Solutions Manual Applied Linear Algebra (Undergraduate Texts in Mathematics) [2 ed.] 9783319910406

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Solu�ons Manual For Applied Linear Algebra (Undergraduate Texts in Mathema�cs), 2e Peter Olver, Chehrzad Shakiban (All Chapters, 100% Original Verified, A+ Grade)

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Instructors’ Solutions Manual for Chapter 1: Linear Algebraic Systems Note: Solutions marked with a

⋆

do not appear in the Students’ Solutions Manual.

1.1.1. (b) Reduce the system to 6 u + v = 5, − 52 v = 52 ; then use Back Substitution to solve for u = 1, v = −1. ⋆ (c) Reduce the system to p + q − r = 0, −3 q + 5 r = 3, − r = 6; then solve for p = 5, q = −11, r = −6. (d) Reduce the system to 2 u − v + 2 w = 2, − 32 v + 4 w = 2, − w = 0; then solve for u = 31 , v = − 43 , w = 0.

⋆ (e)

Reduce the system to 5 x1 + 3 x2 − x3 = 9, x1 = 4, x2 = −4, x3 = −1.

1 5

x2 −

2 5

x3 =

2 5,

2 x3 = −2; then solve for

(f ) Reduce the system to x + z − 2 w = − 3, − y + 3 w = 1, − 4 z − 16 w = − 4, 6 w = 6; then solve for x = 2, y = 2, z = −3, w = 1.

⋆

1.1.2. Plugging in the values of x, y and z gives a + 2 b − c = 3, a − 2 − c = 1, 1 + 2 b + c = 2. Solving this system yields a = 4, b = 0, and c = 1. ♥ 1.1.3. (a) With Forward Substitution, we just start with the top equation and work down. Thus 2 x = −6 so x = −3. Plugging this into the second equation gives 12 + 3y = 3, and so y = −3. Plugging the values of x and y in the third equation yields −3 + 4(−3) − z = 7, and so z = −22. ⋆ (c) Start with the last equation and, assuming the coefficient of the last variable is 6= 0, use the operation to eliminate the last variable in all the preceding equations. Then, again assuming the coefficient of the next-to-last variable is non-zero, eliminate it from all but the last two equations, and so on. ⋆ (d) For the systems in Exercise 1.1.1, the method works in all cases except (c) and (f ). Solving the reduced system by Forward Substitution reproduces the same solution (as it must): 15 15 (a) The system reduces to 23 x = 17 2 , x + 2 y = 3. (b) The reduced system is 2 u = 2 , 3 u − 2 v = 5. (d) Reduce the system to 32 u = 21 , 72 u − v = 52 , 3 u − 2 w = −1. (f ) Doesn’t work since, after the first reduction, z doesn’t occur in the next to last equation.

1.2.1. (a) 3 × 4,

(b) 7,

1.2.2. Examples:

(a)



1

 4 

7

1.2.4. (b) A =

1

6 3

!

(c) 6,

2 5 8

1 , x= −2

(d) ( −2 0 1 2 ),



3  6 , 9

⋆ (b) !

u , b= v

1 1

2 4

(e)

!

3 , 5

   

(c)

!



0  2 . −6



1

 4 

7

2 5 8

3 6 9



4  7 , 3

(e)



1



   2 .  

3

5 ; 5

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Chapter 1: Instructors’ Solutions Manual

⋆ (c)

A=

   



1 2 −1

2  (d) A =  −1  3

⋆ (e)

A=



5

 3 

1  1  2 (f ) A =   0 1

1 −1 −1















−1 p 0       q , b =  3 ; 3 , x =      0 r 6

−1 −1 0







2 u 2       v , b =  1 ; 3 , x =      w 1 −2



3 2 1

−1  −1  , 2 0 1 −1 2 −6 −4 3 2

x=

 x  1  x ,  2 

x3





b=





  



9  5 ; −1





−2 x −3       y  −5  −1  , x =  , b =  .      2  z  2 −1 w 1

1.2.5. (b) u + w = −1, u + v = −1, v + w = 2. The solution is u = −2, v = 1, w = 1. (c) 3 x1 − x3 = 1, −2 x1 − x2 = 0, x1 + x2 − 3 x3 = 1.

⋆

The solution is x1 = 15 , x2 = − 25 , x3 = − 25 . (d) x + y − z − w = 0, −x + z + 2 w = 4, x − y + z = 1, 2 y − z + w = 5. The solution is x = 2, y = 1, z = 0, w = 3.



1.2.6. (a) I =

1

0 1 0 0 0

 0   0   0

0

0 0 1 0 0



0 0 0 1 0



0  0  , 0  0  1

O=

0

 0   0   0

0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0



0  0  . 0  0  0

(b) I + O = I , I O = O I = O. No, it does not. 1.2.7. (b) undefined, (f )



1

 3 

7

1.2.9.



9  −12  , 8

6 4

⋆ (h)

!

0 , 2 

⋆ (e) 9

  −8 

12

−2 6 −3

undefined, 

14  −17  . 28

1, 6, 11, 16. 

1  1.2.10. (a)  0 0 1.2.11. (a) True,

⋆

11 −12 8

(c)

3 −1



0 0 0

0  0 , −1

⋆ (b)

♥ 1.2.12. (a) Let A =

x z

⋆



2  0 (b)   0 0

0 −2 0 0

0 0 3 0



0  0 . 0  −3

true. !

y . Then A D = w

ax az

by bw

!

=

ax bz

ay bw

!

= D A, so if a 6= b these

are equal if and only if y = z = 0. (b) Every 2 × 2 matrix commutes with

a 0

0 a

!

= a I.

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Chapter 1: Instructors’ Solutions Manual

3 



x 0 0  (c) Only 3 × 3 diagonal matrices. (d) Any matrix of the form A = 0 y z  . 0 u v (e) Let D = diag (d1 , . . . , dn ). The (i, j) entry of A D is aij dj . The (i, j) entry of D A is di aij . If di 6= dj , this requires aij = 0, and hence, if all the di ’s are different, then A is diagonal.   

x 0

1.2.14. B =

⋆

y x

!

where x, y are arbitrary.

1.2.15. (a) (A + B)2 = (A + B)(A + B) = AA + AB + BA + BB = A2 + 2AB + B 2 , since ! ! 1 2 0 0 AB = BA. (b) An example: A = ,B= . 0 1 1 0 1.2.17. A On×p = Om×p ,

Ol×m A = Ol×n .

1.2.19. False: for example,

⋆

1 0

0 0

!

0 1

0 0

!

=

0 0

!

0 . 0

1.2.21. Let v be the column vector with 1 in its j th position and all other entries 0. Then A v is the same as the j th column of A. Thus, the hypothesis implies all columns of A are 0 and hence A = O. 1.2.22. (a) A must be a square matrix. (b) By associativity, A A2 = A A A = A2 A = A3 . ⋆ (c) The na¨ıve answer is n − 1. A more sophisticated answer is to note that you can r

compute A2 = A A, A4 = A2 A2 , A8 = A4 A4 , and, by induction, A2 with only r matrix multiplications. More generally, if the binary expansion of n has r + 1 digits, with s nonzero digits, then we need r + s − 1 multiplications. For example, A13 = A8 A4 A since 13 is 1101 in binary, for a total of 5 multiplications: 3 to compute A2 , A4 and A8 , and 2 more to multiply them together to obtain A13 .

1.2.25. The same solution X =

⋆

−1 3

1 −2

!

in both cases. 1 0

1.2.27. (a) X = O. (b) Yes, for instance, A =

2 1

!

, B=

3 −2

2 −1

!

, X=

1 1

!

0 . 1

♦ 1.2.29. (a) The ith entry of A z is 1 ai1 + 1 ai2 + · · · + 1 ain = ai1 + · · · + ain , which is the ith row sum. 1 1−n (b) Each row of W has n − 1 entries equal to n and one entry equal to n and so its 1 1−n row sums are (n − 1) n + n = 0. Therefore, by part (a), W z = 0. Consequently, the row sums of B = A W are the entries of B z = A W z = A 0 = 0, and the result follows.

⋆      





1    (c) z =   1 , and so A z = 1 1

2

2

1

−4

5



−1   −   3  

−1

2 3 1 3 1 3

1 3 − 23 1 3

1 3 1 3 − 23



1   2  −4      

=

     

2 1 5 −





−1 1   1 = 3   −1 1

1 3

− 43

0

1

4

−5

 5 3  −1  , 





2    6 , while B = A W =   0 and so B z =

1



0



   0 .  

0

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⋆

Chapter 1: Instructors’ Solutions Manual

♥ 1.2.30. (a) We need A B and B A to have the same size, (b) A B − B A = O if and only if A B = B A.  ! ! 0 1  −1 2 0 0 (c) (i ) , (ii ) , (iii )   1 0 6 1 0 0 −1 1 (d) (i ) [ c A + d B, C ] = (c A + d B)C − C(c A + d B)

and so this follows from Exercise 1.2.13. 

1  1 ; 0

= c(A C − C A) + d(B C − C B) = c [ A, B ] + d [ B, C ],

[ A, c B + d C ] = A(c B + d C) − (c B + d C)A

= c(A B − B A) + d(A C − C A) = c [ A, B ] + d [ A, C ]. (ii ) [ A, B ] = A B − B A = − (B A − A B) = − [ B, A ]. ♥ 1.2.34. (a) This follows by direct computation. (b) (i ) −2 3

1 2

!

⋆ (ii )

1 1

1 −3

−2 0

!

−2 −1

!

!

−2 1 ( 1 −2 ) + (1 0) = 3 2

=



2 0   −3 2  1 !

=



2 −6

5  0 = −1 5 −15

!

!

1 (2 5) + −3 6 3

+

0 0

!

+

!

−2 3

0 2

4 1 + −6 2 !

0 0

−2 ( −3 0 ) + −1 0 −2

!

=

8 −1

!

!

−1 5

=

4 . −6

!

0 ( 1 −1 ) 2 !

5 . −17

⋆ (c) If we set B = x, where x is an n × 1 matrix, then we obtain (1.14). n ⋆ (d) The (i, j) entry of A B is X aik bkj . On the other hand, the (i, j) entry of ck rk equals k=1

the product of the ith entry of ck , namely aik , with the j th entry of rk , namely bkj . Summing these entries, aik bkj , over k yields the usual matrix product formula.

⋆

♥ 1.2.37. (a) Check that S 2 = A by direct computation. Another example: S =

2 0

!

0 . 2

Or, more generally, 2 times any of the matrices in part (c). (b) S 2 is only defined if S ! ! ±1 0 a b is square. (c) Any of the matrices , , where a is arbitrary and 0 ±1 c −a ! 0 −1 b c = 1 − a2 . (d) Yes: for example . 1 0

1.3.1. (b) u = 1, v = −1;

⋆ (c) ⋆ (e)

1.3.2. (a)

u = 32 , v = − 13 , w = 16 ; p = − 23 , q = 1 −2

7 −9



19 6 ,

4 2

!

r = 52 ;

2R1 +R2

−→

(d) x1 =

11 3 ,

2 x2 = − 10 3 , x3 = − 3 ;

(f ) a = 13 , b = 0, c = 43 , d = − 23 . 1 0

7 5



!

4 . Back Substitution yields x2 = 2, x1 = −10. 10

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Chapter 1: Instructors’ Solutions Manual

(c)

⋆ (d)

   

1 0 −4

4 0 −2

1

  −2 

3

 1 7 R +R 2 3  4 −→  0

0

(e)

⋆ (f )

          



1 2R +R 1 1 2  0 −7  −→   −1 3





−2 2 −3

1 −8 13



0  8  −9



1 −3R +R 1 1 3  0 −5  −→   −1 0



3 2 R2 +R3

−→

Back Substitution yields z = 3, y = 16, x = 29.





0 4R +R 1 1 3  0 8 −→   −9 0

1 −8 9

−2 2 5

5



1 0 0 −4

0 1 −3 0

−1 1 0 4

3 −1 1 −1

−2 −3 2 4 8 0



−2 −7 − 17 4 −2 0 2 0





4 8 −2



−2 −7 2



1

−2 2 0

 0 

0





4 8 −14



−1  2  0  −5



−2  0  7  5





1

 0   0

reduces to

0

0 1 0 0

−2 0 2 0

0 −1 −3 −5





     

reduces to

−1 0 0 0

3 2 0 0

−1 2 −2 0

Back Substitution yields w = 2, z = 0, y = −1, x = 1.

1 0 4 −24



0  8 . 3



1  −5   −4



−1  2 . 6  15

Back Substitution yields x4 = −3, x3 = − 32 , x2 = −1, x1 = −4. 1 −1 4 0

−2 −7 8

1  −5  . Back Substitution yields r = 3, q = 2, p = −1. − 51 4

0 −1 0 7

−1 3 −1 1

1 −8 1







−2  −2  . 8  −48

1.3.3. (a) 3 x + 2 y = 2, − 4 x − 3 y = −1; solution: x = 4, y = −5; ⋆ (c) 2 x − y = 0, − x + 2 y − z = 1, − y + 2 z − w = 1, − z + 2 w = 0; solution: x = 1, y = 2, z = 2, w = 1. !

⋆

!

2 1 2 1 −→ . 1 4 0 72    3 −2 1 3 −2       4 −3  −→  0 10 (c) Regular:  −1 3 3 −2 5 0 0

1.3.4. (a) Regular:

(d) Not regular:



1

  −2 

3

1.3.5. (a)

−i 1− i

1+ i 1

−2 4 −1





3 1   0 −1  −→   2 0

−1 −3 i

!

−→

−i 0



1  − 83  . 4

−2 0 5



3  5 . −7

1+ i 1−2i



−1 1−2i

!

;

use Back Substitution to obtain the solution y = 1, x = 1 − 2 i .

⋆ (c) ⋆

1− i −i



!



!

i 2 1− i 2 i −→ ; 1+ i −1 0 2 i − 32 − 21 i use Back Substitution to obtain the solution y = − 14 +

3 4

i , x = 12 .

♦ 1.3.8. 0 is the (unique) solution since A 0 = 0.

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⋆

Chapter 1: Instructors’ Solutions Manual 1.3.9. Since a11 0

a12 a22

b11 0

b12 b22

!

b11 0

!

b12 b22

a11 0

a12 a22

! !

!

=

a11 b11 0

a11 b12 + a12 b22 , a22 b22

=

a11 b11 0

a22 b12 + a12 b11 , a22 b22

!

the matrices commute if and only if a11 b12 + a12 b22 = a22 b12 + a12 b11 ,  

♦ 1.3.11. (a) Set lij =  

0  (b) L =   1 −2

0 0 0

aij ,

i > j,

0,

i ≤ j,



,



0  0 , 0

3  D= 0 0

or

 

uij =  0 −4 0

(a11 − a22 )b12 = a12 (b11 − b22 ).

aij ,

i < j,

0,

i ≥ j,



 



0  0 , 5

0  U = 0 0

dij =  1 0 0

aij ,

i = j,

0,

i 6= j.



−1  2 . 0

1.3.15. (a) Add −2 times the second row to the first row of a 2 × n matrix. (c) Add −5 times the third row to the second row of a 3 × n matrix.

⋆ (d) ⋆ (e)

Add

1 2

times the first row to the third row of a 3 × n matrix.

Add −3 times the fourth row to the second row of a 4 × n matrix.

1.3.16. (a)



1

 0   0

0

⋆

0 1 0 0

0 0 1 1 



0  0 , 0  1

⋆ (b)



0 1 0 0

1

 0   0

0



0 1



0 0 1 0 

0  0 , −1   1

(c)



1

 0   0

0



0 1 0 0

0 0 1 0



3  0 . 0  1

0 1 0 0      0  , E1 E2 E3 =  −2 1 0  1.3.18. E3 E2 E1 = . The second is easier to predict −1 12 1 −2 12 1 since its entries are the same as the corresponding entries of the Ei . 1

  −2 

1.3.20. (a) Upper triangular; (b) both upper and lower unitriangular; (d) lower unitriangular. 1 0 −1 1  1 0 1 (d) L =  1 2 0 13

1.3.21. (a) L =

⋆

(e) L =

⋆



1

  −2  

−1

1   0  (g) L =   −1  0

!

1 0

!

3 , 3   0 2 0   0 3 0 , U =   1 0 0

, U=

0 1 −1

0 1

3 2 − 12





0 −1    0 0 , U =   1 0 0 0 1 3



(c) L = 3

 − 12  , 7 6



0 1   0 0   , U =  0 0   1

0

0 2 0 0

1



0 1 0

  −1 

1



0 −3 0



⋆ (c) 

lower triangular;

0 −1    0 0 , U =   1 0

1 2 0



−1  0 , 3



0  0 , 2 −1 −1 1 2

0



0  −1   7  2

,

−10

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Chapter 1: Instructors’ Solutions Manual 

⋆ (i)

⋆

L=

       

1

0

0

1 2 3 2 1 2

1

0

− 37

1

1 7

0





  0  ,  0 

5 − 22 

7 2

  0    0 

U=

1

0

1

3

1

7 2

− 32

1 2 5 7 35 22

0

− 22 7

0

0



    .   



1

0 0 0  1 0 0 , (b) (1) Add −2 times first row to second row. 1.3.23. (a) For example, 4 1 0  5 6 7 1 (2) Add −3 times first row to third row. (3) Add −5 times first row to fourth row. (4) Add −4 times second row to third row. (5) Add −6 times second row to fourth row. (6) Add −7 times third row to fourth row. (c) Use the order given in part (b).  2   3

1.3.25. (a) aij = 0 for all i 6= j;

⋆ (e)

(c) aij = 0 for all i > j and aii = 1 for all i;

aij = 0 for all i < j and aii = 1 for all i. !

1 1 1.3.27. False. For instance is regular. Only if the zero appear in the (1, 1) position 1 0 does it automatically preclude regularity of the matrix.

⋆ ⋆

1.3.28. (n − 1) + (n − 2) + · · · + 1 =

n (n − 1) . 2 !

0 1 ♦ 1.3.30. The matrix factorization A = L U is = 1 0 This implies x = 0 and a x = 1, which is impossible.

1.3.32. (a) x =

1.3.33. (a) L =

⋆

−1 2 3

1 −3



1   (b) L =  −1 1 (c) L =

      

1   0 (e) L =    −1 0

2 3 2 9

,

⋆ (b)

0 1

!

x=

, U=



0 1 0 1

−

!

 1  4 , 1 4

−1 0



0 −1     0, U =  0 1 0 0 1 5 3

0 1 3 2 − 12



0

 0 , 

U=

1

0 0 1 −1



9

  0  

!

x 0





3 ; 11 

1 2 0

x1 = 

−1  0 ; 3

−2

− 13

0 

0 1

y z



!



0 1   0 0 , U =   0  0 1 0



−1 

 1 ; 3  − 13

0 2 0 0

−1 3 − 72 0



5 11  , 2 11  

−

⋆ (g)

x2 =

x1 =

1





1 , x3 =  1

x2 =

3



x=



2



  1  .   1

0



   2 ,  

y . ay + z

!

− −1        x1 =  0 , x2 =  −  0 

!

x ax

=

−1 0        (c) x =   1 , (e) x =  −1 , 5 0 2 !

0







1 a



1 6 3 2 5 3

−2





9 11  . 3 11

  .  



   −9 .  

−1



5

 

1 4 1 4



1 

 

1  14  1 2

0  4  14   1    − 5    −4  −1   14  ; x =     . , x = 7 2 1     2

4

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Chapter 1: Instructors’ Solutions Manual 1.4.1. (a) Nonsingular,

⋆ (b)

singular, (c) nonsingular, (e) singular,

⋆ (d)

1.4.2. (a) Regular and nonsingular, (c) nonsingular,

⋆

⋆ (g)

singular.

regular and nonsingular.

1.4.3. Solve the equations −1 = 2 b + c, 3 = − 2 a + 4 b + c, −3 = 2 a − b + c, for a = −4, b = −2, c = 3, giving the plane z = − 4 x − 2 y + 3.

⋆ (c)

1.4.5. (b) x1 = 0, x2 = −1, x3 = 2; (d) x = −

⋆ (e)

13 2

9 2

x1 = −6, x2 = 2, x3 = −2;

, y = − , z = −1, w = −3;

x1 = −11, x2 = − 10 3 , x3 = −5, x4 = −7.

1.4.6. True. All regular matrices are nonsingular.

⋆

1.4.9. By applying the operations # 1 and # 2 to the system Ax = b we obtain an equivalent upper triangular system U x = c. Since A is nonsingular, uii 6= 0 for all i, so by Back Subc 1 stitution each solution component, namely xn = n and xi = unn uii for i = n − 1, n − 2, . . . , 1, is uniquely defined.

⋆



1  0 1.4.10. (a) P1 =   0 0

0 0 0 1

0 0 1 0





0  0 (b) P2 =   0 1

0  1 , 0  0

(c) No, they do not commute.

0 1 0 0



c

i

−

n X

k = i+1



uik xk ,



0 0 1 0

1  0 . 0  0

(d) P1 P2 arranges the rows in the order 4, 1, 3, 2, while

P2 P1 arranges them in the order 2, 4, 3, 1. 

0  1.4.11. (a)  0 1

1.4.13. (b)



0 1  0   0 0  1  0 (c)   0 0 

⋆

0

 1   0



1 0 0

0  1 , 0

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

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

⋆ (b) 

0  0 , 1  0  0  0 , 1  0  0  0 , 0  1



1

 0   0

0 0  1   0 0  0  0   0 1 



0

0

0 0 0 1

0 1 0 0

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

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

0  1 , 0  0  1  0 , 0  0  1  0 . 0  0

 0   1





1  0 , 0  0 

0

(c)

0

1 0 0 0

 1   0

0

0 1 0 0 0  0 1  0 0   1 0 0 0  0   1



0 0 0 1 0 0 0 1

0 0 0 1



0  0 . 1  0



1  0 , 0  0  0  1 ; 0  0

1.4.14. (a) True, since interchanging the same pair of rows twice brings you back to where you   0 0 1    started. ⋆ (b) False; an example is the non-elementary permutation matrix   1 0 0 . 0 1 0 Solutions Manual

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Chapter 1: Instructors’ Solutions Manual

9

−1 0 0 −1 of such matrices, see Exercise 1.2.37.

⋆ (c)

False; for example P =

1.4.16. (a)



0

 0 

1



0 1 0

1  0 . 0

(b) True.

!

⋆ (c)

is not a permutation matrix. For a complete list

False — A P permutes the columns of A accord-

ing to the inverse (or transpose) permutation matrix P −1 = P T .

⋆

♥ 1.4.17. (a) If P has a 1 in position (π(j), j), then it moves row j is enough to establish the correspondence.    1 0 0 0 1 0    0 0 1   (b) (i )   1 0 0  — elementary matrix; (iii )  0 0 0 0 0 1 0 1 0 ! ! 1 2 3 1 2 3 4 (c) (i ) , (iii ) . 4 1 2 3 2 3 1

of A to row π(j) of P A, which 

0  0  — not elementary. 1  0

1.4.19. Let ri , rj denote the rows of the matrix in question. After the first elementary row operation, the rows are ri and rj + ri . After the second, they are ri − (rj + ri ) = − rj and

rj + ri . After the third operation, we are left with − rj and rj + ri + (− rj ) = ri .

1.4.21. (a)

⋆

0 1

1 0



0  (b)  0 1 (c)



0

(e)

0

0

1.4.22. ⋆

(b)





1  (a)  0 0

!

1 0 0 0

0 0 1

=

1 0

0 1

!











1 −3 1   0 2 3 =   0 2 0

0 0 1 0  2 0 3 1   1 4 −1 0  1 7 −1 2 

0 4   −3 1  0 −3

−4 3 1

!

2 0

0 −4 1   0 2 3 =   0 1 7



1 0  0 0  0 1

0 0 1 0

 1   0

1 −1

0 0  1 1  0 0

0 0 1

0

0 2



1 0 0

 1  

!



−1 , 1

x=



0 1 0





0

0

0 1 0 0

1 0 0 0



0 0  0 0   0  1 1 1

1 1 −1 2

−1 1 1 −1





1 1   0 0 =  −3   0 1 1

solution: x = 4, y = 0, z = 1, w = 1.

0 1 1 3



0 0 1 3

0 1 0

solution: x1 = 54 , x2 = 74 , x3 = 32 .

 0   1

x



0 2  1 −3  , 0 9

0 1 0   0 1 0 =  2 2 −5   7 −29 3 

3

2 3  1 7 , 0 −4

0 1  0 0  1 0

 1 2    3  1 = − 4  −2 − 34

 5 2  ; 

0 1  0 0  1 0

0 1 2



0 0 1 5 2

0 1  0 0   0  0 1 0



0  4 

 0  0

1





 5 4  3 ; = 4  − 14       

−1    x=  1 ; 0







−1 4 −1 2     −1  1 0 0 . , x =     0 3 −4   1 3 0 0 1

−4 −2



2

− 12

0

5 2

−1 1 0 0

1 1 −2 0

0

0 1  0 0   0  0 0 1



 ;   

−3  0 ; 1  3 2

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⋆

Chapter 1: Instructors’ Solutions Manual 1.4.24. There are four in all: 

0

0





0  0  1 0 1



0 0

1 0 1



0 0  1 1  0 1



1 0    01 0 1 

0 0 1

 1 

0 0   0 1 1 1

1 0 0 0 1 0

 0 



1 0 0

 1 

1 0  1 0  0 1

1 0 1 1 0 1 1 0 1









2 1    −1   = 0 3 1 2 1   1 = −1    0 3 







2 1     −1  =  1 3 0 2 1   0 −1  =   3 1



0 1 1

0 1   0 0 1 0 

0 1 1

0 1    00 1 0

1 −1 0

0 1  0 0  1 0

The other two permutation matrices are not regular.



0 1 0



0 1 −1

−1  2 , 2

1 0  0 0  1 0 

0 1 −1



0 1 0

−1  4 , −2



3  −4  , −2 

1 1 0

3  2 . −2

1.4.26. False. Changing the permuation matrix typically changes the pivots.

1.5.1. (b)

⋆ (a) 

2

 3 

2

⋆

1.5.2. X =

1.5.3. (a)

!

3 2 −1 −1  1 1 3   −4 2 1  1 2 −1    

16 −8 3

−5 3 −1 !

0 1

1 , 0

(b) C =

⋆

1.5.8. (a) Set



(c)

2 3

!

P1 =

2 , 1

⋆ 

!



1

 0 



0

 1 

0

= P1 ,

1 0

!

16 −8 3 

1  (d)  0 0 2 3 1 3 

!

P2−1

0 1 0 1 0 0



0  0 , 1 

0  0 , 1

= P3 ,

P3−1

P2 =



0

 0 

1

P5 =



0

 0 

1

= P2 ,



6 1   −3  0 1 1 0 1 0

 1 3 . 1 3

0 1 , C −1 = B −1 A−1 =  0 1

P4 = Then



1 , B −1 =  −1 −

0

P1−1

!

6 −5    −3  ; X A =  3 1 −1

−1 2

1.5.6. (a) A−1 =

!

!

−1 −3 1 0 −1 −3 2 3 = = , 1 2 0 1 1 2 −1 −1      −1 −1 1 0 0 3 −1 −1 2 1       0 1 0  =  −4 3 2 2 1 = 2 1      0 1 0 0 1 −1 0 1 2 1

P4−1



0  3 , 1

2 1 −1





1  0 (e)   0 0

 1 3 . − 23

1 0 0 0 1 0



0  1 , 0

1  0 , 0

= P4 ,

P5−1

0 1 −6 0

P3 =



P6 =



0

 1 

0





0 1    3  = 0 −8 0

1

 0 

0

0 0 1 0

0 0 1 0 0 1



1  1 . 2 0 1 0 

0  0 . 0  1



0  0 . 1



1  0 , 0 

0  1 . 0

= P5 , P6−1 = P6 .

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Chapter 1: Instructors’ Solutions Manual 

0  0 1.5.9. (a)   0 1

⋆

0 0 1 0

0 1 0 0



1  0 (c)   0 0

0 0 0 1

0 1 0 0



0  0 , 1  0



⋆ (d)

1

 0   0   0

0

0 0 1 0 0

0 0 0 0 1

0 1 0 0 0



0  0  . 0  1  0

1.5.11. This is true if and only if A2 = I , and so, according to Exercise 1.2.37, A is either of ! ! ±1 0 a b the form or , where a is arbitrary and b c = 1 − a2 . 0 ±1 c −a 1.5.13. Since c is a scalar,

⋆



1  0 , 0  0

11

1 −1 A c

!

(c A) =

1 c A−1 A = I . c

1.5.14. If a = 0 the first row is all zeros, and so A is singular. Otherwise, we make d → 0 by an elementary row operation. If e = 0 then the resulting matrix has a row of all zeros. Otherwise, we make h → 0 by another elementary row operation, and the result is a matrix with a row of all zeros. 1.5.16. If all the diagonal entries are nonzero, then D−1 D = I . On the other hand, if one of diagonal entries is zero, then all the entries in that row are zero, and so D is singular. ♦ 1.5.19. (a) A = I −1 A I . (b) If B = S −1 AS, then A = S B S −1 = T −1 B T , where T = S −1 .

⋆ (c)

If B = S −1 A S and C = T −1 B T , then C = T −1 (S −1 AS)T = (S T )−1 A(S T ).  ! 1 0  0 

⋆

⋆



! −1  1 1 1 0 1.5.21. (a) B A = 1 = .  −1 −1 1 0 1 1 1 (b) A X = I does not have a solution. Indeed, the first column of this matrix equation is     ! 1 −1 1   x     0 , which has no solutions since x − y = 1, y = 0, the linear system  0 1 =     y 1 1 0 and x + y = 0 are incompatible.   −2 y 1 − 2 v   v  1.5.22. The general solution to A X = I is X =   y , where y, v are arbitrary. −1 1 Any of these matrices serves as a right inverse. On the other hand, the linear system Y A = I is incompatible and there is no solution.

1.5.23. (a) No. The only solutions are complex, with a = nonzero complex number. −1 −1

!

1 , B = 0



− 12 ± i

1 0

q

2 3



b, where b 6= 0 is any

!

0 (b) Yes. A simple example is A = . The general solution to the 1 ! x y 2 × 2 matrix equation has the form A = B M , where M = is any matrix with z w tr M = x + w = −1, and det M = x w − y z = 1. To see this, if we set A = B M , then ( I +M )−1 = I +M −1 , which is equivalent to I +M +M −1 = O. Writing this out using the formula (1.39) for the inverse, we find that if det M = x w − y z = 1 then tr M = x + w = −1, while if det M 6= 1, then y = z = 0 and x + x−1 + 1 = 0 = w + w−1 + 1, in which case, as in part (a), there are no real solutions.

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Chapter 1: Instructors’ Solutions Manual 1.5.25. (b)

(f )



−



−



    

1.5.26. (b)

⋆ (c) (f )

⋆ (h) 

1  0   0 0 1.5.28. (a)

⋆



−

 3 8 ; − 18 1 5 8 8 1 1 − 2 2 1 − 38 8

1 8 3 8 5 8 1 2 7 8

1 3

0 1

!

0

1

1 0



3  5

0 −8 0

 

⋆ (c) 

  ;  

1 0

3 1

1

0

!

 4 5 ; 3 5

0

2 −6 −5 2

1 3

3 ; 1

0

 1   0

=



4  3

 

−

i 2 1 2

 1 2 , − 2i

(c)



1

(d) no inverse;

!

− 43





i

 1− 

0 −i −1

i

−1





1 −2 0 0



=

1  −3  . −3   1

3 5 4 5

− 45



; (d) not possible; 3 1 0 1 0 53 0 1 5      1 0 0 1 0 0 1 0 0 1 0 0 1 0 0        3 1 0   0 1 0   0 1 0   0 −1 0   0 1 0        0 0 1 2 0 1 0 3 1 0 0 1 0 0 8       1 2 0 1 2 3 1 0 3 1 0 0        0 1 0   0 1 4   0 1 0  =  3 5 5 ;       0 0 1 2 1 2 0 0 1 0 0 1      1 0 0 0 1 0 0 0 2 0 0 0 1 0 1 0 0 0  1      0 1  0 1 0 0  0 − 1  0 0 1 0  1 0 0 0 0 2  2          1 0 0  0 0 1 0  0  0 1 0 0  0 0 1 0  0 0 0 0 0 1 0 0 0 1 0 0 −2 1 0 0 0 1 0 0     1  1 0 0 2 1 0 0 0 2 1 0 1 0 0 0 1 2 0 0       0 1 0 3  0 1 0 0  0  0 0 1 0 0 1 0 0 1   =      0 1 0 0 1 0 0  0 0 1 0  0 0 1 3  0  1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 −2 



−

⋆ (h)

!



3 5 4 5 



−1  1  , −i

⋆ (d)



3+ i 4i −1 + 2 i

  −4 + 

0 0 1 0



0  0  0  1



1  3 . −1   −5

−1 − i 2− i 1− i



−i 2+ i 1

 . 

e = E E ♥ 1.5.30. (a) If A N N −1 · · · E2 E1 A where E1 , . . . , EN represent the row operations ape = A e B = E E plied to A, then C N N −1 · · · E2 E1 A B = EN EN −1 · · · E2 E1 C, which represents the same sequence of row operations applied to C. (b) 

1  (E A) B =   2 −2

1.5.31. (b)

(d)



5 17 1 17



−

  

9 6 −1





2 −3 −3



−1 1   3 2  −2 −1

  2 17  2  3 12 17

−15 −10 2







−2 8    −9 0 =   1 −9



=

2 2





,

−8 3     −5   −1  = 1 5

⋆ (c) 

2



   3 ,  

0



−3 1    0 −2  =   2 −2 

2

  1  

0

⋆ (e)

− 52 −1 1 2   

0 1 0

 3 2    0   − 12

−4 2 3

3 −1 −1



0 8   −9 0  1 7 

3 

−2 = 2





     





−3  −2   = E (A B). −4 

14  −2

1 3     0  5  = 1 −7



5 , 

   



−4  1 , −3

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Chapter 1: Instructors’ Solutions Manual

(f )



1

 0   2

2

1.5.32. (b)

1.5.33. (b)

⋆ (c)



2

 2 

0

(d)



⋆ (e)



1

 0 

0

2

 1 

1

(f )



1

 1   1

3

1.5.34. (b)

1.6.1. (b)

1.6.2.



0 0 −1 −1

 1  4 ; 1 4

0 1

1 0

1 4 −2 0 0 1

−3 −1 −1



1 −1 −1 −1

0 −7

0 1  1 1  0 2 

1 1 −1 1

−8 , 3

!

0 , 2 

3  AT =  −1  −1

!



− 32



2 1   1 5 =  −1   1 6 3      

−

1 2

⋆ (c) 

1 6 2 3 2 3

0 1 −1 − 34



0

!

2 , 1

!



0 −3 0 0



1    (d)  −2  , 0

0 , 6

−1 2



1  (d)   2 −1 2 0

!

−



  ;  

⋆ (h)



4



   −10  .     −8 

3

1  −1  , 1

1 0



0 1  0 0  0 −7

1 1 0

5

1  0 , 1

0 0 −2 0



0 1  0 0   0  0 4 0



1

−1 1 0 0



−12    (e)  −3  , 7



2  0 , 2

−3 , 4

(B A)T = AT B T

⋆



(f )



1    (e)   2 , −3



−1  =  5 3



7  3 ,



1 0

⋆



1 2

0 −3 0

− 32

0 1  0 0   0  0 1 0

1 8 1 2 5 8

!



0 1  0 0  1 0

    

− 27 , 1

1 0



0 1  0 0  1 0



(f )

0 1   0  0 −1 0



  ,  

BT = −2 2

0

!



0 1 0

1 2

0 0 1 1

0 4

0 3 0



0 2  0 0  1 0



−7 0

0 2   0  0 1 0



0 1 1



1  2 , 1

!

5 1   2 −2  =   3 1



⋆ (c)

0 1



0 1

1 1 −1

(d) singular matrix;

1 0

=



(A B)T = B T AT =

⋆



 7 5 ; 1 5

−

2 1  1  1 =  2 1 2 2

!

1 1



4 2



−1 −4 2 1



2 1    −1   = 1 1 0





1 4 3      11   1 −1   = .    0  −7   4   −1 6 −2

⋆ (c)

!

13

6 −2 −2

1 0 1 0      



2  −1  . 0  1 7  3  2  . 5   − 53

(f )

1 2

3 4

!

5 . 6



−5  11  . 7

1.6.3. If A has size m × n and B has size n × p, then (A B)T has size p × m. Further, AT has size n × m and B T has size p × n, and so unless m = p the product AT B T is not defined. If m = p, then AT B T has size n × n, and so to equal (A B)T , we must have m = n = p,

so the matrices are square. Finally, taking the transpose of both sides, A B = (AT B T )T = (B T )T (AT )T = B A, and so they must commute.

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14

Chapter 1: Instructors’ Solutions Manual 1.6.5. (A B C)T = C T B T AT

⋆

1 0

1.6.6. False. For example,

1 1

!

does not commute with its transpose.

1.6.8. (a) (A B)−T = ((A B)T )−1 = (B T AT )−1 = (AT )−1 (B T )−1 = A−T B −T . ! ! 1 0 1 −2 (b) A B = , so (A B)−T = , 2 1 0 1 ! ! ! 0 −1 1 −1 1 −2 while A−T = , B −T = , so A−T B −T = . 1 1 −1 2 0 1

⋆

!

1 (3 4) = 2

1.6.10. No; for example,

3 6

4 8

!

!

3 (1 2) = 4

while

3 4

!

6 . 8

T ♦ 1.6.13. (a) Using Exercise 1.6.12, aij = eT i A ej = ei B ej = bij for all i, j.

1 0

(b) Two examples: A =

⋆

!

2 1

1 ; 1

0 0

A=

0 0

!

0 1

, B=

!

−1 . 0

♦ 1.6.15. (a) Note that (A P T )T = P AT , which permutes the rows of AT , which are the columns of A, according to the permutation P . Associativity of matrix multiplication implies that it doesn’t matter whether the rows or the columns are permuted first.

⋆ (c)

1.6.17. (b) a = −1, b = 2, c = 3;

⋆

!

1 1

, B=

1.6.18. (a)



1

 

0 1 0

 0 

0

0  0 , 1

0

1 0 0

 1 

0

 

0  0 , 1

a = −2, b = −1, c = −5.

0

 

0 1 0

 0 

1

1  0 , 0

1



0 0 1

 0 

0

0  1 . 0

♦ 1.6.20. True. Invert both sides of the equation AT = A, and use Lemma 1.32.

⋆ ⋆

0 1

♦ 1.6.21. False. For example

1 0

!

2 1

1 3

!

!

1 2

=

3 . 1

1.6.23. (a) Since A is symmetric, (An )T = (A A . . . A)T = AT AT . . . AT = A A . . . A = An . (b) (2 A2 − 3 A + I )T = 2 (A2 )T − 3 AT + I = 2 A2 − 3 A + I ; (c) If p(A) = cn An + · · · +

c1 A + c0 I , then p(A)T = cn An + · · · c1 A + c0 I T = cn (AT )n + · · · c1 AT + c0 I = p(AT ). In particular, if A = AT , then p(A)T = p(AT ) = p(A).

1.6.25. (a)

(c)

⋆ (d)

1 1 

1 4 1

  −1  

−1

1

  −1    0

3

!

=

−1 3 2 −1 2 2 0

1 1

0 1 

!

1 0

0 3



−1 1    −1 = 2   0 −1 0 2 −1 0





!

0 1 1 2

3 1    −1 0 =  0   0 1 3

1 0

1 1

!



,

0 1  0 0  1 0

0 1 2 3

0 0 1

6 5

0 2 0 



0 1  0 0  − 32 0

0 1  0 0   0  0 1 0

0 1 0 0

0 0 −5 0

−1 1 0

−1



1  2 ,

1



0 1  0 0   0  0 − 49 0 5

−1 1 0 0

0 2 1 0



3  3  6 . 5

1

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Chapter 1: Instructors’ Solutions Manual

⋆

⋆

15

♦ 1.6.27. The matrix is not regular, since after the first set of row operations the (2, 2) entry is 0. More explicitly, if       1 0 0 p 0 0 p ap bp         L= D= then L D LT =  a2 p + q abp + cq  a 1 0, 0 q 0,  ap . 2 2 b c 1 0 0 r bp abp + cq b p + c q + r Equating this to A, the (1, 1) entry requires p = 1, and so the (1, 2) entry requires a = 2, but the (2, 2) entry then implies q = 0, which is not an allowed diagonal entry for D. Even if we ignore this, the (1, 3) entry would set b = 1, but then the (2, 3) entry says a b p + c q = 2 6= −1, which is a contradiction. 1.6.29. (a) Let S = 12 (A + AT ), J = 12 (A − AT ). Then S T = S, J T = − J, and A = S + J. ! ! ! 0 − 12 1 2 1 5 (b) = 5 2 + 1 . 3 4 4 0 2 2

eU e , where L e = V T and U e = D L. e Thus, AT ♦ 1.6.30. Write A = L D V , then AT = V T D U T = L e , which are the pivots of AT , are the same as those is regular since the diagonal entries of U of D and U , which are the pivots of A.

19 1.7.1. ⋆ (a) The solution is x = − 10 7 , y = − 7 . Gaussian Elimination and Back Substitution requires 2 multiplications and 3 additions; Gauss–Jordan also uses 2 multiplications and 3 −1

additions; finding A

=

 

1 7 3 7

 2 7 1 7

by the Gauss–Jordan method requires 2 additions      − 1 2 − 10 7 7  4  7  = and 4 multiplications, while computing the solution x =  − 73 17 −7 − 19 7 takes another 4 multiplications and 2 additions.

(b) The solution is x = −4, y = −5, z = −1. Gaussian Elimination and Back Substitution requires 17 multiplications and 11 additions; Gauss–Jordan uses 20 multiplications and 11   0 −1 −1   −1  additions; computing A =   2 −8 −5  takes 27 multiplications and 12 additions, 3 2 −5 −3 while multiplying A−1 b = x takes another 9 multiplications and 6 additions.

⋆

1.7.3. Back Substitution requires about one half the number of arithmetic operations as multiplying a matrix times a vector, and so is twice as fast. ♦ 1.7.4. We begin by proving (1.63). We must show that 1 + 2 + 3 + . . . + (n − 1) = 21 n(n − 1) for n = 2, 3, . . .. For n = 2 both sides equal 1. Assume that (1.63) is true for n = k. Then 1 + 2 + 3 + . . . + (k − 1) + k = 12 k(k − 1) + k = 12 k(k + 1), so (1.63) is true for n = k + 1. Now the first equation in (1.64) follows if we note that 1 + 2 + 3 + . . . + (n − 1) + n =

⋆

1.7.6. Combining (1.62–63), we see that it takes

1 3

n3 + 21 n2 − 65 n multiplications and

additions to reduce the augmented matrix to upper triangular form j th

row by its pivot requires n−j +1 multiplications, for a total of

to produce the upper unitriangular form requires an additional grand total of

1 3

3

n +

solve the system.



1 2





1 2



1 3

n3 − 13 n

U | c . Dividing the

n2 + 12 n multiplications

V | e . To produce the solved form

1 2 1 2 n − 2 n multiplications and the 3 2 5 1 3 2 n − 6 n multiplications and 3 n

n(n + 1).



I | d



same number of additions for a

+

1 2

n2 −

5 6

n additions needed to

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16

⋆

Chapter 1: Instructors’ Solutions Manual 1.7.7. Less efficient, by, roughly, a factor of 1 2

3

1 2

n −

⋆







1 2



n multiplications and



0 1   −1 (b)    0 0

x=



1



  0  .   1

2

1.7.10. Both false. For example, 

1

 1   0

0

1 1 1 0



0 1 1 1

1.7.11.

0 1  1 0   1  0 1 0

⋆

(a)



1 1 1 0

2

  −1 



2

−1 2 −1 0

  −1    0

0

(b)

⋆





3 2 , 2,

4

  1  

♠ 1.7.13.

0 −1 2 −1

1



⋆

4

  1    0 

1



1 1

0

4

1

1

4

0

1







1

   1 = 1  4   1 4 4  

1

4

=

0 1 1 5

1

 1  4    0  1 4







1 2 3 2

0 1

− 23 0

− 23 0 0 1

1

1 1 1 0

 1   0

0



0 1

− 34

2 0   0  0 1 0 

0 2  0 0   0  0 1 0

−1 3 2

0 −1 3 2

0 0





0 −1 1    0 0  =  1   −1 1 1

0 1 1 1



0  −1  ,

0 0 1 −1

−1 1 0 0



1  −1  . 0  1

4 3

0 −1 4 3

0



0  0  , −1   5 4

(c) xi = i(n − i + 1)/2 for i = 1, . . . , n.



0  4 

 0  0

1



0  1 , 2  2



( 2, 3, 3, 2 )T .

  0    1 

2 3 2 1

0 1   1  −1   =  −2 2 0



1

4



0 2   2 0 =  1  1 1 0

0 1   1 − 0 = 2  −1  0   2 0

 3 T , 2 

1



0 1 1 1

−1 2 −1

0

⋆



n3 + n2 −

2 0 1 0 0 1 2 0 −2        −1   0 1 1 ,  3 ; −1 1  = 1 0 x =       −2 3 0 −2 1 0 0 5 0     0 0 1 0 0 0 1 −1 0 0 −1      −1 0 2 1 0 1 0 0 1 1 0 =  ,    −1 4 1   0 −1 1 0  0 0 5 1  0 −1 6 0 0 −1 1 0 0 0 7

1

  −1  

⋆

1 2

. It takes

n additions.



1.7.9. (a)

3 2



1 15 4

0

0

0

1

0

4 15 1 − 15

1 2 7

0 0





1

 3 4  18 5

4

   0  0   0 0 

1



0

1

0

15 4

1

0

56 15

0

0

1



     16   15  24 7

− 14

.



1 −1 −1 1 0 0 1 −1 −1       =  −1 1 0  0 ♥ 1.7.14. (b)  −1 2 −1 1 −2      , −1 −1 3 −1 2 1 0 0 2     1 −1 0 0 −1 1 0 0 0 0 1      −1   −1  0 1 0 0 0 2 −1 0 0         =  0 −1  0 −1  0 3 −1 0 1 0 0      1    0 −1 4 −1   0 0 −2 1 0  0  0 −1 0 0 −1 5 −1 −1 − 12 − 37 1 0

−1 1 0 0 0

0 −1 2 0 0

0 0 −1 7 2

0



−1  −1   . −1   − 32   13 7

The 4 × 4 case is a singular matrix.

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17

!

!

−8 −10 −8.1 , (b) , (c) . (d) Partial pivoting reduces the effect of 4 −4.1 −4.1 round off errors and results in a significantly more accurate answer.

1.7.16. (a)

⋆

1.7.18. (a) x = −2, y = 2, z = 3, (b) x = −7.3, y = 3.3, z = 2.9, (c) x = −1.9, y = 2., z = 2.9

⋆

1.7.19. (a) x = −220, y = 26, z = .91; (b) x = −190, y = 24, z = .84; (c) x = −210, y = 26, z = 1. (d) The exact solution is x = −213.658, y = 25.6537, z = .858586. Full pivoting is the most accurate. Interestingly, partial pivoting fares a little worse than regular elimination.

1.7.20. (a)

1.7.21. (a)

⋆

♠ 1.7.23. (a)



6  5   − 13  5  − 95 

     







1.2   , = −2.6   −1.8

 1 − 13   8 13

=

⋆ (b)

!

−.0769 , .6154

H3−1 =



⋆ (b) 9

  −36 

30

H4−1 =



16



− 45   − 8  15  − 19 15

−360 192 −180

  −120    240 

       

−140

25

1 4 − 54 1 8 1 4

−



    ,         

(c)

0



  1  .   1

0





−.8000    =  −.5333  , (c) −1.2667 

30  −180  , 180

−120 1200 −2700 1680



240 −2700 6480 −4200

        

2 121 38 121 59 242 56 − 121

        

=

     



.0165  .3141  . .2438   −.4628



−140  1680  , −4200   2800



−300 1050 −1400 630  4080 −18900 26880 −12600   . H5−1 = −18900 79380 −117600 56700   26880 −117600 179200 −88200   630 −12600 56700 −88200 44100 (b) The same results are obtained when using floating point arithmetic in either Math  −300    1050    −1400

f H , where K f ematica or Matlab. (c) The product K 10 10 10 is the computed inverse, is fairly close to the 10 × 10 identity matrix; the largest error is .0000801892 in Mathematf H , it is nowhere close to the identity matrix: ica or .000036472 in Matlab. As for K 20 20 in Mathematica the diagonal entries range from −1.34937 to 3.03755, while the largest (in absolute value) off-diagonal entry is 4.3505; in Matlab the diagonal entries range from −.4918 to 3.9942, while the largest (in absolute value) off-diagonal entry is −5.1994.

1.8.1. (a) Unique solution: (− 21 , − 34 )T ;

(c) no solutions;

⋆ (d)

unique solution: (1, −2, 1)T ;

(e) infinitely many solutions: (5 − 2 z, 1, z, 0)T , where z is arbitrary;

⋆ (g)

unique solution: (2, 1, 3, 1)T .

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Chapter 1: Instructors’ Solutions Manual (c) (1, 0)T ;

1.8.2. (b) Incompatible; arbitrary;

⋆

⋆ (f )



1.8.8. (b) (d)



⋆ (e)



1

 0 

  

0



⋆ (h)

(d) 3,

1

 2   1   4

0

T

⋆ (e)

1,

⋆ (h)

(f ) 1,

!





−1

2

−1 1 2 −1 3

2 −1 −3 3 −5



5 ) = ( 1 )( 0

!

1.8.10. Examples: (a)





1 0

0 1

!

−1

2

5 ),

0 1 1 1 1

0 0 1 0 0

0 0 0 1 0

0 1  0 0    0 0   0  0 1 0

1

0 1 0

0 , 0

(b)



 0 

0

( 3 + 2 i , −1 + 2 i , 3 i )T .



0  −1  , 1



1 1   2 0     = 1 −1     2  4 −2 0

⋆ (c)

2.

!

3 0    0 , 0   1 0

0 1 0

(iii ) a 6= −2, b = 0.

, where y is arbitrary;

2 1 3 1 0 2 1 3 = , −2 −1 −3 −1 1 0 0 0     0 0 2 −1 0 1 0 0 2 −1     3 1  =  1 1 0  0 0 1 1 −1   2  2 1 0 2 −1 1 1 0 1 0 0

3 1    0 0 =   −2 − 23

(f ) ( 0

(ii ) a = b 6= 0, or a = −2, b = 0;

1 + i − 12 (1 + i )y, y, − i

1.8.7. (b) 1,

⋆

(−5 − 3 x4 , 19 − 4 x4 , −6 − 2 x4 , x4 )T , where x4 is arbitrary.

1.8.4. (i ) a 6= b and b 6= 0; 1.8.5. (a)

(d) (1 + 3 x2 − 2 x3 , x2 , x3 )T , where x2 and x3 are





0  0 , 0

−1 3 0 0 0

(c)

2 −5 0 0 0 

1

 0 

0



1  −2   . 0  0  0 

0  1 . 0

1.8.11. Example: (a) x2 + y 2 = 1, x2 − y 2 = 2;

⋆ (c)

y = x3 , x − y = 0; solutions: x = y = 0, x = y = −1, x = y = 1.

1.8.13. True. For example, take a matrix in row echelon form with r pivots, e.g., the matrix A with aii = 1 for i = 1, . . . , r, and all other entries equal to 0.

⋆

♥ 1.8.15. (a) Each row of A = v wT is a scalar multiple, namely vi w, of the vector w. If necessary, we use a row interchange to ensure that the first row is non-zero. We then subtract the appropriate scalar multiple of the first row from all the others. This makes all rows below the first zero, and so the resulting matrix is in row echelon form has a single non-zero row, and hence a single pivot — proving that A has rank 1. ! ! −1 2 2 6 −2 (b) (i ) , (iii ) . −3 6 −3 −9 3

(c) The row echelon form of A must have a single nonzero row, say wT . Reversing the elementary row operations that led to the row echelon form, at each step we either interchange rows or add multiples of one row to another. Every row of every matrix obtained in such a fashion must be some scalar multiple of wT , and hence the original matrix A = v wT , where the entries vi of the vector v are the indicated scalar multiples.

1.8.17. 1.

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⋆

1.8.19. Example: A = 0 0

BA =

0 0

!

1 0

0 0

!

, B=

19 0 0

1 0

!

so A B =

0 0

1 0

!

has rank 1, but

has rank 0.

♦ 1.8.21. By Proposition 1.39, A can be reduced to row echelon form U by a sequence of elemen-

tary row operations. Therefore, as in the proof of the L U decomposition, A = E1 E2 · · · EN U

−1 where E1−1 , . . . , EN are the elementary matrices representing the row operations. If A is

singular, then U = Z must have at least one all-zero row.

1.8.22. (a) x = z, y = z, where z is arbitrary; (c) x = y = z = 0; ⋆ (d) x = 31 z − 23 w, y = 56 z − 16 w, where z and w are arbitrary; ⋆ (e) x = 13 z, y = 5 z, w = 0, where z is arbitrary. 1.8.23. (a) (c)

⋆





1 3

y, y

T

, where y is arbitrary;

3 2 6 − 11 5 z + 5 w, 5 z − 5 w, z, w

T

⋆ (b)



− 65 z, 85 z, z

T

, where z is arbitrary;

, where z and w are arbitrary;

( 0, 0, 0 )T .

1.8.25. For the homogeneous case x1 = x3 , x2 = 0, where x3 is arbitrary. For the inhomogeneous case x1 = x3 + 41 (a + b), x2 = 12 (a − b), where x3 is arbitrary. The solution to the homogeneous version is a line going through the origin, while the inhomogeneous solution 

is a parallel line going through the point 14 (a + b), 0, 21 (a − b) free variable x3 is the same as in the homogeneous case. 1.8.27. ⋆ (a) k = 2 or k = −2;

T

(b) Singular matrix, row echelon form U =

⋆

. The dependence on the

(b) k = 0 or k = 12 .

2 0

1.9.1. (a) Regular matrix, reduces to upper triangular form U =    

−1 0 0

0 1 0



!

−1 , so its determinant is 2. 1

3  −2  , so its determinant is 0. 0 



1

2 3  1 2 (c) Regular matrix, reduces to upper triangular form U = , so its determinant is −3. 0 0 −3   −2 1 3    (d) Nonsingular matrix, reduces to upper triangular form U =   0 1 −1  after one row interchange, so its determinant is 6. 0 0 3  0 

⋆ (f )

Nonsingular matrix, reduces to upper triangular form U = after one row interchange, so its determinant is 40.

⋆

⋆ (e)

1.9.2. det A = −2, det B = −11 and det A B =



 det  

5 1 −2

4 5 10



1

 0   0

0



−2 2 0 0

1 −1 −2 0



4  −7   −8   10

4  1  = 22. 0

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Chapter 1: Instructors’ Solutions Manual 1.9.4. (a) True. By Theorem 1.52, A is nonsingular, so, by Theorem 1.18, A−1 exists. ! ! 2 3 0 1 (c) False. For A = and B = , we have −1 −2 0 0 ! 2 4 det(A + B) = det = 0 6= −1 = det A + det B. −1 −2

True. det(A B −1 ) = det A det B −1 = det A/ det B, where the first equality follows from formula (1.85) and the second equality follows from Proposition 1.55. (g) True. Proposition 1.42 says rank A = n if and only if A is nonsingular, while Theorem 1.52 implies that det A 6= 0.

⋆ (e) ⋆

1.9.5. By (1.85, 86) and commutativity of numeric multiplication, 1 det B = det(S −1 A S) = det S −1 det A det S = det A det S = det A. det S

⋆

1.9.7. By Proposition 1.56, det LT = det L. If L is a lower triangular matrix, then LT is an upper triangular matrix. By Theorem 1.50, det LT is the product of its diagonal entries which are the same as the diagonal entries of L. ♦ 1.9.9. det

a c + ka

ka c

d b

!

= c b − a d = −(a d − b c) = − det

kb d

!

= k a d − k b c = k (a d − b c) = k det

b d

!

= a d − b 0 = ad.

a 0

det

⋆

= a d + a k b − b c − b k a = a d − b c = det

c a

det det

!

b d + kb

a c

b d a c

a c

!

b d

!

,

,

b d

!

,

1.9.11. Indeed, by (1.85), det A det A−1 = det(A A−1 ) = det I = 1. 1.9.13. 

a11   a21 det    a31 a41

a12 a22 a32 a42

a13 a23 a33 a43



a14  a24  = a34   a44

a11 a22 a33 a44 − a11 a22 a34 a43 − a11 a23 a32 a44 + a11 a23 a34 a42 − a11 a24 a33 a42

+ a11 a24 a32 a43 − a12 a21 a33 a44 + a12 a21 a34 a43 + a12 a23 a31 a44 − a12 a23 a34 a41

+ a12 a24 a33 a41 − a12 a24 a31 a43 + a13 a21 a32 a44 − a13 a21 a34 a42 − a13 a22 a31 a44

+ a13 a22 a34 a41 − a13 a24 a32 a41 + a13 a24 a31 a42 − a14 a21 a32 a43 + a14 a21 a33 a42 + a14 a22 a31 a43 − a14 a22 a33 a41 + a14 a23 a32 a41 − a14 a23 a31 a42 .

⋆

♦ 1.9.16. The determinant of an elementary matrix of type #2 is −1, whereas all elementary matrices of type #1 have determinant +1, and hence so does any product thereof.

⋆

♦ 1.9.19. Using the L U  1 1   t (b) det  1 t2 t21 t22

factorizations established in Exercise 1.3.24:  1  t3   = (t2 − t1 )(t3 − t1 )(t3 − t2 ). t23 Solutions Manual

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21

♥ 1.9.20. (a) By direct substitution: ax + by = a

aq − pc pd − bq +b = p, ad − bc ad − bc

cx + dy = c

!

⋆

⋆

pd − bq aq − pc +d = q. ad − bc ad − bc !

13 3 1 13 1 1 (b) (i ) x = − det = −2.6, y = − det = 5.2. 10 0 2 10 4 0 ! ! ⋆ (ii ) x = 121 det −24 −26 = 53 , y = 121 det 13 −24 = − 67 .     3 4 0 1 3 0     1 1 7 1 (d) (i ) x = det  1 y = det  1 , 2 2 =− ,  4 2 = 9 9 9 9 0 1 −1 −1 0 −1   1 4 3   1 8  z = det  .  4 2 2 = 9 9 −1 1 0

♦ 1.9.21. (a) We can individually reduce A and B to upper triangular forms U1 and U2 with the determinants equal to the products of their respective diagonal entries. Applying the analo! U1 O gous elementary row operations to D will reduce it to the upper triangular form , O U2 and its determinant is equal to the product of its diagonal entries, which are the diagonal entries of both U1 and U2 , so det D = det U1 det U2 = det A det B. (c) (i )



3

 det  0

0



1   −3 (iii ) det    0 0



2 4 3 2 1 3 0

−2  −5   = det(3) det 7 0 4 1 0



4 3

 4 1   −1   = det  −3  8  0 −3

−5 7 2 1 3

!

= 3 · 43 = 129, 

0  4  det(−3) = (−5) · (−3) = 15. 1

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