∫∫ F ⋅ ndS ˆ dS + ∫∫ F ⋅ ndS ˆ ∫∫ F ⋅ ndS S

S

S

1

S

2

S

3

S

4

5

On the surface S1 , x = 1 and the normal nˆ = iˆ, so � ˆ = ∫∫ F ⋅ ndS S

1

1

∫ ∫

ˆ (4 ziˆ − y 2 ˆj + yzkˆ ) idydz

z =0 y =0 1

=

1

1

1

∫ ∫

4 zdydz =

z =0 y=0

∫ [2

1

2 1 ]0

dy =

y=0

∫

2 dy = 2.

y=0

On the surface S2 , x = 0 and the normal nˆ = iˆ, so 1

�

1

ˆ = ∫ ∫ ∫∫ F ⋅ ndS

d =0 (0iˆ y 2 ˆj yzkˆ ) ⋅ (−iˆ dydz

z =0 y =0

S

2

On the surface S3 , y = 1 and the normal nˆ � ˆ = ∫∫ F ⋅ ndS S

3

1

1

∫ ∫

(4 xziˆ

ˆj + zkkˆ ) ⋅ ( ˆj dxdz

z =0 x =0 1

=

ˆj , so

1

∫ ∫

(−1) 1) dxdz = 1

z =0 x =0

On the surface S4 , y = 0 and the normal nˆ � ˆ = ∫∫ F ⋅ ndS S

4

1

ˆj , so

1

∫ ∫

ziˆ 0 ˆj + zkˆ ) ⋅ (− ˆj dxdz = 0 (4 xzi

z =0 x =0

On the surface S5 , z = 1 and the normal nˆ = kˆ, so � ˆ = ∫∫ F ⋅ ndS S

5

1

∫ ∫

(4 xiˆ − y 2 ˆj + ykˆ ) ⋅ (kkˆ dxdy

y =0 x =0 1

=

1

1

∫ ∫

y=0 x =0

ydxdy d =

1 2

S

6

9.52

Engineering Mathematics-I

On the surface S6 , z = 0 and the normal nˆ = kˆ, so � ˆ = ∫∫ F ⋅ ndS S

6

1

1

∫ ∫

(4 xiˆ − y 2 ˆj + 0 kˆ ) ( kˆ ) dxdy = 0

y =0 x =0

Therefore, � 1 3 ˆ = 2 −1+ = ∫∫ F ⋅ ndS 2 2 S So, we have � � ∫∫∫ ∇ ⋅ FdV V

�

ˆ ∫∫ F ⋅ ndS S

Hence the theorem is verified.

9.20 STOKE’S THEOREM � Let F be a continuously differentiable vector point function and S be the surface bounded by a closed curve C. Then Stoke’s theorem states that � � � ˆ F dr cu l F ◊ nds

�Ú

ÚÚÚÚ

C

S

where the curve C is described in the anticlockwise sense and nˆ being the unit normal at any point of S is drawn with a similar sense, in which a right-handed screw would move when rotated in the sense of description of C. Example 17

Verify Stoke’s theorem for

� F = ( x 2 + y 2 )iˆ 2 xyjˆ taken around the rectangle bounded by the lines x = ± a, y = 0, y

b b. [WBUT 2003]

Sol. Y

y =b

D

B

x = −a

x=a

E

y =0

Figure 9.16

X A

9.53

Vector Analysis

Stoke’s theorem states that � � � �∫ F dr = ∫∫ curl F ⋅ nˆ ds C

S

Here, � F = ( x 2 + y 2 )iˆ 2 xyjˆ then � curl F =

iˆ ∂ ∂x x 2 + y2

ˆj ∂ ∂y 2 xyy

kˆ ∂ = 4 ykˆ ∂z 0

Now for the surface S, nˆ kˆ � curl F nˆ = − 4 ykˆ ⋅ kˆ y Therefore, � ∫∫ curl F nˆ ⋅ ds = S

Again � � �∫ F d

�∫ {( x

C

2

b

a

∫ ∫

ab2 .

− 4 ydxdy

y =0 x =− a

+ y 2 )iˆ − 2 xyj yjˆ}{dx d ⋅ iˆ dy ˆj}

C

= ∫ {(

2

y 2 )dx ) dx − 2 xydy}

C

Now from Fig. 9.16, it is obvious that

�∫ {( x

2

+ y2 )d

2 xydy d }=

C

�∫ {( x

2

+ y2 )d

E EA

+ �∫ {( x + y ) d − 2 xydy d }+ 2

2

BD

∫ {( x

2 xydy d}

∫ {( x

2 xydy d }=

2

+ y ) dx − 2 xydy}

DE

∫x

2

dx (since y = 0, dy = 0)

−a

EA

=

2a3 3 b

2 2 �∫ {( x + y )d

AB

2 xydy d } = ∫ ( 2 ay) dy 0

= −ab2

y 2 ) dx − 2 xyddy}

AB 2

a 2 2 �∫ {( x + y )d

2

x = a, dx = 0)

...(1)

9.54

Engineering Mathematics-I −a

2 2 �∫ {( x + y )d

2 xydy d }=

BD

∫ (x

2

b2 ) dx

y = b, dy = 0)

a

=−

2a3 − 2 ab2 3

0

2 2 �∫ {( x + y )d

2 xydy d } = ∫ 2 ay dy

DE

x = −a, dx = 0)

b

= −ab2 So from (1), we have

�∫ {( x

2

+ y2 )d

2 xydy d } = 4 ab2

C

Therefore, � � �∫ F d

�∫

b

C

� F ⋅ nˆ ds

S

Hence Stoke’s theorem is verified. Example 18

Apply Stoke’s theorem to evaluate

is the curve of intersection of x 2 + y 2 + z 2

�∫ ( ydx

zdy + xdz ) where C

C

a 2 and x + z = a.

[WBUT-2001]

Sol. Y

A (a, 0, 0 ) B (0, 0, a)

X R C

Z

Figure 9.17 Since the intersection of x 2 + y 2 + z 2 a 2 and x + z = a is a circle, here the curve C is a circle with diameter AB where A and B have coordinates ( , 0, 0) and (0, 0, ) respectively. a . Therefore, the radius of the circle is 2

9.55

Vector Analysis

We can write

�∫ ( ydx C

� � zdy + xdz ) = ∫ ( yiˆ z ˆj xkˆ )(dxiˆ + dyj yˆ dzkˆ ) = �∫ F dr C

...(1)

C

� � where F = ( yiˆ + z ˆj + xkˆ ) and d d = (dxiˆ dyjˆ + dzkˆ ) By Stoke’s theorem, we have � � � ˆ �∫ Fdr ∫∫ ∫∫ curl F ⋅ nds C

…(2)

S

The unit normal to the surface is nˆ =

kˆ

1 iˆ 2

iˆ kˆ + 2 2

=

By (1) and (2), we can write

�∫ ( ydx C

zdy + xdz ) = ∫∫ ∫ S

⎛ iˆ kˆ ⎞ ( yiˆ z ˆj xkˆ ) ⋅ ⎜ + ⎟ ds 2⎠ ⎝ 2

...(3)

Now, i ∂ curll ( yiˆ z ˆj xkˆ ) = ∂x y

� curl F

j ∂ ∂y z

k ∂ = (iˆ ∂z x

ˆj + kˆ )

Therefore, from (3) and (4) we have

�∫ ( ydx

zdy + xdz )

C

=

⎛ iˆ kˆ ⎞ + ⎟ ds 2⎠ ⎝ 2

∫∫ (iˆ + ˆj + kˆ) ⎜ S

=

−2 2

since

∫∫ ds, S

⎛ a ⎞ ∫∫ ds = area of circle bounded by C = π ⎜⎝ 2 ⎟⎠ S

−2 ⎛ a ⎞ = π⎜ ⎟ 2 ⎝ 2⎠ =

−π a 2 2

2

2

...(4)

9.56

Engineering Mathematics-I

WORKED-OUT EXAMPLES Example 9.27

Verify Green’s theorem for

�∫ [(3 x − 8 y

2

) dx + (4 y 6 xy) dy]

C

where C is the boundary of the region bounded by x = 0, y = 0, and x + y = 1. [WBUT-2002] Sol. Y

B(0, 1) x +y =1 R X

O A(1, 0)

Figure 9.11 Green’s theorem states that ⎛ ∂N ∂M ⎞ �∫ ( d dy) = ∫∫ ⎜⎝ ∂x − ∂y ⎟⎠ dxdy C R where R is the region bounded by the closed curve C constisting of the lines OA, AB and BO, where A and B are the points (1, 0) and (0, 1) respectively. Here, ∂N ∂M M ( x, y) (3 x − 8 y 2 ), N ( x, y) = (4 y 6 xy x ) d = −6 y, = 16 y ∂x ∂y Now,

�∫ (

dy) = �∫ [(3 x 8 y 2 ) dx + (4y (4 y − 6 xy) dy]

d

C

=

C

∫ [(3 x − 8 y

2

) dx + (4 y 6 xy) dy]

OA

+

∫ [(3 x − 8 y

BO

∫ [(3 x − 8 y

AB 2

) dx + (4 y − 6 xy) dy]

2

) dx (4 y − 6 y) dy]

9.57

Vector Analysis

On OA, y = 0, so dy = 0. On AB, x + y = 1, so dy = so dx = 0. Therefore,

�∫ (

dx. On BO, x = 0,

dy)

d

C

1

∫

=

0

0

∫ [{3

3xdx

x =0

8(1 − x )2} dx + {4(1 x ) 6 x(1 x )}( d )]

x =1

d ∫ 4 ⋅ ydy

y=1

0

⎛1⎞ = 3 ⎜ ⎟ + ∫ ( 14 ⎝ 2 ⎠ x=1

2

⎛ −1 ⎞ 5 29 x − 12) dx + 4 ⎜ ⎟ = ⎝ 2 ⎠ 3

Again, ⎛ ∂N

∫∫ ⎜⎝ ∂x − R

∂M ⎞ ⎟ dxdy = ∫∫ (−6 y + 16 y) dxdy ∂y ⎠ R 1 1− x

=

1

1 5 (1 x )2 d = 2 3 x =0

∫ ∫ 10 ydxdy = ∫

x =0 y=0

Hence Green’s theorem is verified. Example 9.28

Verify Green’s theorem in the plane for

�∫ (

2

d

xydy)

C

where C is the square in the xy plane given by x = 0, y = 0, x = a, y = a (a > 0). [WBUT-2005] y =a

Sol.

B

C

R x =a

A

O

Figure 9.12 Green’s theorem states that

�∫ ( C

d

⎛ ∂N ∂M ⎞ dy) = ∫∫ ⎜ − ⎟ dxdy ∂x ∂y ⎠ R ⎝

9.58

Engineering Mathematics-I

where R is the region bounded by the closed curve C. Here, x 2 , N ( x, y )

M ( x, y )

xy x and

∂N ∂M = y, =0 ∂x ∂y

Now,

�∫ (

dy) = ∫ ( x 2 dx + xydy d )

d

C

C

=

∫ (x

2

OA

∫(

∫ (x

2

dx + xydy d )+

∫ (x

2

d

dx + xydy) d )

AB

2

xydy)) +

d

BC

xydy)

CO

On OA, y = 0, so dy = 0. On AB, x = a, so dx = 0. On BC, y dy = 0. On CO, x = 0, so dx = 0

a so a,

Therefore, a

�∫ ( d

dy) =

∫

x =0

C

0

a

∫

x 2 dx

aydy

y=0 a

∫

0

x 2 dx +

x=a

∫

0 ⋅ ydy

y= a 0

a

⎡ x3 ⎤ ⎡ y2 ⎤ ⎡ x3 ⎤ = ⎢ ⎥ +a⎢ ⎥ + ⎢ ⎥ ⎣ 3 ⎦0 ⎣ 2 ⎦0 ⎣ 3 ⎦a = Again, ⎛ ∂N

∫∫ ⎜⎝ ∂x − R

a3 a3 a3 a3 + − = 3 2 3 2

∂M ⎞ ⎟ dxdy = ∫∫ ( y − ∂y ⎠ R

dxdy a

⎡ y2 ⎤ = ∫ ∫ ydxdy d = ∫ ⎢ ⎥ dx 2⎦ x =0 ⎣ x =0 y=0 0 a

a

a

a

=

a2 a3 dx = 2 2 x =0

∫

Hence, Green’s theorem is verified. Example 9.29

� Verify Gauss’s divergence theorem for F

cylindrical region bounded by x 2 + y 2 = 9, z = 0, z = 2.

yiˆ xxjˆ z 2 kˆ over the [WBUT-2003, 2007]

9.59

Vector Analysis

Sol.

Z z =2

V

Y O x 2 +y 2 =9 X

Figure 9.14 Gauss’s divergence theorem states that � � � ∫∫∫ ∇ ⋅ F dv ∫∫ F ⋅ nˆ ds V

S

where the volume V bounded by a closed surface S and nˆ is the outward drawn unit normal vector to the surface S. Here, V is the volume bounded by surface S x 2

y 2 = 9 z = 0, z = 2.

Now � F yiˆ xxjˆ z 2 kˆ So � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ ˆ ˆ 2 ˆ ∇ ⋅ F = ⎜ iˆ + j + k ⎟ yi + xj + z k ) ∂y ∂z ⎠ ⎝ ∂x

z

For a particular z, x 2 + y 2 = 9 is a circle. Therefore, −3 ≤ ≤ 3. and for a particular value of x, − 9 − x 2 ≤ y ≤ 9 − x 2 Therefore, � � ∫∫ ∇ ⋅ F dv = V

9− x2

3

∫

x = −3

∫

2

2 zdxdydz =

z =0

y=

9 x

∫

x= 3

3

∫

4 dxdy = 2

9− x2

3

9− x2

∫

x= 3

∫

y − 9− x

3

=

2

∫

x= 3

4[ y]

∫

y

9 x2

− 9− x2

9 x

dx

2

⎡ z2 ⎤ 2 ⎢ ⎥ dxxdy ⎣2⎦ 2 0

9.60

Engineering Mathematics-I 3

⎡ x 9 − x2 9 x⎤ = 4 ∫ 2 9 − x dx = 16 ⎢ + sin −1 ⎥ 2 2 3⎥ ⎢⎣ x= 3 ⎦0 3

2

⎡9 ⎤ = 16 ⎢ sin −1 1 = 36π 2 ⎣ ⎦

…(1)

Now � � � ˆ d = ∫∫ F ⋅ nds ˆ + ∫∫ Fˆ ⋅ nds ˆ d + ∫∫ F ⋅ nds ˆ ∫∫ F ⋅ nds S

S 1

S 2

…(2)

S 3

where S1 is the circular base in the plane z = 0, S2 is the circular top in the plane z = 2 and S3 is the curved surface of the cylinder, given by x 2 + y 2 = 9. � ˆ , nˆ is normal to S1 , so nˆ kˆ and z = 0. Therefore In the integral ∫∫ F ⋅ nds S 1

� ˆ = ∫∫ yi + xj + ∫∫ F ⋅ nds

(

S 1

S 1

) ⋅ kˆ

ˆ kdxdy = 0.

�

ˆ , ∫∫ F ⋅ nds

In the integral

nˆ is normal to S2 , so nˆ

…(3) kˆ and z = 2. Therefore

S 2

� ˆ ∫∫ F ⋅ nds

∫∫ ( yiˆ + xjˆ + 2

S 2

2

)

ˆ ⋅ kˆ kdxdy

S 2

=4

ÚÚ dxdy = 4 ◊ (Area of S ) 2

S 2

= 4 ⋅ π (3)2 = 36π

…(4)

� Again S3 is represented by x + y − 9 = 0. So, ∇( vector on S3 . So the unit normal vector, 2

� 2 ∇(x (x ˆn = � ∇(x ( x2 =

y 2 9) y 2 9)

=

2

2 xiˆ 2 yjˆ 4 x2

4 y2

2 xiˆ 2 yjˆ [since x 2 + y 2 = 9] 36

xiˆ + yjˆ 3 Therefore, � ˆ = ∫∫ ∫∫ F ⋅ nds =

S 3

2

S 3

=

ˆ

ˆ

( yiˆ + xjˆ + z kˆ ) xi 3+ yj ds

2 xyds 3 ∫∫ S 3

2

2

+ y − 9) is normal

9.61

Vector Analysis

On S3 , x 2 + y 2 = 9, let x = 3 cos θ , y = 3sin θ and ds = 3dzddθ . So, for the entire surface, θ varies from 0 to 2π and z varies from 0 to 2 � 2 ˆ = ∫∫ xyds ∫∫ F ⋅ nds 3S S 3

3

2π

=

2

2 ∫ 27 sinθ cos θ dθ dz 3 θ ∫=0 z=0 2π

= 18

∫

i 2θ dθ z ]20

θ =0 2π

⎡ cos 2θ ⎤ = 36 ⎢ − =0 2 ⎥⎦ 0 ⎣

…(5)

Using (3), (4) and (5) in (1), � ˆ = 0 + 36π + 0 = 36π . ∫∫ F ⋅ nds

…(6)

S

From (1) and (6), it is clear that Gauss’s divergence theorem is verified.

Example 9.30

∫∫ {x

2

Evaluate by divergence theorem

dydz y 2 dzdx + 2 z(( y

y)dxdy}

S

where S is the surface of the cube 0

x 1, 0 ≤ y ≤ 1, 0

z 1.

[WBUT-2005]

Sol. Z C (0, 0, 1)

D

F E

O

Y

A (1, 0, 0)

B (0, 1, 0) G

X

Figure 9.15

9.62

Engineering Mathematics-I

� If F

F1iˆ F2 ˆj F3 kˆ then the divergence theorem can be written as

⎛ ∂F1

∫∫∫ ⎜⎝ ∂x

∂F2 ∂F3 ⎞ + ⎟ dxdydz = ∫∫∫∫ ( F1dydz + F2 dzdx + F3 dxdy) ∂y ∂z ⎠ S

+

V

where V is the volume enclosed by the closed surface S. Here, F1 = x 2 , F2 and

y 2 , F3

2 z ( xyy x y)

∂F1 ∂F ∂F = 2 x, 2 = 2 y, 3 = 2( xyy − x − y) ∂x ∂y ∂z

Therefore,

∫∫ {x

2

dydz y 2 dzdx + 2 z(( y

y)dxdy}

S

∫∫∫ = {2 x V

=

1

1

22yy + 2( xy − x y)} dxdydz 1

∫ ∫ ∫ {2 x

2 y + 2( xy − x y)} dxdydz 2y

x =0 y=0 z =0 1

=

1

1

∫ ∫ ∫

1

2 xydxdydz =

x =0 y=0 z =0

1

∫ ∫

2 xy[ z ]10 dxdy

x =0 y=0

2 ⎤1

1

1 ⎡y ⎡ x2 ⎤ 1 = ∫ 2 x ⎢ ⎥ dx = ∫ xdx = ⎢ ⎥ = ⎣ 2 ⎦0 ⎣ 2 ⎦0 2 x =0 x =0 1

Example 9.31

� � � Using divergence theorem, evaluate ∫∫ u ⋅ nds where u

S � and S is the sphere x 2 + y 2 + z 2 = 9 and n is outward normal to S.

Sol.

xiˆ yj yˆ zkˆ

[WBUT-2006]

Let V be the volume of the sphere x 2 + y 2 + z 2 = 9 with a radius of 3. By the divergence theorem, � � � � ∫∫ u ⋅ nds ∫∫∫ ∇ ⋅ udv S

V

Now, � � ⎛ ∂ ∂ ˆ ∂ ˆ⎞ ˆ ˆ ˆ ∇ ⋅ u = ⎜ iˆ + j + k ⎟ ( xi + yj + zk ) ∂ x ∂ y ∂z ⎠ ⎝ ∂ ∂ ∂ = ( x) ( y) ( z) = 3 ∂x ∂y ∂z

9.63

Vector Analysis

Therefore, � �

� �

∫∫ u ⋅ nds ∫∫∫ ∇ ⋅ udv = ∫∫∫ 3dv = 3∫∫∫ dxdydz S

V

V

V

= 3 × (volume of the sphere of radius = 3

i )

⎛4 ⎞ = 3 × ⎜ π 33 = 108π ⎝3 ⎠ Example 9.32

� Use Stoke’s theorem to prove div curl F = 0.

[WBUT-2002]

Sol.

S1

S2

V

Figure 9.18 Let V be any volume enclosed by a closed surface S. Let us divide V by a plane into two surfaces S1 and S2 , and let C denote the common closed curve bounding both the portions. Therefore, by Stoke’s theorem � � ˆ ∫∫∫ div curl Fdv ∫∫ curl F ⋅ nds V

S

� � ˆ + ∫∫ curl F ⋅ nds ˆ = ∫∫ curl F ⋅ nds S 1

� � = ∫ F dr C

S 2

� � ∫ F ⋅ dr = 0

C

where negative sign is taken in the second integral as it is traversed in the direction opposite to that of the first. Since the above result is true for every volume element V, we have � div curl F = 0 Example 9.33

� Verify Stoke’s theorem for A 2 yiˆ 3 x ˆj z 2 kˆ where S is the

upper half surface of the sphere x 2 + y 2 + z 2 = 9 and C is its boundary. [WBUT-2004]

9.64

Engineering Mathematics-I Z

Sol.

S X C

R

Y

Figure 9.19 � Stoke’s theorem states that for any vector function A � � � ˆ , �∫ Adr ∫∫ curl A ⋅ nds C

S

where nˆ being the outward drawn unit normal at any point of S. The boundary of C of S is a circle in xy plane whose equation is x2 + y2 = 9, z = 0. Let the parametric equation of C be x = 3 Therefore, � � Ad (2 yi y ˆ 3 xj xˆ z 2 kˆ )(dxiˆ + dyj yˆ dzkˆ )

�∫ C

t, y

3 i t, 0

t

2π .

�∫ C

= ∫ (2 ydx 3 xdy − z 2 dz ) C

2π

=9

∫(

2 si

2

t +3

2

) dt = 9π

…(1)

t =0

Now, iˆ � ∂ curl A = ∂x 2y

ˆj ∂ ∂y 3x

kˆ ∂ = iˆ(0) ˆj(0) j (0) kˆ(3 ∂z −z2

) kˆ

Therefore, � dxdy ˆ = ∫∫ kˆ nˆ = ∫∫ dxdy ∫∫ curl A ⋅ nds kˆ nˆ S R R where R is the region enclosed by the circle x 2 + y 2 = 9. So,

∫∫ dxdy = (Area of the circle x R

=

(3)2 = 9π .

2

+ y 2 = 9)

9.65

Vector Analysis

Therefore, � ˆ = 9π . ∫∫ curl A ⋅ nds S

From (1) and (2), it is clear that Stoke’s theorem is verified. � Example 9.34 Verify Stokes theorem for F = (2 x y)iˆ − yz 2 ˆj y 2 zkˆ where S is the upper half surface of the sphere x 2 + y 2 + z 2 = 1 and C is its boundary. [WBUT-2006] Sol. Z

S X C

R

Y

Figure 9.20 � Stoke’s theorem states that for any vector function F � � � ˆ , �∫ Fdr ∫∫ curl F ⋅ nds C

S

where nˆ being the outward drawn unit normal at any point of S. Here, the boundary C of S is a circle in the x y plane whose equation is x 2 + y 2 = 1, z = 0. Let the parametric equation of C is x Now � � �∫ Fd C

t, y

i t , z = 0, 0 ≤ t ≤ 2π

�∫ {(2 x − y)iˆ − yz

2ˆ

j

y 2 zkˆ}(dxiˆ dyjˆ + dzkˆ )

C

= ∫ {(2 x y)dx − yz 2 dy − y 2 zdz} C

= ∫ (2 C

y) dx [since, on C , z = 0, dz = 0]

9.66

Engineering Mathematics-I 2π

=

∫

2π

(2

i )(

i t ) dt =

0

⎛

∫⎝ 0

si 2 t +

1 cos 2t ⎞ ⎟ dt 2 ⎠

2π

⎡ cos 2t t sin 2t ⎤ + − =⎢ =π 2 4 ⎥⎦ 0 ⎣ 2

…(1)

Again, � curl F =

ˆj iˆ ∂ ∂ ∂x ∂y 2 x y − yz 2

kˆ ∂ = kˆ ∂z − y2 z

Therefore, � ˆ ∫∫ curl F ⋅ nds S

� dxdy r C = ∫∫ curl F nˆ where R is the region bounded by the circle nˆ kˆ R dxdy = ∫∫ (kˆ nˆ ) = dxdy nˆ k ∫∫ R R = (1)2 (area of the circle x 2 + y 2 = 1) = π

…(2)

Therefore by (1) and (2), Stoke’s theorem is verified. Example 9.35

� Verify Stoke’s theorem for A ( y − z 2)iˆ + (yz ( yz y 4) ˆj xxzkˆ over

the surface of the cube x

y = z = 0 and x

y = z = 2 above x y plane. [WBUT-2007]

Sol. Z D (0, 0, 2)

C (0, 2, 2)

A (2, 0, 2) B (2, 2, 2) G (0, 2, 0) O Y E (2, 0, 0) F(2, 2, 0) X

Figure 9.21

9.67

Vector Analysis

Let us denote S1 , S2 S3 S4 and S5 as the five faces ABFE, DCGO, ABCD C , BCGF and ADOE respectively above XOY plane. Therefore, the boundary of the surface is the square EFGO. Stoke’s theorem states that � � � ˆ = ∫ A dr curl A ⋅ nds ∫∫ S S +S S +S 1 2 3 4 5

EFGO

Now, iˆ � ∂ curl A = ∂x ( y − z + 2)

ˆj kˆ ∂ ∂ ∂y ∂z ( yz 4 ) − xz

= y ˆ ( z −1) 1) ˆj − kˆ Therefore, � ˆ = ∫∫ {− yi y ˆ + ( z 1) ˆj − kˆ} iˆ ⋅ ds d , since on S1 , nˆ = iˆ ∫∫ curl A ⋅ nds S 1

S 1

22

= ∫∫ 00

2

2 ⎡ y2 ⎤ = ∫ − y[ z ]20 dy = 2 ⎢ ⎥ = 4 ⎣ 2 ⎦0 0

Similarly, � ˆ = ∫∫ {− yiˆ + ( z 1) ˆj − kˆ} ( iˆ ) ⋅ ds d , since on S2 , nˆ = −iˆ ∫∫ curl A ⋅ nds S 2

S 2

22

= ∫ ∫ ydydz = 4 00

� ˆ = ∫∫ {− yiˆ + ( z 1) ˆj − kˆ} (kˆ ) ⋅ ds, since on S3 , nˆ = kˆ ∫∫ curl A ⋅ nds S 3

S 3

22

= ∫ ∫ dxdy = − 4 00

� ˆ = ∫∫ {− yiˆ + ( z 1) ˆj − kˆ ∫∫ curl A ⋅ nds S

4

S

4

22

= ∫ ∫ ( z 1)) dxdz = 0 00

ˆj ) ⋅ ds, since on S , nˆ = ˆj 4

…(1)

9.68

Engineering Mathematics-I

�

ˆ = ∫∫ {− yiˆ + ( z ∫∫ curl A ⋅ nds S 5

1) ˆj − kˆ

ˆj ) ⋅ ds since on S , nˆ = − ˆj 5

S 5

22

= ∫ ∫ ( z − 1)) dxdz = 0 00

Therefore,

∫∫

S S +S S +S 1 2 3 4 5

� � � ˆ = ∫∫ curl A ⋅ nds ˆ + ∫∫ curl A ⋅ nds ˆ curl A ⋅ nds S 1

S 2

� � � ˆ + ∫∫ curl A ⋅ nds ˆ + ∫∫ curl A ⋅ nds ˆ + ∫∫ curl A ⋅ nds S 3

S

S 5

4

= 4+4 4 0 0= 4

…(2)

Now, � � A dr

∫

∫

EFGO

EFGO

∫

=

{( y − z 2)iˆ + ( yz + 4) ˆj xzk x ˆ}{dxiˆ dyjˆ + dzkˆ )

{( y

((yz yz + 4) dy − xzdz}

2) d

EFGO

=

∫ {( y

2) d

((yz yz + 4) dy − xzdz}

OE

+ ∫ {( y − z + 2) dx + ( y + 4) dy −

d}

EF

+

∫ {( y − z + 2)dx + ( y

+ 4) dy −

d}

∫ {( y − z + 2)dx + ( y

+ 4) dy −

d}

FG

+

GO 2

=

∫

2

2 dx

x =00

∫

y=0

0

4 dy

∫

x =22

0

4 dx +

∫

4 dy = − 4

…(3)

y=2

Therefore, From (1) and (2) we conclude that Stoke’s theorem (1) is satisfied. Example 9.36

� If A = (3 x 2 6 y)iˆ 14 yz y ˆj 20 xz x 2 kˆ, evaluate

�

∫A

� dr from (0, 0, 0)

C

to (1, 1, 1) along the path C given by x = t , y Sol.

� Let r

t 2 and z

� xiˆ yyjˆ z kˆ then dr = dxiˆ dyjˆ + dzkˆ.

t 3.

[WBUT-2002].

9.69

Vector Analysis

Therefore, � � 2 2 ∫ A dr ∫ [(3 x + 6 y)iˆ − 14 yz ˆj 20 xxz kˆ](dxiˆ dyjˆ + dzkˆ)

C

C

= ∫ [(3 x 2 + 6 y) dx − 14 yzdy + 20 xz 2 dz ] C 1

= ∫ [(3t 2 6 2 ) dt − 14t 5 (2tdt ) 20 7 (3 2 d )] 0

1

= ∫ [9

2

28

6

60 9 ]dt

0

1

⎡ t3 t7 t10 ⎤ = ⎢9 − 28 + 60 ⎥ 7 10 ⎦ ⎣ 3 0 = 3 − 4 6 = 5.

EXERCISES

Short and Long Answer Type Questions 1. Show that the vector ˆ

ˆj

kˆ is perpendicular to the vector 4iˆ 6 ˆj + 5 ˆ.

2. Determine λ so that λiˆ 4 ˆj + 3kˆ and 3� λ ˆj 2 kˆ are perpendicular. [Ans : λ = 6] 3. Find the angle between the vectors iˆ 2 ˆj − 2 kˆ and 2iˆ

ˆj 2 kˆ. ⎡ −1 ⎛ 4 ⎞ ⎤ ⎢ Ans : cos ⎜ 3 ⎟ ⎥ ⎝ ⎠⎦ ⎣

4. Find a vector of magnitude 5, perpendicular to both the vectors 2 ˆ ˆj 3kˆ and −5 ⎡ ⎤ iˆ 2 ˆj + kˆ. ( + j + k )⎥ ⎢ Ans : 3 ⎣ ⎦ � � � � � � 5. If a, b , c be three vectors such that a b + c = 0, show that � � � � � � −3 a) a b + b c + c a = 2 � � � � � � b) a b b c = c × a � � � � � � � � � 6. If a, b , c be three vectors such that a b + c = 0, a = 3, b = 4, c = 5, show that � � � � � � a b + b c + c a = −25.

9.70

Engineering Mathematics-I

7. If

�

� b = 4, find the values of

for which the vectors a

� b and a Ans :

will be perpendicular to each other. 8. Prove that the following vectors are coplanar: ˆ, 3ˆ ˆ + kˆ ˆ + kˆ iˆ + ˆ a) ˆ b) ˆ

ˆ+ ˆ

ˆ + 3ˆ

ˆ,

ˆj

� b 3 4

ˆ

9. Determine the value of the constant λ so that the vectors and + k are coplanar. 10. Prove that � � � � � � � � � � � a) c , ]=[ e c, , ] [ f ][c , , e] � � b) b [ , ] � � � � � � c) b ×c × a + c a×b = 0 11. If a and c are perpendicular to each other then show that � � � ( b × c are also perpendicular to each other.

+

�

�

×c

and

ˆ ˆ+ ˆ 12. Find the volume of the parallelepiped whose edges are along � � ˆ and c ˆ. ˆ ˆ iˆ iˆ � � � � � � � � � � � � 13. If × γ α and γ α then show that , γ are mutually perpendicular. � 14. If = ˆ ˆ (

� d � d ˆ =( , find × at t = 2. dt dt � � d r � dr � � � 15. If a os nt + b nt then prove that r × = n(a × b ) and 2 + n2r = 0. dt dt � � � � � � � � � � 16. If c d be vectors such that b c then show that ( ) and ( ) are collinear. ˆ + ˆ a × kˆ . 17. Show that iˆ a × iˆ ˆ × � 18. If r x x ˆ e x ˆ x y ˆ then show that the value of ∂ 2r r × 2 at (1, 0) is − ˆ + 12 ˆ + 12 ˆ ). 2 x ∂y � � ˆ ˆ z kˆ and 19. If zi + − x kˆ then show that the value of � ∂2 ˆ +2ˆ × at (1, 0, 2) is 1) ˆ and

�

20. Find the directional derivative of the direction of 2 ˆ ˆ 2 k.

ˆ

,

ˆ

=

xz at the point (1, 2, 1) in

21. Find the angle of intersection of the spheres 2 x z 7 = 0 at (4, 3, 2).

z 2 = 29

and

9.71

Vector Analysis

22. Find the equations of the tangent line and normal plane to the curve of intersection of the surfaces x 2 + y 2 + z 2 = 9 and z = x 2 y 2 3 at the point (2, 1, 2). 23. Show that the vector yiˆ + x ˆj is both solenoidal and irrotational. � 24. If f x, y, z ) be a scalar point function such that ∇ 2 f ( x, y, z ) = 0, then show that � ∇ 2 f is irrotational as well as solenoidal. � � 25. If r xi yj y zk and r = r , prove that � � 1 a) ∇( ⋅ −2 ) = 2 r � � � b) div(a × r ) = 0 where a is a constant vector. c) div grad n ) = (n + 1)r n−2 � � ⎛ a×r ⎞ d) div ⎜ � n ⎟ = 0 ⎝ r ⎠ � e) curl( 1 ) = 0 � f) curl( 3 ) = 0 26. Verify Green’s theorem in the plane for ∫ [( y

i x) x)d

xdy] where C is the

C

⎛π ⎞ ⎛π ⎞ triangle whose vertices are (0, 0), ⎜ , 0 ⎟ and ⎜ , 1 ⎟ . ⎝2 ⎠ ⎝2 ⎠ 27. Evaluate by Green’s theorem in the plane for ∫ (e − x sin ydx + e − x cos ydy) where C C

⎛ π⎞ ⎛ π⎞ is the rectangle with vertices (0, 0), ( , 0), π , ⎟ and ⎜ 0, ⎟ ⎝ 2⎠ ⎝ 2⎠ [Ans : 2(e −π − 1) ] � 2 2 ˆ 28. Verify Stoke’s theorem for F = ( x + y )i 2 xyjˆ taken round the rectangle bounded x = ± a, y = 0, y = b. 29. Evaluate by Stoke’s theorem x 2 + y 2 = 1, z = y 2 .

d ), ∫ ( yzdx + z xdy + xydz

where C is the curve

C

[Ans : 0] � 30. Verify Gauss’s divergence theorem for F yiˆ x ˆj z 2 kˆ over the cylindrical region bounded by x 2 + y 2 = a 2 , z = 0 and z h. h � � 2ˆ ˆ ˆ ˆ 31. Evaluate ∫∫ F ⋅ nds where F = 3 xzi x + y j − 3 yzk and S is the surface of the cube S

bounded by x = 0, x = 2; y 0 y 2; z = 0, z = 2 and nˆ is the outward drawn unit normal to the surface S. � � 32. Evaluate ∫∫∫ Fdv where F 2 ziˆ x ˆj ykˆ and V is the region bounded by the V

surfaces x = 0, y = 0, x = 2, y = 4, z

x 2 and z = 2. 32 ˆ ⎡ ˆ ⎤ ⎢ Ans : 15 (3i 5 j ) ⎥ ⎣ ⎦

9.72

Engineering Mathematics-I

�

33. Evaluate

�

∫∫∫ Fdv where F = (2 x

2

3z )i 2 xyj − 4 xk and V is the region bounded

v

by the surfaces x = 0, y = 0, z = 0, and 2 x 22yy + z = 4. 8⎤ ⎡ ⎢ Ans : 3 ⎥ ⎣ ⎦ 34. Evaluate

∫ (sin zdx

cos xdy + sin ydz ) by Stoke’s theorem where C is the

C

boundary of the rectangle 0

x

,0

y 1, = 3. [Ans : 2]

35. Evaluate

∫∫ {(z

2

x )dydz − xydxdz + 3zdxdz} where S is the surface of the closed

S

region bounded by y 2

4 z and the planes x = 0, x = 3, z = 0. [Ans : 16]

Multiple-Choice Questions � � 1. The value of for which the vectors a λiˆ 4 ˆj kˆ and b 3iˆ + λ ˆj kˆ are perpendicular to each other is a) –6 b) 6 c) 3 d) 2 2. A unit vector perpendicular to each of the vectors and j k is 1 1 1 1 ( j k) ( j k ) c) ( j k ) d) ( j k) b) a) 3 3 3 3 � � � � � � � � 3. If for the vectors a and b , a + b = a − b then a and b are a) parallel b) collinear c) perpendicular d) none of these � � � � 4. If a and b are two mutually perpendicular vectors then a b = � � � � � � a) a b b) a b c) a b d) 0 � � � � 5. ( b ) × ( b ) = � � � � � � a) 2(a b ) b) a b c) 2(a b ) d) 0 � ˆ ˆ ˆ � � ˆ 6. The value of λ for which the vectors a = i + j k , b i − 4 ˆj and c = iˆ λ ˆj are coplanar is 3 1 2 5 b) c) d) a) 5 5 5 3 � � � � � 7. If a = 3iˆ − 2 ˆj kˆ, b iˆ − kˆ, then ( b ) ⋅ a = b) iˆ kˆ c) 0 d) 2 8. The vector xiˆ + yjˆ + zkˆ is perpendicular to the vector 2iˆ 5 ˆj + 11kˆ when a) x = 2, y = 3, z = 11 b) x = 2, y = −3, z = 11 c) x = 2, y = 3, z = 11 d) x = 2, y = −3, z = 1 a) iˆ

ˆj + kˆ

kˆ

9.73

Vector Analysis

9. The angle between the vectors 2iˆ 2 ˆj − kˆ and 3iˆ 4 kˆ is ⎛1⎞ ⎛2⎞ ⎛2⎞ b) cos−1 ⎜ ⎟ c) cos−1 ⎜ ⎟ d) none of these a) cos−1 ⎜ ⎟ ⎝ 15 ⎠ ⎝5⎠ ⎝ 15 ⎠ �2 �2 � � � � 10. If a = b , then the vectors a b and a b are a) parallel b) perpendicular c) inclined at an angle 30� with each other

d) none of these

� � d π � (a b ) at t = is 11. If a = 3t 2iˆ t ˆj t 3 kˆ and b = sin tiˆ 2 cos t ˆj , then dt 2 π3 ˆ 3 2ˆ ˆ π3 ˆ 3 2ˆ ˆ π j − k b) π j+k a) i i 4 4 4 4 π3 ˆ 3 2ˆ ˆ π j + k d) none of these c) i 4 4 � � 1 3ˆ dr d 2r � t k then the value of 12. If r 2t iˆ t 2 ˆj × 2 at t = 1 is 3 dt dt ˆ ˆ ˆ ˆ ˆ ˆ a) 2i 4 j + 4 k b) 2i 4 j + 4 k c) −2iˆ + 4 ˆj + 4 kˆ d) 2iˆ 4 ˆj − 4 kˆ 13. If f x, y, z ) = x 3 3 yyz 2 then grad f at (1, 1, 1) is a) 3iˆ 3 ˆj + 3kˆ

b) iˆ

ˆj + 2 kˆ

c) 3iˆ 3 ˆj + 6 kˆ

14. A normal vector to the plane x + 2 y + 3z − 1 = 0 is a) 2 ˆ ˆj 3kˆ b) iˆ 2 ˆj + 3kˆ c) iˆ 2 ˆj − 3kˆ � 15. If φ = x 3 3 yz 2 then ∇ 2ϕ is a) x + 6 y b) 6x + y � � 16. If r xiˆ yj yˆ z kˆ then curl r = a) 3iˆ b) 0

c) 6 x 6 y c) iˆ

ˆj + kˆ

d) none of these d) iˆ 2 ˆj − 3kˆ d) x + y

d) none of these

17. The magnitude of the vector drawn in the direction perpendicular to the surface x 2 + 2yy 2 z 2 = 7 at (1, –1, 2) is 2 3 b) c) 3 d) 6 a) 3 2 � 18. If r 2 x 2 + y 2 + z 2 then ∇ 2 (l ) = 1 1 b) 2 r r � � 19. If F = xyiˆ + yz ˆj z xkˆ then curl F = a)

a) xiˆ + yjˆ + zkˆ

b) yiˆ + z ˆj + kˆ

c) r2

d) r

c) − yiˆ − z ˆj − xkˆ d) yiˆ + z ˆj + xkˆ

9.74

Engineering Mathematics-I

� 20. The value of t for which F = (txy t − z 3 )iˆ (t 2) x 2 ˆj (1 t ) xz 2 kˆ is irrotational is a) 0

b) 4

c) –4 d) 2 � ˆ ˆ 3 y)i + ( y 2 z ) j ( x + az )kˆ is solenoidal is 21. The value of for which F = ( x + 3y a) 2 b) –2 c) 0 d) none of these 22. The directional derivative of = xyz at (1, 1, 1) in the direction ˆj is a)

3 b)1 c) 2 d) 4 � � � � � � 23. If v w × r where w is a constant vector and r = xiˆ yyjˆj z kˆ, then div v = a) 1 b) 0 c) 2 d) none of these � 24. The value of grad ( y z ) d r from (0, 1, 1) to (1, 2, 0) is a) 3 b) –1 c) 1 d) 2 25. If ( x, y, ) = c represent the equation of a surface then normal to this surface is b) div grad φ ) c) curl(grad φ ) d) none of these a) grad φ Answers: 1. (a) 2. (b) 9. (c) 10. (b) 17. (d) 18. (a) 25. (a)

3. (c) 11. (a) 19. (c)

4. (c) 12. (a) 20. (b)

5. (c) 13. (c) 21. (b)

6. (d) 14. (b) 22. (b)

7. (c) 15. (c) 23. (b)

8. (d) 16. (b) 24. (a)

SOLUTION OF UNIVERSITY QUESTIONS (W.B.U.T.) B.TECH SEM-1 (NEW) 2010 MATHEMATICS-I (M 101) Time Alloted: 3 Hours

Full Marks: 70

GROUP-A (Multiple-Choice Type) 1. Choose the correct alternative for any ten of the following: *(i) If a, b are the roots of the equation x2 – 3x + 2 = 0 then

(a) 6

(b)

0

a

b 1

0 0 is -a a

3 2

(10 ¥ 1 = 10)

b

(c) – 6

(d) 3

Solution: Since a, b are the roots of the equation, we have a + b = 3 and Now

ab = 2.

a b 0 0 = – ab (a + b) = – 6 -a a

0 b 1

Hence, the correct alternative is (c) - 6 . *(ii) If y = eax + b then (y5)0 = (a) aeb

(b) a5eb

Solution: Here, y =

eax + b

fi y5

(c) abeax = a5e ax + b;

(d) none of these

therefore (y5) 0 = a5eb.

Hence, the correct alternative is (b) a 5eb . *(iii) If Rolle’s theorem is applied to f (x) = x (x2 – 1) in [0, 1] then c = 1 1 (d) 3 3 2 Solution: If Rolle’s theorem is applied to f (x) = x (x – 1), we have 1 f ¢(c) = 0 fi 3c2 – 1 = 0 fi c = ± . 3 (a) 1

(b) 0

Level of difficulty:- *Low, **Medium, ***High.

(c) –

SQP1.2

Engineering Mathematics-I

Since by the conditions of Rolle’s theorem 0 < c < 1, we take c = Hence, the alternative option is (d) **(iv) If u + v = x, uv = y then (a)

1 u-v

1 . 3

1 . 3

∂(u, v) = ∂ ( x, y )

(b) uv

(c) u + v

(d)

u . v

(d)

2.(6.4.2) 7.5.3.1

Solution: We know ∂(u, v) 1 = ∂ ( x, y ) ∂ ( x, y ) ∂ (u, v)

Now

∂x ∂ ( x, y ) ∂u = ∂y ∂ (u, v) ∂u So,

∂x 1 1 ∂v = = u – v. ∂y v u ∂v

∂ (u, v) 1 is (a) . u-v ∂ ( x , y) p 2

*(v) The value of

Ú sin q dq is 7

-

(a)

p 2

6.4.2 7.5.3.1

(b)

6! 7!

(c) 0

Solution: The correct alternative is (c) 0 , since sin7 q is an odd function of q.

{

*(vi) The sequence ( - 1)n (a) convergent

}

1 is n

(b) oscillatory

(c) divergent

(d) none of these

Solution: The correct alternative is (b) oscillatory .      **(vii) If a = 3iˆ – 2 ˆj + kˆ , b = 2iˆ – kˆ then (a ¥ b ). a is equal to (a) iˆ + ˆj + kˆ (b) iˆ + kˆ (c) iˆ – kˆ Solution: Here,

(d) 0 ˆj iˆ kˆ   a ¥ b = 3 - 2 1 = 2iˆ + 5ˆj + 4 kˆ . 2 0 -1

SQP1.3

Solutions of University Questions (W.B.U.T.)

Therefore,    (a ¥ b ) ◊ a = 2.3 + 5.(–2) + 4.1 = 0. Hence the correct alternative is (d) 0 . È cos q ***(viii) The matrix Í Î - sin q

sin q ˘ ˙ is cos q ˚

(a) symmetric (c) singular

(b) skew-symmetric (d) orthogonal

Solution: Here, È cos q Í Î - sin q

sin q ˘ È cos q ˙ cos q ˚ ÍÎ - sin q

sin q ˘ ˙ cos q ˚

T

cos q = ÍÈ Î sin q

sin q ˘ È cos q - sin q ˘ ˙ Í ˙ cos q ˚ Î sin q cos q ˚

È1 0˘ = Í ˙ Î0 1˚ Hence, the correct alternative is (d) orthogonal . **(ix) The value of t for which  f = (x + 3y) iˆ + (y – 2z) ˆj + (x + tz) kˆ is solenoidal is (a) 2 (b) – 2 (c) 0 (d) 1 Solution: The correct alternative is (b) - 2 . See Example 9.18. ***(x) The distance between two parallel planes x + 2y – z = 4 and 2x + 4y – 2z = 3 is (a)

5 24

(b)

5 24

11 24

(c)

Solution: The correct alternative is (a)

(d) none of these

5 . 24

**(xi) In the MV Theorem, f (h) = f (0) + hf ¢(qh);

0 < q < 1.

1 and h = 3 then value of q is 1+ x 1 1 (a) 1 (b) (c) (d) none of these 3 2 1 Solution: Here, f (x) = and h = 3, so by given relation 1+ x If f (x) =

f (h) = f (0) + hf ¢(qh);

0 1 .

GROUP-B (Short-Answer type Questions) (3 ¥ 5 = 15)

Answer any three of the following: **2. If y = (x 2 – 1)n then show that (x 2 – 1) yn + 2 + 2xyn + 1 – n (n + 1)yn = 0. Hence, find yn (0). Solution: See Example 3.14. **3. Using MVT, prove that x > tan–1 x >

x p ,0 10) (a) a10

(b) 10!a10

Solution: The correct alternative is (c) � � (viii) If for any two vectors a and b, � � a+b = � � then a and b are (a) parallel (c) perpendicular

(c) 0

(d) 10!

0 � � a-b (b) collinear (d) none of these

Solution: The correct alternative is (d ) orthogonal (ix) If A–1 =

1 È 3 1˘ Í ˙ then A = 7 Î-1 2 ˚

È 3 1˘ (a) Í ˙ Î-1 2 ˚

È 2 1˘ (b) Í ˙ Î-1 3˚

(c)

1 È2 -1˘ Í ˙ 7 Î 1 3˚

È 2 1˘ Solution: The correct alternative is (b) Í ˙ Î-1 3˚

È2 -1˘ (d) Í ˙ Î 1 3˚

Solutions of University Questions (W.B.U.T.)

SQP2.3

p 2

Ú

(x) The reduction formula of In = cosn xdx is 0

(a) In = ÊÁ n - 1ˆ˜ In – 1 Ë n ¯

(b) In = ÊÁ n ˆ˜ In Ë n - 1¯

n - 1ˆ (c) In = ÊÁ In – 2 Ë n ˜¯

(d) none of these

Ê n - 1ˆ Solution: The correct alternative is (c) I n = Á I Ë n ˜¯ n - 2 (xi) The series

n2

 2n n =1

2

+1

is

(a) convergent (c) oscillatory

(b) divergent (d) none of these

Solution: The correct alternative is (b) divergent (xii) Lagrange’s Mean value Theorem is obtained from Cauchy’s Theorem for two functions f (x) and g (x) by putting g (x) = 1 (a) 1 (b) x2 (c) x (d) x Solution: The correct alternative is (c) x

Group-B (Short-answer Type Questions) (3 ¥ 5 = 15)

Answer any three of the following:

2. Prove that every square matrix can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix. Solution: See Theorem 1.2 of Page 1.10. 3. By Laplace’s method, prove that a b -b a -c - d -d

c d d -c = (a2 + b2 + c2 + d2)2 a b

c -b

(consider minors of order 2).

a

SQP2.4

Engineering Mathematics-I

Solution: Here, we expand the given determinant by Laplace’s method of expansion in terms of a minor of order 2 considering the first two rows as follows: a

b

c

-b a -c - d -d

d

d -c a b

c -b

a

=

a b a b a c -d b ¥ (–1)(1 + 2) + (1 + 2) + ¥ (–1)(1 + 2) + (1 + 3) -b a -b a -b d c a

+

-c b a d -d a b c ¥ (–1)(1 + 2) + (1 + 4) + ¥ (–1)(1 + 2) + (2 + 3) - d a -b - c c -b a d

+

b

d

a -c

¥ (–1)(1 + 2) + (2 + 4)

-c

a

-d

-b

+

c

d

d -c

¥ (–1)(1 + 2) + (3 + 4)

-c - d -d

c

= (a2 + b2) (a2 + b2) + (ad + bc) (ad + bc) + (–ac + bd) (–ac + bd) + (bd – ac) (bd – ac) + (bc + ad) (bc + ad) + (c2 + d2) (c2 + d2) = (a2 + b2)2 + 2(ad + bc)2 + 2(– ac + bd)2 + (c2 + d2)2 = (a2 + b2)2 + 2[(ad + bc)2 + (– ac + bd)2] + (c2 + d2)2 = (a2 + b2)2 + 2[a2 d2 + b2 c2 – 2adbc + a2 c2 + b2 d2 – 2acbd] + (c2 + d2)2 = (a2 + b2)2 + 2(a2 + b2) (c2 + d2) + (c2 + d2)2 = (a2 + b2 + c2 + d2)2

4. If 2x =

1 m y

-

+y

1 m

then prove that

(x2 – 1)yn + 2 + (2n + 1)xyn +1 + (n2 – m2)yn = 0

SQP2.5

Solutions of University Questions (W.B.U.T.)

Solution: Here, we have 1

-

2x = y m + y 1

or,

y m – 2x +

1

1 m

=0

1 ym

2

Ê 1ˆ Ê 1ˆ Á y m ˜ – 2x Á y m ˜ + 1 = 0 ÁË ˜¯ ÁË ˜¯

or,

Applying the rule for finding solution of the above quadratic equation, we get 1

ym = fi

2 x ± (2 x )2 - 4.1.1 2 = x ± x -1 2

(

(

y = x ± x2 - 1

)

)

m

(i)

Differentiating (i) w.r.t. x, we have m -1

(

)

(

)

y1 = m x ± x 2 - 1

= m x ± x2 - 1

(

m -1

2

= ±m x ± x - 1

i.e.,

y1 = ± m

(x ±

Ê 1 ◊ Á1 ± ◊ 2 ÁË

x2 - 1

) )

ˆ ◊ 2 x˜ ˜¯ x2 - 1 1

Ê x2 - 1 ± x ˆ ◊Á ˜ ÁË x 2 - 1 ˜¯

m -1

(x ± ◊

x2 - 1

)

2

x -1

m

=±

x2 - 1

my x2 - 1

(ii)

Squaring (ii) and simplifying, we get (y1)2 (x2 – 1) = m2 y2

(iii)

Again differentiating (iii) w.r.t. x, we have 2y1 y2(x2 – 1) + (y1)2 2x = m2 2y ◊ y1 fi

y2(x2 – 1) + y1x – m2y = 0

(iv)

SQP2.6

Engineering Mathematics-I

Now applying Leibnitz’s theorem, we differentiate (iv) n times w.r.t. x, {y2(x2 – 1)}n + {y1x}n – {m2y}n = 0 fi

[{y2}n ◊ (x2 – 1) + nC1{y2}n – 1 ◊ (2x) + nC2{y2}n – 2 ◊ (2)] + [{y1}n ◊ x + nC1{y1}n – 1 ◊ 1] – m2yn = 0

fi

(x2 – 1)yn + 2 + 2nxyn + 1 + n(n – 1)yn + xyn + 1 + nyn – m2yn = 0

fi

(x2 – 1)yn + 2 + (2n + 1)xyn + 1 + (n2 – m2)yn = 0

Ê yˆ 5. If u = x f Á ˜ + gÊÁ y ˆ˜ then show that Ë x¯ Ë x¯ x2

∂2 u ∂x 2

+ 2xy

∂ 2u ∂ 2u + y2 2 = 0. ∂x ∂y ∂y

Solution: See Example 21 of Page 6.19. 6. Show that the area bounded by a simple closed curve C is given by 1 (xdy – ydx). 2

�Ú C

Solution: We know that Green’s theorem states the following:

�Ú {M(x, y) dx + N (x, y) dy} = ÚÚ ÊÁË ∂∂Nx - ∂∂My ˆ˜¯ dx dy C

(i)

R

where the region R on the two-dimensional xy plane is bounded by a simple closed curve C and the line integral along the curve C is taken in the anticlockwise direction. Here, comparing LHS of (i) with

�Ú (xdy – ydx), we have C

∂M ∂N = –1, =1 ∂y ∂x

M = –y, N = x fi

(ii)

Therefore using (ii) in (i), we get

�Ú (xdy – ydx) = ÚÚ [1 – (–1)] dx dy C

R

=2

ÚÚ dx dy R

= 2 ¥ [Area bounded by C] fi Area bounded by C =

1 2

�Ú (xdy – ydx) C

Hence, the area bounded by a simple closed curve C is given by 1 2

�Ú (xdy – ydx). C

SQP2.7

Solutions of University Questions (W.B.U.T.)

Group-c (long-answer Type Questions) (3 ¥ 15 = 45)

Answer any three of the following: 7. (i) If

verify

Ê yˆ Ê xˆ f (x, y) = x2 tan–1 Á ˜ – y2 tan–1 Á ˜ , Ë x¯ Ë y¯ fxy = fyx.

Solution: See Example 6.6 of Page 6.34. (ii) State Rolle’s theorem and examine if you can apply the same for f (x) = tan x in [0, p]. Solution: See Example 4.3 of Page 4.33. (iii) Find the value of l and m for which x+y+z =3 2x – y + 3z = 4 5x – y + lz = m has (a) a unique solution, (b) many solutions (c) no solution. Solution: If we write the system of linear equations in the matrix form as AX = B then the coefficient matrix of the system of linear equations is Ê 1 1 1ˆ A = Á 2 -1 3˜ Á ˜ Ë 5 -1 l ¯ and the augmented matrix is Ê 1 1 1 3ˆ A = Á 2 -1 3 4˜ Á ˜ Ë 5 -1 l m ¯ The system of equations has a unique solution when the determinant of the coefficient matrix is not equal to zero. 1 1 1 det A = 2 -1 3 5 -1 l = 1(–l + 3) – 1(2l – 15) + 1(– 2 + 5) = – 3l + 21 Therefore, for det A π 0 fi –3l + 21 π 0 fi l π 7, the system of equations have unique solution.

SQP2.8

Engineering Mathematics-I

When l = 7, the augmented matrix becomes Ê 1 1 1 3ˆ A = Á 2 -1 3 4˜ Á ˜ Ë 5 -1 7 m ¯ Applying elementary row operations on the matrix A, we have 1 1 3 ˆ Ê 1 1 1 3ˆ Ê1 Á 0 -3 1 A = Á 2 -1 3 4˜ �R������������������ 2 R , R 5 R 2 ˜ �R�������� - 2R2� 1 3 �1 Á Á ˜ 2 ˜ 3 Ë 5 -1 7 m ¯ Ë 0 - 6 2 m - 15¯ 1 1 3 ˆ Ê1 Á 0 -3 1 -2 ˜ Á ˜ 0 0 m - 11¯ Ë0

1 Ê1 1 Á 1 Ê 1ˆ ÁË - ˜¯ R2 Á 0 1 3 �Á 3 ��������� ÁË 0 0 0

3 ˆ 2 ˜ ˜ 3 ˜ m - 11˜¯

4 Ê 1 0 Á 3 Á 1 1 - R2� Á 0 1 �R������ Á 3 ÁË 0 0 0

7 ˆ 3 ˜ ˜ 2 ˜ 3 ˜ m - 11˜¯

The system of equations is consistent when Rank A = Rank A and this is possible for m – 11 = 0 fi m = 11. In this case, Rank A = Rank A = 2, which is less then number of unknowns (= 3) and the system has infinitely many solutions. Again, if m – 11 π 0 fi m π 11. then Rank A = 2 and Rank A = 3, i.e., Rank A π Rank A, and so the system of equations is inconsistent and correspondingly the system has no solution. Summarizing the above, the system of equations has (a) a unique solution when l π 7 (b) infinitely many solutions when l = 7 and m = 11 (c) no solution when l = 7 and m π 11.

SQP2.9

Solutions of University Questions (W.B.U.T.)

8. (i) Find the maxima and minima of the function f (x, y) = x3 + y3 – 63(x + y) + 12xy Find also the saddle points. Solution: See Example 6.24 of Page 6.51. (ii) State Leibnitz’s test for alternating series and apply it to examine the convergence of 1 1 1 1- 2 + 2 - 2 + 2 3 4 Solution: See Example 8.27 of Page 8.42. (iii) Applying Lagrange’s Mean value Theorem, prove that x £ log(1 + x) £ x, for all x > 0. 1+ x Solution: See Example 7 of Page 4.11. 9. (i) If y = em sin–1 x, show that (1 – x2)yn + 2 – (2n + 1)xyn + 1 – (n2 + m2)yn = 0. Hence, find yn when x = 0. Solution: Here, we are given that y = em sin–1 x Differentiating (i) w.r.t. x, we have y1 = em sin–1 x ◊ m i.e.,

y1 =

(i) Ê ÁË

1 1- x

2

ˆ ˜¯

my

(ii)

1 - x2

Squaring (ii) and simplifying, we get (y1)2 (1 – x2) = m2 y2

(iii)

Again differentiating (iii) w.r.t. x, we have 2y1 y2(1 – x2) + (y1)2 (–2x) = m2 2y ◊ y1 fi

y2(1 – x2) – y1x – m2y = 0

(iv)

Now applying Leibnitz’s theorem, we differentiate (iv) n times w.r.t. x, {y2(1 – x2)}n – {y1x}n – {m2y}n = 0. fi

[{y2}n ◊ (1 – x2) + nC1{y2}n – 1 ◊ (– 2x) + nC2{y2}n – 2 ◊ (–2)] – [{y1}n ◊ x + nC1{y1}n – 1 ◊ 1] – m2 yn = 0

SQP2.10

Engineering Mathematics-I

fi

(1 – x2)yn + 2 – 2nxyn +1 – n(n – 1)yn – xyn +1 – nyn – m2yn = 0

fi

(1 – x2)yn + 2 – (2n + 1)xyn +1 – (n2 + m2)yn = 0

(v)

Calculation of yn when x = 0, i.e., (yn)0: Putting x = 0 in (v), we have (yn + 2)0 = (n2 + m2) (yn)0 Replacing n by n – 2, we get (yn)0 = [(n – 2)2 + m2] (yn – 2)0

(vi)

Replacing n by n – 2 in (vi), we get (yn – 2)0 = [(n – 4)2 + m2] (yn – 4)0

(vii)

Using (vii) in (vi), (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] (yn – 4)0 Similarly, we have (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] [(n – 6)2 + m2] (yn – 6)0

(viii)

Proceeding in a similar manner we have from (viii), when n is odd as the following: (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [32 + m2] [12 + m2] (y1)0

(ix)

From (ii), we have (y1)0 = m. Using this in (ix), we get (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [32 + m2] [12 + m2] m, when n is odd. Also proceeding in a similar manner we have from (viii), when n is even as the following: (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [42 + m2] [22 + m2] (y2)0

(x)

From (iv), we have (y2)0 = m2. Using this in (x), we get (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [42 + m2] [22 + m2]m2, when n is even. � � � � � � ��� � � � (ii) Prove that ÈÎa + b b + c c + a ˘˚ = 2 ÈÎa b c ˘˚ , where a b c are three vectors. Solution: Using the definition of scalar triple product, we write � � � � � � � � � � � � ÈÎa + b b + c c + a ˘˚ = ( a + b ) ◊ ÈÎ(b + c ) ¥ (c + a )˘˚ � � � � � � � � = ( a + b ) ◊ ÈÎ(b + c ) ¥ c + (b + c ) ¥ a ˘˚ � � � � � � � � � � = ( a + b ) ◊ ÈÎ(b ¥ c ) + (c ¥ c ) + (b ¥ a ) + (c ¥ a )˘˚

Solutions of University Questions (W.B.U.T.)

SQP2.11

� � � � � � � � = ( a + b ) ◊ ÈÎ(b ¥ c ) + (b ¥ a ) + (c ¥ a )˘˚ ,

� � � since c ¥ c = 0 � � � � � � � � � � � � = a ◊ (b ¥ c ) + a ◊ (b ¥ a ) + a ◊ (c ¥ a ) + b ◊ (b ¥ c ) � � � � � � + b ◊ (b ¥ a ) + b ◊ (c ¥ a ) � � �� = ÈÎa� b c� ˘˚ + ÈÎa� b a� ˘˚ + [ a� c� a� ] + ÈÎb b c� ˘˚ ��� ��� + ÎÈb b a ˚˘ + ÎÈb c a ˚˘

��� By the property of scalar triple product of vectors, we have ÈÎa b a ˘˚ = ��� ��� ��� 0, [ a c a ] = 0, ÈÎb b c ˘˚ = 0, ÈÎb b a ˘˚ = 0 (since two vectors in the product ��� ��� are same) and ÈÎb c a ˘˚ = ÈÎa b c ˘˚ . Using this in the above, we get � � � � � � ��� ��� ��� ÈÎa + b b + c c + a ˘˚ = ÈÎa b c ˘˚ + 0 + 0 + 0 + 0 + ÈÎa b c ˘˚ = 2 ÈÎa b c ˘˚ (iii) Find the directional derivative of f = xyz at (1, 1, 1) in the direction 2iˆ - ˆj - 2 kˆ . Solution: Here, it is given that f = xyz. Then � ∂ ∂ ∂ —f = Ê iˆ + ˆj + kˆ ˆ f ÁË ∂x ∂y ∂z ˜¯ ∂f ∂f ˆ Ê ∂f = Á iˆ + ˆj + kˆ ˜ ∂y ∂z ¯ Ë ∂x = yziˆ + xzjˆ + xykˆ So,

� ÈΗf ˘˚ = iˆ + ˆj + kˆ (1,1,1)

Here we are to find the directional derivative in the direction 2iˆ - ˆj - 2 kˆ . The unit vector in the direction of 2iˆ - ˆj - 2 kˆ is given by aˆ =

2iˆ - ˆj - 2 kˆ

=

22 + (-1)2 + (-2)2

2iˆ - ˆj - 2 kˆ 3

Then the required directional derivative of f = xyz at (1, 1, 1) in the direction of 2iˆ - ˆj - 2 kˆ is given by � Ê 2iˆ - ˆj - 2 kˆ ˆ 1 ÈΗf ˘˚ ◊ aˆ = iˆ + ˆj + kˆ ◊ Á ˜¯ = (1,1,1) Ë 3 3

(

)

SQP2.12

Engineering Mathematics-I

10. (i) Prove that b2 + c 2

a2

a2

b2

c2 + a2

b2

c2

c2

a 2 + b2

= 4a2 b2

2

Solution: See Example 1.16 of Page 1.43. � (ii) State the Divergence Theorem of Gauss. verify divergence theorem for F = yiˆ + xjˆ + z 2 kˆ over the cylindrical region bounded by x2 + y2 = 9, z = 0, z = 2. Solution: See Example 9.29 of Page 9.58. (iii) Test the series for convergence: 1p

2p +

3p +

2q

3q

+º 4q

Solution: Let us consider the given series as

Â

an =

1p

2p +

2q

3p +

3q

+º 4q

n=1

Then an =

np (n + 1)q

np

=

1ˆ Ê nq Á 1 + ˜ Ë n¯

= q

1 1ˆ Ê nq - p Á 1 + ˜ Ë n¯

q

Let us consider another series

Âb = Â n 1 n

n=1

q- p

n =1

which is convergent for q – p > 1 and divergent for q – p £ 1. Now we have lim

n

Since

Âb

n

an 1 = lim = 1. q bn n 1ˆ Ê ÁË 1 + ˜¯ n

is convergent for q – p > 1 and divergent for q – p £ 1, by

n=1

comparison test,

Âa

q – p £ 1.

n=1

n

is convergent for q – p > 1 and divergent for

SQP2.13

Solutions of University Questions (W.B.U.T.) p 2

11. (i) Obtain a reduction formula for

p 2

Ú sin x dx. Hence obtain Ú sin x dx. 9

n

0

0

Solution: See Section 5.2 of Page 5.1. � � � � � � � � (ii) Given two vectors a = 3i – j , b = 2i + j – 3k . Express b in the form � � � � � � b1 + b2, where b1 is parallel to a and b2 is perpendicular to a . Solution: See Example 9.1 of Page 9.17. � (iii) Show that A = (6xy + z3)iˆ + (3x2 – z) ˆj + (3xz2 – y) kˆ is irrotational. Find � � the scalar function f, such that A = —f. Solution: See Example 9.11 of Page 9.36.

SOLUTIONS OF UNIVERSITY QUESTIONS (W.B.U.T.) B.TECH SEM-1 (NEW) 2012 MATHEMATICS-II (M 101) Time Alloted: 3 Hours

Full Marks: 70

GROUP-A (Multiple Choice Type Questions) 1. Choose the correct alternatives for any ten of the following: (10 ¥ 1 = 10) 1 *(i) The sequence ( -1) n is n (a) convergent (b) oscillatory

{

}

(c) divergent cos q *(ii) The matrix ÈÍ Î - sin q

(d) none of these sin q ˘ is cos q ˙˚

(a) symmetric

(b) skew-symmetric

(c) singular

(d) orthogonal

**(iii) The value of t for which  f = (x + 3y) i + (y − 2z) j + (x + tz)  k is solenoidal is (a) 2

(b) −2

*(iv) The series Â

1 is convergent if np (b) p £ 1

(a) p ≥ 1

**(v) The two eigenvalues of the matrix È 2 -2 2˘ Í 1 1˙ A = Í1 ˙ ÍÎ 1 2 -1˙˚

Level of difficulty:- *Low, **Medium, ***High.

(c) 0

(d) 1

(c) p > 1

(d) p < 1

Engineering Mathematics-I

SQP3.2

are 2 and −2. The third eigenvalue is (a) 1 (b) 0

(c) 3

(d) 2

*(vi) If Rolle’s theorem is applied to f(x) = x(x2 − 1) in [0, 1] then c = 1 1 (a) 1 (b) 0 (c) (d) 3 3 x3 + y 3

*(vii) If u =

x2 + y 2

, find the value of n so that xux + yuy = nu.

(a) 0 (b) 2 1 (c) (d) none of these 2 **(viii) The nth derivative of sin(5x + 3) is (a) 5n cos(5x + 3) (b) 5n sin( np + 5x + 3) 2 np (c) 5n cos( + 5x + 3) (d) none of these 2 **(ix) The value of

Ú ( xdx - dy) where C is a line joining (0, 1) to (1, 0) is

C

(a) 0

(b)

3 2

(c)

1 2

(d)

2 3

p 2

*(x) The value of

Úp sin 7 q dq is

-

2

6.4.2 7.5.3.1 (d) none of these

(a) 0 (c)

(b)

6! 7!

*(xi) If the characteristic equation of a matrix A is X3 + 3X2 + 5X + 9 = 0 then determinant of the matix is (a) 7

(b) 5

(c) 6

(d) 9

*(xii) Let A and B be two square matrices and A−1, B−1 exist. Then (AB)−1 is a) A−1B−1

b) B−1A−1

c) AB

d) none of these

Answers (i) (b)

(ii) (d)

(iii) (b)

(iv) (c)

(v) (d)

(vi) (d)

(vii) (b)

(viii) (b)

(ix) (b)

(x) (a)

(xi) (d)

(xii) (b)

Solutions of University Questions (W.B.U.T.)

SQP3.3

GROUP B (Short Answer Type Questions) Answer any three of the following:

(3 × 15 = 45)

**2. Verify Rolle’s theorem for the function f (x) = |x|, −1 £ x £ 1 Solution: See Example 2 of Page 4.4. ***3. A and B are orthogonal matrices and |A| + |B| = 0. Prove that A + B is singular. Solution: Since A and B are orthogonal, |A| = ±1 π 0, |B| = ±1 π 0. Also, ATA = AAT = I and BTB = BBT = I. Let us consider C = A + B. Then CT = (A + B)T = AT + BT fi CTA = ATA + BTA fi

CTA = I + BTA

fi

BCTA = BI + BBT A

fi

BCTA = B + IA

fi

BCTA = B + A = A + B

Now, |A + B| = |BCTA| = |B||CT||A| = −|A| |C| |A| , since |A| + |B| = 0 and |CT| = |C| = −{|A|}2 |C| = −|C|, since |A| = ±1 = −|A + B| \

2 |A + B| = 0

\

|A + B| = 0.

Hence, A + B is singular. **4. Find the nth derivative of

x2 + 1 ( x - 1)( x - 2)( x - 3)

Solution: Let us consider y=

x2 + 1 A B C = + + ( x - 1)( x - 2)( x - 3) ( x - 1) ( x - 2) ( x - 3)

x2 + 1 A( x - 2)( x - 3) + B( x - 1)( x - 3) + C ( x - 1)( x - 2) = ( x - 1)( x - 2)( x - 3) ( x - 1)( x - 2)( x - 3) fi

x2 + 1 = A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2)

Engineering Mathematics-I

SQP3.4

Substituting x = 1, 2, 3 we have respectively 2 = A(1 − 2)(1 − 3) i.e., A=1 5 = B(2 − 1)(2 − 3) i.e., B = −5. and 10 = C(3 − 1)(3 − 2) i.e., C=5 Therefore, x2 1 5 5 y= = – + ( x - 1)( x - 2)( x - 3) ( x - 1) ( x - 2) ( x - 3) since yn =

( -1) n .n !.a n (ax + b)

n +1

when y =

yn =

1 , we get from the above, ax + b

( -1) n .n ! ( x - 2)

-5

n +1

( -1) n .n ! ( x - 2)

n +1

+5

( -1) n .n ! ( x - 3) n +1

**5. Let Ï xy , ( x, y ) π (0, 0)¸ Ô Ô 2 f(x, y) = Ì x + y ˝ Ô 0, ( x, y ) = (0, 0)Ô˛ Ó Evaluate fxy(0, 0) and fyx(0, 0). Solution: We have fxy(0, 0) = lim

f y (h, 0) - f y (0, 0)

hÆ0

(1)

h

Now, hk h.0 2 f (h, k ) - f (h, 0) h + 02 = lim h + k fy(h, 0) = lim k Æ0 k Æ0 k k = lim

k Æ0

h h + k2

=1

and fy(0, 0) = lim

k Æ0

f (0, k ) - f (0, 0) 0-0 = lim =0 k Æ0 k k

Using the above two results in (1), we obtain fxy(0, 0) = lim

f y (h, 0) - f y (0, 0)

hÆ0

which does not exist. Again, we have

h

= lim

hÆ0

1- 0 1 = lim h Æ 0 h h

Solutions of University Questions (W.B.U.T.)

fyx(0, 0) = lim

k Æ0

SQP3.5

f x (0, k ) - f x (0, 0) k

(2)

Now, hk -0 2 f (h, k ) - f (0, k ) = lim h + k fx(0, k) = lim hÆ0 hÆ0 h h k 1 = = lim hÆ 0 h + k 2 k and f (h, 0) - f (0, 0) 0-0 = lim =0 hÆ0 h h Using the above two results in (2), we obtain f (0, k ) - f x (0, 0) fyx(0, 0) = lim k Æ0 k 1 -0 1 k = lim 2 = lim k Æ0 k k Æ0 k which also does not exist.   **6. Find div F, and curl F where  F = grad(x3 + y3 + z3 − 3xyz) fx(0, 0) = lim

hÆ0

Solution: See Example 9.9 of Page 9.35.

GROUP C (Long Answer Type Questions) Answer any three of the following:

(3 × 15 = 45)

***7. (a) If u = x2 − 2y, v = x + y + z, w = x − 2y + 3z, find Solution: Here, ∂u ∂x ∂v ∂(u , v, w) = ∂x ∂ ( x, y , z ) ∂w ∂x

∂u ∂y ∂v ∂y ∂w ∂y

∂u ∂z ∂v ∂z ∂w ∂z

2 x -2 0 = 1 1 1 = 10x + 4 1 -2 3

∂(u , v, w) ∂ ( x, y , z )

Engineering Mathematics-I

SQP3.6

1 a a 2 - bg **(b) Prove that 1 b 1 g

b 2 - ga = 0 g 2 - ab

Solution: See Example 1.14 of Page 1.41. **(c) If v = f (x2 + 2yz, y2 + 2zx), prove that ∂v ∂v (y2 − zx) ∂v + (x2 – yz) + (z2 – xy) =0 ∂y ∂z ∂x Solution: See Example 6.7 of Page 6.35. n

***8. (a) If q = t

-r2 e 4t

, find what value of n will make 1 ∂ Ê 2 ∂q ˆ ∂q . Ár ˜= r 2 ∂r Ë ∂r ¯ ∂t

Solution: Since q = tn e

-r2 4t

, we have

∂q = t ne ∂r

-r2 4t

Now,

qr -2r ˆ =◊ ÊÁ . ˜ Ë 4t ¯ 2t

1 ∂ Ê 2 ∂q ˆ 1 ∂ Ê Ê q r ˆ ˆ 1 ∂ Ê q r3 ˆ ÁË r ˜¯ = 2 Ár ◊Á - ˜˜ = 2 ∂r r ∂r r ∂r Ë Ë 2t ¯ ¯ r 2 ∂r ÁË 2t ˜¯ = =

1 r 3 ∂q 1 3r 2q 1 r 3 Ê q r ˆ 1 3r 2q = Á- ˜ r 2 2t Ë 2t ¯ r 2 2t r 2 2t ∂r r 2 2t

qr2 4t

2

-

3q 2t

Again, ∂q t -1 = nt e ∂t

-r2 4t

+t

n

-r2 e 4t

+t

By the given condition, we have 1 ∂ Ê 2 ∂q ˆ ∂q Ár ˜= r 2 ∂r Ë ∂r ¯ ∂t q r 2 3q nq q r 2 = + 2 t 4t 2 2t 4t which implies n = -

3 2

n

-r2 e 4t

Ê r 2 ˆ nq q r 2 ◊Á 2 ˜ = + 2 t Ë 4t ¯ 4t

Solutions of University Questions (W.B.U.T.)

SQP3.7

**(b) Using the mean-value theorem, prove that 0
0 2 ! 3! 4 !

Solution: See Example 8.26 of Page 8.42. **(b) Show that

         Èa + b , b + c , c + a ˘ = 2 Èa, b , c ˘ Î ˚ Î ˚ Solution: See Problem 9(ii) of Page SQP 2.10.

**(c) If f(v2 − x2, v2 − y2, v2 − z2) = 0, where v is a function of x, y, z then show that 1 ∂v 1 ∂v 1 ∂v 1 + + = x ∂x y ∂y z ∂z v Solution: See Example 17 of Page 6.16. ***9.

(a) Determine the conditions under which the system of equations x+y+z=1 x + 2y − z = k 5x + 7y + az = k2 admits (i) only one solution, (ii) no solution, and (iii) many solutions. Solution: See Problem 8(a) of Page SQP1.5.

=

� ***(b) Verify the divergence theorem for the vector function F = 4 xzi�- y 2 � j �- y � � j + yzk taken over a cube bounded by x = 0, x = 1; y = 0, y = 1; z = 0, z = 1.

Solutions of University Questions (W.B.U.T.)

SQP5.7

Solution: See Example 16 of Page 9.50. **(c) If In =

p 2 0

Ú

x n sin xdx(n > 1) then prove that

p In + n(n − 1)In−2 = n ÊÁ ˆ˜ Ë 2¯

n -1

Solution: See Example 5.7 of Page 5.27. **10. (a) Verify Lagrange’s mean-value theorem at [−1, 1] for f(x) = x sin 1 , x π 0 x = 0, x = 0 Solution: See Example 4.4 of Page 4.34. y y ***(b) If u = xf ÊÁ ˆ˜ + g ÊÁ ˆ˜ then show that Ë x¯ Ë x¯ x

∂u ∂u y ∂ 2u ∂ 2u ∂ 2u +y + y2 2 = 0 = xf ÊÁ ˆ˜ and x 2 2 + 2 xy Ë x¯ ∂x ∂y ∂ x∂ y ∂x ∂y

Solution: See Example 21 of Page 6.19. È 2 3 16 5 ˘ Í4 5 6 7 ˙ **(c) Find the rank of the following matrix Í ˙. Í2 0 1 3 ˙ Í ˙ Î8 8 23 15˚ Solution: Let us apply elementary row operations on the matrix; then È2 Í4 Í Í2 Í Î8

5˘ È2 ˙ Í 7 4 1 + R2 + R3 ˙ æRææææ ÆÍ 3˙ Í8 ˙ Í 8 23 15˚ Î8 3 16 5 6 0 1

È2 Í4 R4 - R3 ææææ ÆÍ Í8 Í Î0 È2 Í0 ( -1) R2 Í æ(ææ Æ -1) R3 Í0 Í Î0

3 16 5 ˘ 5 6 7˙ ˙ 8 23 15˙ ˙ 8 23 15˚

5˘ 5˘ È 2 3 16 ˙ Í 5 6 7 0 -1 -26 -3˙ 2 - 2 R1 ˙ æRæææ Í ˙ Æ 8 23 15˙ R3 - 4 R1 Í 0 -4 -41 -5˙ ˙ Í ˙ 0 0 0˚ 0 0˚ Î0 0 3 16

3 16 5˘ È2 ˙ Í0 1 26 3 R1 - 3 R2 ˙ ææææ Í Æ 4 41 5˙ R3 - 4 R2 Í 0 ˙ Í 0 0 0˚ Î0

0 -62 -4 ˘ 1 26 3˙ ˙ 0 -63 -7 ˙ ˙ 0˚ 0 0

Engineering Mathematics-I

SQP5.8

È1 0 -13 0˘ È1 0 -31 -2˘ 1 26 3 ˙ R1 + 2 R3 Í0 1 -1 0˙ ˙ ææææ ˙ ÆÍ 0 9 1 ˙ R2 - 3 R3 Í0 0 9 1 ˙ Í ˙ Í ˙ 0˚ Î0 0 0 Î 0 0 0 0˚ Since the number of nonzero rows of the reduced matrix is 3, therefore, the rank of the matrix is 3.

Ê 1ˆ Í0 ÁË ˜¯ R1 2 Í æÊæææ -1ˆ Æ Í0 R ËÁ 7 ¯˜ 3

** 11. (a) Find the extremum of the following function: x3 + y3 − 3axy Solution: See Example 6.23 of Page 6.50.  **(b) Show that — f is irrotational, where f = x2y + 2xy + z2. Solution: Now,  Ê ∂f  ∂f  ∂f ˆ — f = Ái +j +k ˜ Ë ∂x ∂y ∂z ¯ 2  = (2xy + 2y) i + (x + 2x) j + 2zk and  curl —f =

( )

i

j

k

∂ ∂x

∂ ∂y

∂ ∂z

(2 xy + 2 y ) ( x 2 + 2 x) 2 z = i (0 - 0) -  j ( 0 - 0) +  k (2 x + 2 - 2 x - 2) =0  Therefore, — f is irrotational. **(c) Evaluate

a

x

y

Ú0 Ú0 Ú0 x3 y 2 zdxdydz

Solution: See Example 7.4 of Page 7.19.

SOLUTIONS OF UNIVERSITY QUESTIONS (W.B.U.T.) B.TECH SEM-1 (NEW) 2011 MATHEMATICS-I (M 101) Time Alloted: 3 Hours

Full Marks: 70

GROUP-A (Multiple Choice Type Questions) 1. Choose the correct alternatives for any ten of the following: Ï n ¸ *(i) The least upper bound of the sequence Ì ˝ is Ó n + 1˛ 1 (a) 0 (b) (c) 1 2

(10 ¥ 1 = 10)

(d) 2

Solution: The correct alternative is (c) 1 2000 2001 2002 *(ii) The value of 2003 2004 2005 is 2006 2007 2008 (a) 2000 (c) 45

(b) 0 (d) none of these

Solution: The correct alternative is (b) 0 **(iii) If l3 – 6l2 + 9l – 4 = 0 is the characteristic equation of a square matrix A then A–1 is equal to 1 9 3 (a) A2 – 6A + 9I (b) A2 – A + I 4 4 2 1 2 3 9 2 (c) A – 6A + 9 (d) A – A + 4 4 2 Solution: The correct alternative is (b) *(iv) If x = r cos q, y = r sin q, then

1 2 3 9 A - A+ I 4 2 4

∂(r , q ) is ∂ ( x, y )

(a) r 1 (c) r Level of difficulty:- *Low, **Medium, ***High.

(b) 1 (d) none of these

SQP2.2

Engineering Mathematics-I

Solution: The correct alternative is (c) *(v) f (x, y) = (a)

1 r

y+ x is a homogeneous function of degree y+x

1 2

(b) –

1 2

(c) 1

Solution: The correct alternative is (b) -

(d) 2

1 2

      **(vi) If a ◊ (b ¥ g ) = 0, then a , b , g are (a) coplanar (c) collinear

(b) independent (d) none of these

Solution: The correct alternative is (a ) coplanar **(vii) The nth derivative of (ax + b)10 is (where n > 10) (a) a10

(b) 10!a10

Solution: The correct alternative is (c)   ***(viii) If for any two vectors a and b,   a+b =   then a and b are (a) parallel (c) perpendicular

(c) 0

(d) 10!

0   a-b (b) collinear (d) none of these

Solution: The correct alternative is (d ) orthogonal **(ix) If A–1 =

1 È 3 1˘ Í ˙ then A = 7 Î-1 2 ˚

È 3 1˘ (a) Í ˙ Î-1 2 ˚

È 2 1˘ (b) Í ˙ Î-1 3˚

(c)

1 È2 -1˘ Í ˙ 7 Î 1 3˚

È 2 1˘ Solution: The correct alternative is (b) Í ˙ Î-1 3˚

È2 -1˘ (d) Í ˙ Î 1 3˚

Solutions of University Questions (W.B.U.T.)

SQP2.3

p 2

Ú

*(x) The reduction formula of In = cosn xdx is 0

(a) In = ÊÁ n - 1ˆ˜ In – 1 Ë n ¯

(b) In = ÊÁ n ˆ˜ In Ë n - 1¯

n - 1ˆ (c) In = ÊÁ In – 2 Ë n ˜¯

(d) none of these

Ê n - 1ˆ Solution: The correct alternative is (c) I n = Á I Ë n ˜¯ n - 2 •

**(xi) The series

n2

 2n n =1

2

+1

is

(a) convergent (c) oscillatory

(b) divergent (d) none of these

Solution: The correct alternative is (b) divergent *(xii) Lagrange’s Mean Value Theorem is obtained from Cauchy’s Theorem for two functions f (x) and g (x) by putting g (x) = 1 (a) 1 (b) x2 (c) x (d) x Solution: The correct alternative is (c) x

GROUP-B (Short-Answer Type Questions) Answer any three of the following:

(3 ¥ 5 = 15)

*2. Prove that every square matrix can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix. Solution: See Theorem 1.2 of Page 1.10. ***3. By Laplace’s method, prove that a b c d -b a d -c = (a2 + b2 + c2 + d2)2 -c - d a b -d c -b a (consider minors of order 2).

SQP2.4

Engineering Mathematics-I

Solution: Here, we expand the given determinant by Laplace’s method of expansion in terms of a minor of order 2 considering the first two rows as follows: a

b

c

-b

a

d -c

-c - d a -d c -b

d b a

=

a b a b a c -d b ¥ (–1)(1 + 2) + (1 + 2) + ¥ (–1)(1 + 2) + (1 + 3) c a -b a -b a -b d

+

-c b a d -d a b c ¥ (–1)(1 + 2) + (1 + 4) + ¥ (–1)(1 + 2) + (2 + 3) - d a -b - c c -b a d

+

b d -c a c d -c - d ¥ (–1)(1 + 2) + (2 + 4) + ¥ (–1)(1 + 2) + (3 + 4) a -c - d -b d -c -d c = (a2 + b2) (a2 + b2) + (ad + bc) (ad + bc) + (–ac + bd) (–ac + bd) + (bd – ac) (bd – ac) + (bc + ad) (bc + ad) + (c2 + d2) (c2 + d2) = (a2 + b2)2 + 2(ad + bc)2 + 2(– ac + bd)2 + (c2 + d2)2 2 2 2 2 = (a + b ) + 2[(ad + bc) + (– ac + bd)2] + (c2 + d2)2 = (a2 + b2)2 + 2[a2 d2 + b2 c2 – 2adbc + a2 c2 + b2 d2 – 2acbd] + (c2 + d2)2 2 2 2 2 = (a + b ) + 2(a + b2) (c2 + d2) + (c2 + d2)2 = (a2 + b2 + c2 + d2)2 1

***4. If 2x = y m + y

-

1 m

then prove that

(x2 – 1)yn + 2 + (2n + 1)xyn +1 + (n2 – m2)yn = 0

SQP2.5

Solutions of University Questions (W.B.U.T.)

Solution: Here, we have 2x = 1 ym

or,

– 2x +

1

1 m y

+y

-

1 m

=0

1 m y

2

Ê 1ˆ Ê 1ˆ Á y m ˜ – 2x Á y m ˜ + 1 = 0 ÁË ˜¯ ÁË ˜¯

or,

Applying the rule for finding solution of the above quadratic equation, we get 1

ym = fi

(

2 x ± (2 x )2 - 4.1.1 2 = x ± x -1 2

(

y = x ± x2 - 1

)

)

m

(i)

Differentiating (i) w.r.t. x, we have

(

)

(

)

y1 = m x ± x 2 - 1

= m x ± x2 - 1

m -1

m -1

(

i.e.,

y1

= ± m x ± x2 - 1

)

(x ± = ±m

)

x2 - 1 x2 - 1

Ê 1 ◊ Á1 ± ◊ 2 ÁË

ˆ ◊ 2 x˜ ˜¯ x2 - 1 1

Ê x2 - 1 ± x ˆ ◊Á ˜ ÁË x 2 - 1 ˜¯

m -1

◊

(x ±

x2 - 1

)

x2 - 1

m

=±

my x2 - 1

(ii)

Squaring (ii) and simplifying, we get (y1)2 (x2 – 1) = m2 y2

(iii)

Again differentiating (iii) w.r.t. x, we have 2y1 y2(x2 – 1) + (y1)2 2x = m2 2y ◊ y1 fi

y2(x2 – 1) + y1x – m2y = 0

(iv)

SQP2.6

Engineering Mathematics-I

Now applying Leibnitz’s theorem, we differentiate (iv) n times w.r.t. x, {y2(x2 – 1)}n + {y1x}n – {m2y}n = 0 fi

[{y2}n ◊ (x2 – 1) + nC1{y2}n – 1 ◊ (2x) + nC2{y2}n – 2 ◊ (2)] + [{y1}n ◊ x + nC1{y1}n – 1 ◊ 1] – m2yn = 0

fi

(x2 – 1)yn + 2 + 2nxyn + 1 + n(n – 1)yn + xyn + 1 + nyn – m2yn = 0

fi

(x2 – 1)yn + 2 + (2n + 1)xyn + 1 + (n2 – m2)yn = 0

Ê yˆ **5. If u = x f Á ˜ + gÊÁ y ˆ˜ then show that Ë x¯ Ë x¯ x2

∂2 u ∂x 2

+ 2xy

∂ 2u ∂ 2u + y2 = 0. ∂x ∂y ∂y 2

Solution: See Example 21 of Page 6.19. **6. Show that the area bounded by a simple closed curve C is given by 1 (xdy – ydx). 2

Ú C

Solution: We know that Green’s theorem states the following:

Ú {M(x, y) dx + N (x, y) dy} = ÚÚ ÊÁË ∂∂Nx - ∂∂My ˆ˜¯ dx dy C

(i)

R

where the region R on the two-dimensional xy plane is bounded by a simple closed curve C and the line integral along the curve C is taken in the anticlockwise direction. Here, comparing LHS of (i) with

Ú (xdy – ydx), we have C

∂M ∂N = –1, =1 ∂y ∂x

M = –y, N = x fi

(ii)

Therefore using (ii) in (i), we get

Ú (xdy – ydx) = ÚÚ [1 – (–1)] dx dy C

R

=2

ÚÚ dx dy R

= 2 ¥ [Area bounded by C] fi Area bounded by C =

1 2

Ú (xdy – ydx) C

Hence, the area bounded by a simple closed curve C is given by 1 2

Ú (xdy – ydx). C

SQP2.7

Solutions of University Questions (W.B.U.T.)

GROUP-C (Long-Answer Type Questions) (3 ¥ 15 = 45)

Answer any three of the following: ***7. (i) If

verify

Ê yˆ Ê xˆ f (x, y) = x2 tan–1 Á ˜ – y2 tan–1 Á ˜ , Ë x¯ Ë y¯ fxy = fyx.

Solution: See Example 6.6 of Page 6.34. *(ii) State Rolle’s theorem and examine if you can apply the same for f (x) = tan x in [0, p]. Solution: See Example 4.3 of Page 4.33. ***(iii) Find the value of l and m for which x+y+z =3 2x – y + 3z = 4 5x – y + lz = m has (a) a unique solution, (b) many solutions (c) no solution. Solution: If we write the system of linear equations in the matrix form as AX = B then the coefficient matrix of the system of linear equations is Ê 1 1 1ˆ A = Á 2 -1 3˜ Á ˜ Ë 5 -1 l ¯ and the augmented matrix is Ê 1 1 1 3ˆ A = Á 2 -1 3 4˜ Á ˜ Ë 5 -1 l m ¯ The system of equations has a unique solution when the determinant of the coefficient matrix is not equal to zero. 1 1 1 det A = 2 -1 3 5 -1 l = 1(–l + 3) – 1(2l – 15) + 1(– 2 + 5) = – 3l + 21 Therefore, for det A π 0 fi –3l + 21 π 0 fi l π 7, the system of equations have unique solution.

SQP2.8

Engineering Mathematics-I

When l = 7, the augmented matrix becomes Ê 1 1 1 3ˆ A = Á 2 -1 3 4˜ Á ˜ Ë 5 -1 7 m ¯ Applying elementary row operations on the matrix A, we have 1 1 3 ˆ Ê1 Ê 1 1 1 3ˆ Á Á ˜ A = 2 -1 3 4 �R������������������ - 2R1, R3 - 5 R�1 0 - 3 1 - 2 ˜ �R�������� - 2R2� Á ˜ 3 Á ˜ 2 Ë 0 - 6 2 m - 15¯ Ë 5 -1 7 m ¯ 1 1 3 ˆ Ê1 Á 0 -3 1 -2 ˜ Á ˜ 0 0 m - 11¯ Ë0

1 Ê1 1 Á 1 1 Ê ˆ ÁË - ˜¯ R2 Á 0 1 3 �Á 3 ��������� ÁË 0 0 0

3 ˆ 2 ˜ ˜ 3 ˜ m - 11˜¯

4 Ê 1 0 Á 3 Á 1 1 - R2� Á 0 1 �R������ Á 3 ÁË 0 0 0

7 ˆ 3 ˜ ˜ 2 ˜ 3 ˜ m - 11˜¯

The system of equations is consistent when Rank A = Rank A and this is possible for m – 11 = 0 fi m = 11. In this case, Rank A = Rank A = 2, which is less then number of unknowns (= 3) and the system has infinitely many solutions. Again, if m – 11 π 0 fi m π 11. then Rank A = 2 and Rank A = 3, i.e., Rank A π Rank A, and so the system of equations is inconsistent and correspondingly the system has no solution. Summarizing the above, the system of equations has (a) a unique solution when l π 7 (b) infinitely many solutions when l = 7 and m = 11 (c) no solution when l = 7 and m π 11.

SQP2.9

Solutions of University Questions (W.B.U.T.)

**8. (i) Find the maxima and minima of the function f (x, y) = x3 + y3 – 63(x + y) + 12xy Find also the saddle points. Solution: See Example 6.24 of Page 6.51. **(ii) State Leibnitz’s test for alternating series and apply it to examine the convergence of 1 1 1 1- 2 + 2 - 2 +º• 2 3 4 Solution: See Example 8.27 of Page 8.42. *(iii) Applying Lagrange’s Mean Value Theorem, prove that x £ log(1 + x) £ x, for all x > 0. 1+ x Solution: See Example 7 of Page 4.11. *9. (i) If y = em sin–1 x, show that (1 – x2)yn + 2 – (2n + 1)xyn + 1 – (n2 + m2)yn = 0. Hence, find yn when x = 0. Solution: Here, we are given that y = em sin–1 x Differentiating (i) w.r.t. x, we have y1 = em sin–1 x ◊ m i.e.,

y1 =

(i) Ê ÁË

1 1- x

2

ˆ ˜¯

my

(ii)

1 - x2

Squaring (ii) and simplifying, we get (y1)2 (1 – x2) = m2 y2

(iii)

Again differentiating (iii) w.r.t. x, we have 2y1 y2(1 – x2) + (y1)2 (–2x) = m2 2y ◊ y1 fi

y2(1 – x2) – y1x – m2y = 0

(iv)

Now applying Leibnitz’s theorem, we differentiate (iv) n times w.r.t. x, {y2(1 – x2)}n – {y1x}n – {m2y}n = 0. fi

[{y2}n ◊ (1 – x2) + nC1{y2}n – 1 ◊ (– 2x) + nC2{y2}n – 2 ◊ (–2)] – [{y1}n ◊ x + nC1{y1}n – 1 ◊ 1] – m2 yn = 0

SQP2.10

Engineering Mathematics-I

fi

(1 – x2)yn + 2 – 2nxyn +1 – n(n – 1)yn – xyn +1 – nyn – m2yn = 0

fi

(1 – x2)yn + 2 – (2n + 1)xyn +1 – (n2 + m2)yn = 0

(v)

Calculation of yn when x = 0, i.e., (yn)0: Putting x = 0 in (v), we have (yn + 2)0 = (n2 + m2) (yn)0 Replacing n by n – 2, we get (yn)0 = [(n – 2)2 + m2] (yn – 2)0

(vi)

Replacing n by n – 2 in (vi), we get (yn – 2)0 = [(n – 4)2 + m2] (yn – 4)0

(vii)

Using (vii) in (vi), (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] (yn – 4)0 Similarly, we have (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] [(n – 6)2 + m2] (yn – 6)0

(viii)

Proceeding in a similar manner we have from (viii), when n is odd as the following: (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [32 + m2] [12 + m2] (y1)0

(ix)

From (ii), we have (y1)0 = m. Using this in (ix), we get (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [32 + m2] [12 + m2] m, when n is odd. Also proceeding in a similar manner we have from (viii), when n is even as the following: (yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [42 + m2] [22 + m2] (y2)0 From (iv), we have (y2)0 =

m2.

(x)

Using this in (x), we get

(yn)0 = [(n – 2)2 + m2] [(n – 4)2 + m2] … [42 + m2] [22 + m2]m2, when n is even.           *(ii) Prove that ÈÎa + b b + c c + a ˘˚ = 2 ÈÎa b c ˘˚ , where a b c are three vectors. Solution: Using the definition of scalar triple product, we write             ÈÎa + b b + c c + a ˘˚ = ( a + b ) ◊ ÈÎ(b + c ) ¥ (c + a )˘˚         = ( a + b ) ◊ ÈÎ(b + c ) ¥ c + (b + c ) ¥ a ˘˚           = ( a + b ) ◊ ÈÎ(b ¥ c ) + (c ¥ c ) + (b ¥ a ) + (c ¥ a )˘˚

Solutions of University Questions (W.B.U.T.)

SQP2.11

        = ( a + b ) ◊ ÈÎ(b ¥ c ) + (b ¥ a ) + (c ¥ a )˘˚ ,

   since c ¥ c = 0             = a ◊ (b ¥ c ) + a ◊ (b ¥ a ) + a ◊ (c ¥ a ) + b ◊ (b ¥ c )       + b ◊ (b ¥ a ) + b ◊ (c ¥ a )    = ÈÎa b c ˘˚ + ÈÎa b a ˘˚ + [ a c a ] + ÈÎb b c ˘˚   + ÈÎb b a ˘˚ + ÈÎb c a ˘˚  By the property of scalar triple product of vectors, we have ÈÎa b a ˘˚ =    0, [ a c a ] = 0, ÈÎb b c ˘˚ = 0, ÈÎb b a ˘˚ = 0 (since two vectors in the product   are same) and ÈÎb c a ˘˚ = ÈÎa b c ˘˚ . Using this in the above, we get          ÈÎa + b b + c c + a ˘˚ = ÈÎa b c ˘˚ + 0 + 0 + 0 + 0 + ÈÎa b c ˘˚ = 2 ÈÎa b c ˘˚ *(iii) Find the directional derivative of f = xyz at (1, 1, 1) in the direction 2iˆ - ˆj - 2 kˆ . Solution: Here, it is given that f = xyz. Then  ∂ ∂ ∂ —f = Ê iˆ + ˆj + kˆ ˆ f ÁË ∂x ∂y ∂z ˜¯ ∂f ∂f ˆ Ê ∂f = Á iˆ + ˆj + kˆ ˜ ∂y ∂z ¯ Ë ∂x ˆ ˆ ˆ = yzi + xzj + xyk So,

 ÈΗf ˘˚ = iˆ + ˆj + kˆ (1,1,1)

Here we are to find the directional derivative in the direction 2iˆ - ˆj - 2 kˆ . The unit vector in the direction of 2iˆ - ˆj - 2 kˆ is given by aˆ =

2iˆ - ˆj - 2 kˆ

=

22 + (-1)2 + (-2)2

2iˆ - ˆj - 2 kˆ 3

Then the required directional derivative of f = xyz at (1, 1, 1) in the direction of 2iˆ - ˆj - 2 kˆ is given by  ˆ ˆ ˆ ˆ + ˆj + kˆ ◊ Ê 2i - j - 2 k ˆ = - 1 ÈΗf ˘˚ ˆ a i ◊ = Á ˜¯ (1,1,1) Ë 3 3

(

)

SQP2.12

Engineering Mathematics-I

**10. (i) Prove that b2 + c 2

a2

a2

b2

c2 + a2

b2

c2

c2

a 2 + b2

= 4a2 b2

2

Solution: See Example 1.16 of Page 1.43.  ***(ii) State the Divergence Theorem of Gauss. Verify divergence theorem for F = yiˆ + xjˆ + z 2 kˆ over the cylindrical region bounded by x2 + y2 = 9, z = 0, z = 2. Solution: See Example 9.29 of Page 9.58. **(iii) Test the series for convergence: 1p 2

q

2p

+

q

3

3p

+

+º

4q

Solution: Let us consider the given series as •

Âa

=

n

n=1

1p 2

q

+

2p q

3

+

3p 4q

+º

Then an =

np (n + 1)

q

np

= qÊ

1ˆ n Á1 + ˜ Ë n¯

q

=

1 1ˆ Ê nq - p Á 1 + ˜ Ë n¯

q

Let us consider another series •

 n=1

•

bn =

Ân n =1

1 q- p

which is convergent for q – p > 1 and divergent for q – p £ 1. Now we have an 1 = lim = 1. q n Æ• bn n Æ• Ê 1ˆ ÁË 1 + ˜¯ n lim

•

Since

Âb

n

is convergent for q – p > 1 and divergent for q – p £ 1, by

n=1

comparison test, q – p £ 1.

•

Âa n=1

n

is convergent for q – p > 1 and divergent for

SQP2.13

Solutions of University Questions (W.B.U.T.) p 2

*11. (i) Obtain a reduction formula for

p 2

Ú sin x dx. Hence obtain Ú sin x dx. 9

n

0

0

Solution: See Section 5.2 of Page 5.1.         **(ii) Given two vectors a = 3i – j , b = 2i + j – 3k . Express b in the form       b1 + b2, where b1 is parallel to a and b2 is perpendicular to a . Solution: See Example 9.1 of Page 9.17.  **(iii) Show that A = (6xy + z3)iˆ + (3x2 – z) ˆj + (3xz2 – y) kˆ is irrotational. Find   the scalar function f, such that A = —f. Solution: See Example 9.11 of Page 9.36.