Linear-Algebra-9E-Sol--Bernard-Kolman--David-Hill

Instructor’s Solutions Manual

Elementary Linear
Algebra with
Applications
Ninth Edition

Bernard Kolman
Drexel University

David R. Hill
Temple University


Editorial Director, Computer Science, Engineering, and Advanced Mathematics: Marcia J. Horton
Senior Editor: Holly Stark
Editorial Assistant: Jennifer Lonschein
Senior Managing Editor/Production Editor: Scott Disanno
Art Director: Juan L´
opez
Cover Designer: Michael Fruhbeis
Art Editor: Thomas Benfatti
Manufacturing Buyer: Lisa McDowell
Marketing Manager: Tim Galligan
Cover Image: (c) William T. Williams, Artist, 1969 Trane, 1969 Acrylic on canvas, 108 × 84 .
Collection of The Studio Museum in Harlem. Gift of Charles Cowles, New York.

c 2008, 2004, 2000, 1996 by Pearson Education, Inc.
Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
Earlier editions c 1991, 1986, 1982, by KTI;

1977, 1970 by Bernard Kolman

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in
writing from the publisher.

Printed in the United States of America
10

9

8

7

6

5

4

3

2

1

ISBN 0-13-229655-1

Pearson
Pearson

Pearson
Pearson
Pearson
Pearson
Pearson
Pearson

Education, Ltd., London
Education Australia PTY. Limited, Sydney
Education Singapore, Pte., Ltd
Education North Asia Ltd, Hong Kong
Education Canada, Ltd., Toronto
Educaci´
on de Mexico, S.A. de C.V.
Education—Japan, Tokyo
Education Malaysia, Pte. Ltd

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Contents
Preface

iii

1 Linear Equations and Matrices
1.1 Systems of Linear Equations . . . . . . . . . . . . . .
1.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Matrix Multiplication . . . . . . . . . . . . . . . . .
1.4 Algebraic Properties of Matrix Operations . . . . . .

1.5 Special Types of Matrices and Partitioned Matrices .
1.6 Matrix Transformations . . . . . . . . . . . . . . . .
1.7 Computer Graphics . . . . . . . . . . . . . . . . . . .
1.8 Correlation Coefficient . . . . . . . . . . . . . . . . .
Supplementary Exercises . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . .

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1
1
2
3
7
9
14
16
18
19

24

2 Solving Linear Systems
2.1 Echelon Form of a Matrix . .
2.2 Solving Linear Systems . . . .
2.3 Elementary Matrices; Finding
2.4 Equivalent Matrices . . . . .
2.5 LU -Factorization (Optional) .
Supplementary Exercises . . .
Chapter Review . . . . . . . .

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27
27
28
30
32
33
33
35

3 Determinants
3.1 Definition . . . . . . . . . . . . . . .
3.2 Properties of Determinants . . . . .
3.3 Cofactor Expansion . . . . . . . . . .
3.4 Inverse of a Matrix . . . . . . . . . .
3.5 Other Applications of Determinants
Supplementary Exercises . . . . . . .
Chapter Review . . . . . . . . . . . .

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37
37
37
39
41
42
42
43

4 Real Vector Spaces
4.1 Vectors in the Plane and in 3-Space
4.2 Vector Spaces . . . . . . . . . . . . .
4.3 Subspaces . . . . . . . . . . . . . . .
4.4 Span . . . . . . . . . . . . . . . . . .
4.5 Span and Linear Independence . . .
4.6 Basis and Dimension . . . . . . . . .
4.7 Homogeneous Systems . . . . . . . .
4.8 Coordinates and Isomorphisms . . .
4.9 Rank of a Matrix . . . . . . . . . . .

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45
45
47
48
51
52
54
56
58
62

. . .
. . .
A−1
. . .
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. . .

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ii

CONTENTS
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 Inner Product Spaces
5.1 Standard Inner Product on R2 and
5.2 Cross Product in R3 (Optional) . .
5.3 Inner Product Spaces . . . . . . . .
5.4 Gram-Schmidt Process . . . . . . .
5.5 Orthogonal Complements . . . . .
5.6 Least Squares (Optional) . . . . .
Supplementary Exercises . . . . . .
Chapter Review . . . . . . . . . . .

64
69

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71
71
74
77
81
84
85
86
90

6 Linear Transformations and Matrices
6.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Kernel and Range of a Linear Transformation . . . . . . . . . . . . . . . . . . . .
6.3 Matrix of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Vector Space of Matrices and Vector Space of Linear Transformations (Optional)
6.5 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6 Introduction to Homogeneous Coordinates (Optional) . . . . . . . . . . . . . . .
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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93
93
96
97
99
102
103
105
106

7 Eigenvalues and Eigenvectors
7.1 Eigenvalues and Eigenvectors . . . . .
7.2 Diagonalization and Similar Matrices .
7.3 Diagonalization of Symmetric Matrices
Supplementary Exercises . . . . . . . .
Chapter Review . . . . . . . . . . . . .

R3
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109
109

115
120
123
126

8 Applications of Eigenvalues and Eigenvectors (Optional)
8.1 Stable Age Distribution in a Population; Markov Processes
8.2 Spectral Decomposition and Singular Value Decomposition
8.3 Dominant Eigenvalue and Principal Component Analysis .
8.4 Differential Equations . . . . . . . . . . . . . . . . . . . . .
8.5 Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . .
8.6 Real Quadratic Forms . . . . . . . . . . . . . . . . . . . . .
8.7 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . .
8.8 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . .

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129
129
130
130
131
132
133
134
135

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10 MATLAB Exercises

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137

Appendix B Complex Numbers
163
B.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
B.2 Complex Numbers in Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

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Preface
This manual is to accompany the Ninth Edition of Bernard Kolman and David R.Hill’s Elementary Linear
Algebra with Applications. Answers to all even numbered exercises and detailed solutions to all theoretical
exercises are included. It was prepared by Dennis Kletzing, Stetson University. It contains many of the
solutions found in the Eighth Edition, as well as solutions to new exercises included in the Ninth Edition of
the text.

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Chapter 1

Linear Equations and Matrices
Section 1.1, p. 8
2. x = 1, y = 2, z = −2.
4. No solution.
6. x = 13 + 10t, y = −8 − 8t, t any real number.
8. Inconsistent; no solution.
10. x = 2, y = −1.
12. No solution.
14. x = −1, y = 2, z = −2.

16. (a) For example: s = 0, t = 0 is one answer.

(b) For example: s = 3, t = 4 is one answer.
(c) s = 2t .
18. Yes. The trivial solution is always a solution to a homogeneous system.
20. x = 1, y = 1, z = 4.
22. r = −3.
24. If x1 = s1 , x2 = s2 , . . . , xn = sn satisfy each equation of (2) in the original order, then those
same numbers satisfy each equation of (2) when the equations are listed with one of the original ones
interchanged, and conversely.
25. If x1 = s1 , x2 = s2 , . . . , xn = sn is a solution to (2), then the pth and qth equations are satisfied.
That is,
ap1 s1 + · · · + apn sn = bp

aq1 s1 + · · · + aqn sn = bq .

Thus, for any real number r,
(ap1 + raq1 )s1 + · · · + (apn + raqn )sn = bp + rbq .
Then if the qth equation in (2) is replaced by the preceding equation, the values x1 = s1 , x2 = s2 , . . . ,
xn = sn are a solution to the new linear system since they satisfy each of the equations.

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2

Chapter 1
26. (a) A unique point.
(b) There are infinitely many points.
(c) No points simultaneously lie in all three planes.
C2

28. No points of intersection:

C1

One point of intersection:

C1

Two points of intersection:

C2

C1

C2

C2

C1

C1 = C2

Infinitely many points of intersection:

30. 20 tons of low-sulfur fuel, 20 tons of high-sulfur fuel.
32. 3.2 ounces of food A, 4.2 ounces of food B, and 2 ounces of food C.
34. (a)

p(1) = a(1)2 + b(1) + c = a + b + c = −5
p(−1) = a(−1)2 + b(−1) + c = a − b + c = 1

p(2) = a(2)2 + b(2) + c = 4a + 2b + c = 7.

(b) a = 5, b = −3, c = −7.

Section 1.2, p. 19


0
1


2. (a) A =  0

0
1

1
0
1
1
1

0
1
0
0
0

0
1

0
0
0


1
1


0

0
0



0
1


(b) A =  1

1
1

1
0
1
0
0

1
1
0
1
0

1
0
1
0
0


1
0


0 .

0
0

4. a = 3, b = 1, c = 8, d = −2.


5 −5
8
7 −7
6. (a) C + E = E + C =  4

(b) Impossible.
(c)
.
2
9 .
0
1
5
3
4




−9
3 −9
0
10 −9
(d)  −12 −3 −15 .
(e)  8 −1 −2 .
(f) Impossible.
−6 −3 −9
−5 −4
3


1
T

8. (a) A = 2

3


2
1
1 , (AT )T =
2
4

2
1

3
.
4



5

(b) −5
8

www.pdfgrip.com

4
2
9



5
3 .
4

(c)

−6
11

10
.
17


3

Section 1.3


(d)


4
3 .
10

3

(e) 6
9

0 −4
.
4
0

(f)

17
−16

2
.
6

1 0
1 0
3 0
+1
=
.
0 1
0 0
0 2


λ−1
−2
−3
12.  −6

λ+2
−3 .
−5
−2
λ−4

10. Yes: 2

14. Because the edges can be traversed in either direction.
 
x1
 x2 
 
16. Let x =  .  be an n-vector. Then
 .. 
xn


   
  
x1
0
x1 + 0
x1
 x2   0   x2 + 0   x2 
    
  
x + 0 =  .  +  .  =  .  =  .  = x.
 ..   ..   ..   .. 
0

xn
xn
xn + 0

n

m

18.
i=1 j=1

aij = (a11 + a12 + · · · + a1m ) + (a21 + a22 + · · · + a2m ) + · · · + (an1 + an2 + · · · + anm )
= (a11 + a21 + · · · + an1 ) + (a12 + a22 + · · · + an2 ) + · · · + (a1m + a2m + · · · + anm )
m

n

=

aij .
j=1 i=1

n

n

19. (a) True.

n

(ai + 1) =
i=1
n

(b) True.
i=1



(c) True. 




m
j=1

n

i=1



ai +
i=1

ai  

m

j=1

i=1

ai + n.
i=1

n

1 =



n

1=

m = mn.
i=1



m

bj  = a1

m

m

bj + a2
j=1

j=1

bj + · · · + an
m

= (a1 + a2 + · · · + an )
n

=

m

m
j=1

i=1

bj =

ai
i=1

bj
j=1
n

j=1

ai bj

20. “new salaries” = u + .08u = 1.08u.

Section 1.3, p. 30
2. (a) 4.

(b) 0.

(c) 1.

(d) 1.

4. x = 5.

www.pdfgrip.com

bj
j=1


4

Chapter 1
√
6. x = ± 2, y = ±3.
8. x = ±5.

10. x = 65 , y =

12
5 .




0 −1 1
(b)  12
5 17 .
19
0 22

12. (a) Impossible.

14. (a)

58 12
.
66 13

(b) Same as (a).

(d) Same as (c).

16. (a) 1.


9
(f)  0

−3

(c)


0 −3
0
0 .
0
1

15 −7
(c)  23 −5
13 −1
(c)

28 32
; same.
16 18

(e)

(b) −6.



−3 0 1 .

28 8 38
.

34 4 41


14
29 .
17



8
(d)  14
13


8
13 .
9

(e) Impossible.

−16 −8 −26
.
−30
0 −31


−1
4
2
(d)  −2

(e) 10.
8
4 .
3 −12 −6
(f)

(g) Impossible.

18. DI2 = I2 D = D.
0
.
0
 
1
 14 

22. (a) 
 0 .

20.

0
0

13




0

 18 

(b) 
 3 .

13
 
 
 
 
 
 
1
−2
−1
1
−2
−1
24. col1 (AB) = 1  2  + 3  4  + 2  3 ; col2 (AB) = −1  2  + 2  4  + 4  3 .
3
0
−2
3
0
−2
26. (a) −5.

(b) BAT

28. Let A = aij be m × p and B = bij be p × n.

(a) Let the ith row of A consist entirely of zeros, so that aik = 0 for k = 1, 2, . . . , p. Then the (i, j)
entry in AB is
p

aik bkj = 0

for j = 1, 2, . . . , n.

k=1

(b) Let the jth column of A consist entirely of zeros, so that akj = 0 for k = 1, 2, . . . , m. Then the
(i, j) entry in BA is
m

bik akj = 0

for i = 1, 2, . . . , m.

k=1



2
3
30. (a) 
2
0

3 −3

1
0
2
0
3
0 −4
0
1
1


1
3
.
0
1



2
3
(b) 
2
0

3 −3
1
0
2
0

3
0 −4
0
1
1

www.pdfgrip.com

 
 
 x1
7
1  
x
2




3     −2 
.
 x3  =
0    3
 x4 
5
1
x5


5

Section 1.3

2
3
(c) 
2
0
32.

3 −3
1
0
2
0
3
0 −4
0
1
1

−2
3
1 −5

x1
5
=
.
x2

4


1
7
3 −2 

0
3
1
5

2x1 + x2 + 3x3 + 4x4 = 0
34. (a) 3x1 − x2 + 2x3
=3
−2x1 + x2 − 4x3 + 3x4 = 2
36. (a) x1

38. (a)

3
1

1 2
2 5

39. We have

(b) same as (a).



   
−1
1
3
2
1
4
+ x2
+ x3
=
.
(b) x1  2  + x2  −1  =  −2 .
−1
4
−2
3
1
1
 

   
x1
1 2 1
x1
0
0  
1
.

(b)  1 1 2   x2  =  0 .
x2 =
3
1
2 0 2
0
x3
x3

n

u·v =

i=1

ui vi = u1 u2 · · · un


v1
 v2 
 
 .  = uT v.
 .. 


vn





1 0 0
40. Possible answer:  2 0 0 .
3 0 0
42. (a) Can say nothing.

(b) Can say nothing.

n

43. (a) Tr(cA) =

n

caii = c
i=1

aii = c Tr(A).
i=1

n

(b) Tr(A + B) =

n

n

(aii + bii ) =
i=1

aii +
i=1

bii = Tr(A) + Tr(B).
i=1

(c) Let AB = C = cij . Then
n

n

Tr(AB) = Tr(C) =
i=1
n

(d) Since aTii = aii , Tr(AT ) =

n

n

cii =

n

aik bki =
i=1 k=1

bki aik = Tr(BA).
k=1 i=1

n

aTii =
i=1

aii = Tr(A).
i=1

(e) Let AT A = B = bij . Then
n

bii =

n

aTij aji =
j=1

n

a2ji

=⇒

Tr(B) = Tr(AT A) =

j=1

i=1

Hence, Tr(AT A) ≥ 0.

www.pdfgrip.com

n

n

bii =
i=1 j=1

a2ij ≥ 0.


6

Chapter 1
44. (a) 4.

(b) 1.

(c) 3.
1 0
0 1

45. We have Tr(AB − BA) = Tr(AB) − Tr(BA) = 0, while Tr


b1j

 b2j 


be m × n and n × p, respectively. Then bj =  .  and the ith
 .. 
bnj


46. (a) Let A = aij and B = bij
n

entry of Abj is

= 2.

aik bkj , which is exactly the (i, j) entry of AB.
k=1

(b) The ith row of AB is
we have

k

aik bk1

ai b =

k

k

aik bk2 · · ·

aik bk1

k

k

aik bkn . Since ai = ai1 ai2 · · · ain ,

aik bk2 · · ·

k

aik bkn .

This is the same as the ith row of Ab.
47. Let A = aij and B = bij be m × n and n × p, respectively. Then the jth column of AB is

a11 b1j + · · · + a1n bnj


..
(AB)j = 

.
am1 b1j + · · · + amn bnj





a11
a1n




= b1j  ...  + · · · + bnj  ... 
am1
amn


= b1j Col1 (A) + · · · + bnj Coln (A).

Thus the jth column of AB is a linear combination of the columns of A with coefficients the entries in
bj .
48. The value of the inventory of the four types of items.
50. (a) row1 (A) · col1 (B) = 80(20) + 120(10) = 2800 grams of protein consumed daily by the males.
(b) row2 (A) · col2 (B) = 100(20) + 200(20) = 6000 grams of fat consumed daily by the females.

51. (a) No. If x = (x1 , x2 , . . . , xn ), then x · x = x21 + x22 + · · · + x2n ≥ 0.
(b) x = 0.

52. Let a = (a1 , a2 , . . . , an ), b = (b1 , b2 , . . . , bn ), and c = (c1 , c2 , . . . , cn ). Then
n

(a) a · b =

i=1

n

ai bi and b · a =

i=1

n

(b) (a + b) · c =

n

n

(ai + bi )ci =
i=1

n

(c) (ka) · b =

bi ai , so a · b = b · a.

i=1

bi ci = a · c + b · c.

n

(kai )bi = k
i=1

ai ci +
i=1

i=1

ai bi = k(a · b).

www.pdfgrip.com


7

Section 1.4
53. The i, ith element of the matrix AAT is
n

n

n

aik aTki =
k=1

aik aik =
k=1

(aik )2 .

k=1

n

Thus if AAT = O, then each sum of squares
k=1

and k. Thus A = O.
54. AC =

(aik )2 equals zero, which implies aik = 0 for each i

17 2 22
. CA cannot be computed.
18 3 23

55. B T B will be 6 × 6 while BB T is 1 × 1.

Section 1.4, p. 40
1. Let A = aij , B = bij , C = cij . Then the (i, j) entry of A + (B + C) is aij + (bij + cij ) and
that of (A + B) + C is (aij + bij ) + cij . By the associative law for addition of real numbers, these two
entries are equal.
2. For A = aij , let B = −aij .
n

(aik + bik )ckj and that of

4. Let A = aij , B = bij , C = cij . Then the (i, j) entry of (A + B)C is
n

AC + BC is

n

aik ckj +
k=1

k=1

bik ckj . By the distributive and additive associative laws for real numbers,
k=1

these two expressions for the (i, j) entry are equal.
6. Let A = aij , where aii = k and aij = 0 if i = j, and let B = bij . Then, if i = j, the (i, j) entry of
n

AB is

n

ais bsj = kbij , while if i = j, the (i, i) entry of AB is
s=1

ais bsi = kbii . Therefore AB = kB.
s=1
n

7. Let A = aij and C = c1 c2 · · · cm . Then CA is a 1 × n matrix whose ith entry is

a1j

 a2j 


Since Aj =  . , the ith entry of
 .. 
amj


n

cj aij .
j=1

m

cj Aj is
j=1

cj aij .
j=1

cos 2θ sin 2θ
cos 3θ sin 3θ
cos kθ sin kθ
.
(b)
.
(c)
.
− sin 2θ cos 2θ

− sin 3θ cos 3θ
− sin kθ cos kθ
(d) The result is true for p = 2 and 3 as shown in parts (a) and (b). Assume that it is true for p = k.
Then

8. (a)

Ak+1 = Ak A =
=
=

cos kθ sin kθ
− sin kθ cos kθ

cos θ sin θ
− sin θ cos θ

cos kθ cos θ − sin kθ sin θ cos kθ sin θ + sin kθ cos θ
− sin kθ cos θ − cos kθ sin θ cos kθ cos θ − sin kθ sin θ
cos(k + 1)θ sin(k + 1)θ
.
− sin(k + 1)θ cos(k + 1)θ

Hence, it is true for all positive integers k.

www.pdfgrip.com


8

Chapter 1


√1
√1
1 0
0 1
2 .
10. Possible answers: A =
;A=
;A= 2
1
0 1
1 0
√1
√
− 2
2
12. Possible answers: A =

1
1
0 0
0 1
;A=
;A=
.
−1 −1
0 0
0 0

13. Let A = aij . The (i, j) entry of r(sA) is r(saij ), which equals (rs)aij and s(raij ).
14. Let A = aij . The (i, j) entry of (r + s)A is (r + s)aij , which equals raij + saij , the (i, j) entry of
rA + sA.
16. Let A = aij , and B = bij . Then r(aij + bij ) = raij + rbij .
n

n

18. Let A = aij and B = bij . The (i, j) entry of A(rB) is
k=1

(i, j) entry of r(AB).
20.

1
6 A,

aik (rbkj ), which equals r

aik bkj , the
k=1

k = 16 .

22. 3.
24. If Ax = rx and y = sx, then Ay = A(sx) = s(Ax) = s(rx) = r(sx) = ry.
26. The (i, j) entry of (AT )T is the (j, i) entry of AT , which is the (i, j) entry of A.
27. (b) The (i, j) entry of (A + B)T is the (j, i) entry of aij + bij , which is to say, aji + bji .
(d) Let A = aij and let bij = aji . Then the (i, j) entry of (cA)T is the (j, i) entry of caij , which

is to say, cbij .




5 0
−4 −8
28. (A + B)T =  5 2 , (rA)T =  −12 −4 .
1 2
−8 12




−34
−34
30. (a)  17 .
(b)  17 .
(c) B T C is a real number (a 1 × 1 matrix).
−51
−51
1 −3
1 2
−1 2
;B= 2
;C=
.
0
0
1

0 1
3
2 0
0 0
0 0
A=
;B=
;C=
.
3 0
1 0
0 1

32. Possible answers: A =

33. The (i, j) entry of cA is caij , which is 0 for all i and j only if c = 0 or aij = 0 for all i and j.
34. Let A =

a b
be such that AB = BA for any 2 × 2 matrix B. Then in particular,
c d
a b
c d

1 0
1 0
=
0 0
0 0
a 0

a b
=
c 0
0 0

so b = c = 0, A =

a 0
.
0 d

www.pdfgrip.com

a b
c d


9

Section 1.5
Also
a 0
0 d

1 1
1 1
=
0 0
0 0

a 0
0 d

a a
a d
=
,
0 0
0 0
which implies that a = d. Thus A =

a 0
for some number a.
0 a

35. We have
(A − B)T = (A + (−1)B)T

= AT + ((−1)B)T
= AT + (−1)B T = AT − B T

by Theorem 1.4(d)).

36. (a) A(x1 + x2 ) = Ax1 + Ax2 = 0 + 0 = 0.
(b) A(x1 − x2 ) = Ax1 − Ax2 = 0 − 0 = 0.
(c) A(rx1 ) = r(Ax1 ) = r0 = 0.

(d) A(rx1 + sx2 ) = r(Ax1 ) + s(Ax2 ) = r0 + s0 = 0.
37. We verify that x3 is also a solution:
Ax3 = A(rx1 + sx2 ) = rAx1 + sAx2 = rb + sb = (r + s)b = b.

38. If Ax1 = b and Ax2 = b, then A(x1 − x2 ) = Ax1 − Ax2 = b − b = 0.

Section 1.5, p. 52
1. (a) Let Im = dij so dij = 1 if i = j and 0 otherwise. Then the (i, j) entry of Im A is
m

dik akj = dii aij

(since all other d’s = 0)

k=1

= aij

(since dii = 1).

2. We prove that the product of two upper triangular matrices is upper triangular: Let A = aij with
n

aij = 0 for i > j; let B = bij with bij = 0 for i > j. Then AB = cij where cij =

aik bkj . For
k=1

i > j, and each 1 ≤ k ≤ n, either i > k (and so aik = 0) or else k ≥ i > j (so bkj = 0). Thus every
term in the sum for cij is 0 and so cij = 0. Hence cij is upper triangular.
3. Let A = aij and B = bij , where both aij = 0 and bij = 0 if i = j. Then if AB = C = cij , we
n

have cij =

aik bkj = 0 if i = j.
k=1



9 −1
4. A + B =  0 −2
0
0




1
18 −5 11
7  and AB =  0 −8 −7 .
3
0
0
0

5. All diagonal matrices.

www.pdfgrip.com


10

Chapter 1

6. (a)

7 −2
−3 10

(b)

−9 −11
22
13

(c)

20 −20
4
76
q summands

8. Ap Aq = (A · A · · · A) (A · A · · · A) = Ap+q ;
p factors

(Ap )q = Ap Ap Ap · · · Ap = Ap + p + · · · + p = Apq .
q factors

q factors

p + q factors

9. We are given that AB = BA. For p = 2, (AB)2 = (AB)(AB) = A(BA)B = A(AB)B = A2 B 2 .
Assume that for p = k, (AB)k = Ak B k . Then

(AB)k+1 = (AB)k (AB) = Ak B k · A · B = Ak (B k−1 AB)B
= Ak (B k−2 AB 2 )B = · · · = Ak+1 B k+1 .

Thus the result is true for p = k + 1. Hence it is true for all positive integers p. For p = 0, (AB)0 =
In = A0 B 0 .
10. For p = 0, (cA)0 = In = 1 · In = c0 · A0 . For p = 1, cA = cA. Assume the result is true for p = k:
(cA)k = ck Ak , then for k + 1:
(cA)k+1 = (cA)k (cA) = ck Ak · cA = ck (Ak c)A = ck (cAk )A = (ck c)(Ak A) = ck+1 Ak+1 .
11. True for p = 0: (AT )0 = In = InT = (A0 )T . Assume true for p = n. Then
(AT )n+1 = (AT )n AT = (An )T AT = (AAn )T = (An+1 )T .
12. True for p = 0: (A0 )−1 = In−1 = In . Assume true for p = n. Then
(An+1 )−1 = (An A)−1 = A−1 (An )−1 = A−1 (A−1 )n = (A−1 )n+1 .
13.

1 −1
kA

k = 0.

(kA) =

1
k

· k A−1 A = In and (kA)

1 −1
kA

= k·

1
k

AA−1 = In . Hence, (kA)−1 = k1 A−1 for

14. (a) Let A = kIn . Then AT = (kIn )T = kInT = kIn = A.
(b) If k = 0, then A = kIn = 0In = O, which is singular. If k = 0, then A−1 = (kA)−1 = k1 A−1 , so A
is nonsingular.
(c) No, the entries on the main diagonal do not have to be the same.
16. Possible answers:

a b
. Infinitely many.
0 a

17. The result is false. Let A =

1 2
5 11
10 14
. Then AAT =
and AT A =
.
3 4
11 25
14 20

18. (a) A is symmetric if and only if AT = A, or if and only if aij = aTij = aji .
(b) A is skew symmetric if and only if AT = −A, or if and only if aTij = aji = −aij .

(c) aii = −aii , so aii = 0.

19. Since A is symmetric, AT = A and so (AT )T = AT .
20. The zero matrix.
21. (AAT )T = (AT )T AT = AAT .
22. (a) (A + AT )T = AT + (AT )T = AT + A = A + AT .

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11

Section 1.5
(b) (A − AT )T = AT − (AT )T = AT − A = −(A − AT ).
23. (Ak )T = (AT )k = Ak .
24. (a) (A + B)T = AT + B T = A + B.

(b) If AB is symmetric, then (AB)T = AB, but (AB)T = B T AT = BA, so AB = BA. Conversely, if
AB = BA, then (AB)T = B T AT = BA = AB, so AB is symmetric.
25. (a) Let A = aij be upper triangular, so that aij = 0 for i > j. Since AT = aTij , where aTij = aji ,
we have aTij = 0 for j > i, or aTij = 0 for i < j. Hence AT is lower triangular.
(b) Proof is similar to that for (a).
26. Skew symmetric. To show this, let A be a skew symmetric matrix. Then AT = −A. Therefore
(AT )T = A = −AT . Hence AT is skew symmetric.
27. If A is skew symmetric, AT = −A. Thus aii = −aii , so aii = 0.
k

28. Suppose that A is skew symmetric, so AT = −A. Then (Ak )T = (AT )k = (−A) = −Ak if k is a
positive odd integer, so Ak is skew symmetric.
29. Let S = 12 (A + AT ) and K = 12 (A − AT ). Then S is symmetric and K is skew symmetric, by

Exercise 18. Thus
S + K = 12 (A + AT + A − AT ) = 12 (2A) = A.

Conversely, suppose A = S + K is any decomposition of A into the sum of a symmetric and skew
symmetric matrix. Then
AT = (S + K)T = S T + K T = S − K

A + AT = (S + K) + (S − K) = 2S,

A − AT = (S + K) − (S − K) = 2K,


2
1
30. S =
7
2
3

K=

w x
1 0
=
. Since the linear systems
y z
0 1
2w + 3y = 1
4w + 6y = 0

and

2x + 3z = 0
4x + 6z = 1

have no solutions, we conclude that the given matrix is singular.


1
0
0

4


32. D−1 =  0 − 12
0 .


1
0
0
3


34. A = 

36. (a)

− 12

1
2

2 −1

1 2
1 3

1
2
1
2




3
0 −1 −7
1
3  and K =  1
0
1 .
2
6
7 −1
0

7
12

3

2 3
4 6

31. Form

S=



.

4
16
=
.
6
22

(b)

38
.
53

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(A + AT ),
(A − AT )

12

Chapter 1

38.

−9
.
−6

40.

8
.
9

42. Possible answer:

1 0
0 0
1 0
+
=
.
0 0
0 1
0 1

43. Possible answer:

1 2
−1 −2
0 0
+
=
.
3 4
3
4
6 8

44. The conclusion of the corollary is true for r = 2, by Theorem 1.6. Suppose r ≥ 3 and that the
conclusion is true for a sequence of r − 1 matrices. Then
−1
−1 −1
−1
(A1 A2 · · · Ar )−1 = [(A1 A2 · · · Ar−1 )Ar ]−1 = A−1
= A−1
r (A1 A2 · · · Ar−1 )
r Ar−1 · · · A2 A1 .

45. We have A−1 A = In = AA−1 and since inverses are unique, we conclude that (A−1 )−1 = A.
46. Assume that A is nonsingular, so that there exists an n × n matrix B such that AB = In . Exercise 28
in Section 1.3 implies that AB has a row consisting entirely of zeros. Hence, we cannot have AB = In .
47. Let



a11
 0

A=

0

0
a22
0

where aii = 0 for i = 1, 2, . . . , n. Then

A−1



1
a11



 0
=



0



0 ···
0
0 ···
0 

,
..

.
· · · · · · ann

0

0

1
a22

0
..
.

0

···
···

··· ···

as can be verified by computing AA−1 = A−1 A = In .




16 0
0
48. A4 =  0 81
0 .
0 0 625


ap11
 0

49. Ap = 

0

0
ap22
0


0 ···
0
0 ···
0 

.
..


.
p
· · · · · · ann

50. Multiply both sides of the equation by A−1 .
51. Multiply both sides by A−1 .

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0





0 





1
ann


13

Section 1.5
52. Form

a b
c d

w x
1 0
=
. This leads to the linear systems
y z
0 1
aw + by = 1
cw + dy = 0

and

ax + bz = 0
cx + dz = 1.

A solution to these systems exists only if ad − bc = 0. Conversely, if ad − bc = 0 then a solution to
these linear systems exists and we find A−1 .
53. Ax = 0 implies that A−1 (Ax) = A0 = 0, so x = 0.
54. We must show that (A−1 )T = A−1 . First, AA−1 = In implies that (AA−1 )T = InT = In . Now
(AA−1 )T = (A−1 )T AT = (A−1 )T A, which means that (A−1 )T = A−1 .


4
5
0
1  is one possible answer.
4
55. A + B =  0

6 −2
6




2×2 2×2 2×1
2×2 2×3
56. A =  2 × 2 2 × 2 2 × 1  and B =  2 × 2 2 × 3 .
2×2 2×2 2×1
1×2 1×3
3×3 3×2
3×3 3×2

21 48 41
 18 26 34


 24 26 42
AB = 
 28 38 54

 33 33 56
34 37 58
A=

and B =
48
33
47

70
74
79


40
5


16 
.
35 

42 
54

3×3 3×2
.
2×3 2×2

57. A symmetric matrix. To show this, let A1 , . . . , An be symmetric matrices and let x1 , . . . , xn be scalars.
Then AT1 = A1 , . . . , ATn = An . Therefore
(x1 A1 + · · · + xn An )T = (x1 A1 )T + · · · + (xn An )T
= x1 AT1 + · · · + xn ATn

= x1 A1 + · · · + xn An .
Hence the linear combination x1 A1 + · · · + xn An is symmetric.
58. A scalar matrix. To show this, let A1 , . . . , An be scalar matrices and let x1 , . . . , xn be scalars. Then
Ai = ci In for scalars c1 , . . . , cn . Therefore
x1 A1 + · · · + xn An = x1 (c1 I1 ) + · · · + xn (cn In ) = (x1 c1 + · · · + xn cn )In

which is the scalar matrix whose diagonal entries are all equal to x1 c1 + · · · + xn cn .
59. (a) w1 =

5
19
65
214
, w2 =
, w3 =
, w4 =
; u2 = 5, u3 = 19, u4 = 65, u5 = 214.
1
5
19
65

(b) wn−1 = An−1 w0 .
60. (a) w1 =

4
8
16
, w2 =
, w3 =
.
2
4
8

(b) wn−1 = An−1 w0 .

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14

Chapter 1

63. (b) In Matlab the following message is displayed.
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate.
RCOND = 2.937385e-018
Then a computed inverse is shown which is useless. (RCOND above is an estimate of the condition
number of the matrix.)
(c) In Matlab a message similar to that in (b) is displayed.
64. (c) In Matlab, AB − BA is not O. It is a matrix each of whose entries has absolute value less than
1 × 10−14 .
65. (b) Let x be the solution from the linear system solver in Matlab and y = A−1 B. A crude measure
of difference in the two approaches is to look at max{|xi − yi | i = 1, . . . , 10}. This value is
approximately 6 × 10−5 . Hence, computationally the methods are not identical.
66. The student should observe that the “diagonal” of ones marches toward the upper right corner and
eventually “exits” the matrix leaving all of the entries zero.
67. (a) As k → ∞, the entries in Ak → 0, so Ak →

0 0
.
0 0

(b) As k → ∞, some of the entries in Ak do not approach 0, so Ak does not approach any matrix.

Section 1.6, p. 62
2.

y

3
1
−3

f(u) = (3, 0)

−1 O

4.

x

1
3
u = (1, −2)

y

3
1
O

−2 −1

x

1

2

f(u) = (6.19, −0.23)

u = ( − 2, −3)

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15

Section 1.6
6.

y

( − 6, 6)

f(u) = − 2 u

6
4

u = ( − 3, 3)

−6

−4

2

O

−2

8.

x

1

z

u = (0, −2, 4)
f(u) = (4, −2, 4)

1
O

1

y

1

x

10. No.
12. Yes.
14. No.
16. (a) Reflection about the line y = x.
(b) Reflection about the line y = −x.


  
2
0



18. (a) Possible answers: −1 , 0 .
0
1
   
0
1
(b) Possible answers:  4 ,  2 .
4
0

20. (a) f (u + v) = A(u + v) = Au + Av = f (u) + f (v).
(b) f (cu) = A(cu) = c(Au) = cf (u).
(c) f (cu + dv) = A(cu + dv) = A(cu) + A(cv) = c(Au) + d(Av) = cf (u) + df (v).
21. For any real numbers c and d, we have
f (cu + dv) = A(cu + dv) = A(cu) + A(dv) = c(Au) + d(Av) = cf (u) + df (v) = c0 + d0 = 0 + 0 = 0.

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16

Chapter 1


   
0 ··· 0
u1
0





.
.
.
..
22. (a) O(u) = 
  ..  =  ..  = 0.
0 ··· 0
0
un



(b) I(u) = 

1

0

0

0

   
··· 0
u1
u1
  ..   .. 
..
  .  =  .  = u.
.
··· 1
un
un

Section 1.7, p. 70
2.

y

4
2

O

x
2

4. (a)

4

6

8

10 12 14 16

y
(12, 16)

(4, 16)

16

12

8

4

(12, 4)

(4, 4)

3
1
O

x

1

3

4

8

12

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17

Section 1.7
(b)

y

2
1
1
4

O

6.

1
4

3
4

x

1

2

y

1
1
2

O

x

1

8. (1, −2), (−3, 6), (11, −10).

10. We find that
(f1 ◦ f2 )(e1 ) = e2

(f2 ◦ f1 )(e1 ) = −e2 .

Therefore f1 ◦ f2 = f2 ◦ f1 .
12. Here f (u) =

2 0
u. The new vertices are (0, 0), (2, 0), (2, 3), and (0, 3).
0 3
y

(2, 3)

3

O

x

2

14. (a) Possible answer: First perform f1 (45◦ counterclockwise rotation), then f2 .
(b) Possible answer: First perform f3 , then f2 .
cos θ − sin θ
. Then A represents a rotation through the angle θ. Hence A2 represents a
sin θ
cos θ
rotation through the angle 2θ, so

cos 2θ − sin 2θ
A2 =
.
sin 2θ
cos 2θ

16. Let A =

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18

Chapter 1
Since
A2 =

cos θ − sin θ
sin θ
cos θ

cos θ − sin θ
cos2 θ − sin2 θ
=
sin θ
cos θ
2 sin θ cos θ

we conclude that

−2 sin θ cos θ
,
cos2 θ − sin2 θ

cos 2θ = cos2 θ − sin2 θ
sin 2θ = 2 sin θ cos θ.

17. Let
A=

cos θ1 − sin θ1
sin θ1
cos θ1

and B =

cos(−θ2 ) − sin(−θ2 )
cos θ2 sin θ2
=
sin(−θ2 )
cos(−θ2 )
− sin θ2 cos θ2

.

Then A and B represent rotations through the angles θ1 and −θ2 , respectively. Hence BA represents
a rotation through the angle θ1 − θ2 . Then
cos(θ1 − θ2 ) − sin(θ1 − θ2 )
.
sin(θ1 − θ2 )

cos(θ1 − θ2 )

BA =
Since
BA =

cos θ2 sin θ2
− sin θ2 cos θ2

we conclude that

cos θ1 − sin θ1
cos θ1 cos θ2 + sin θ1 sin θ2 cos θ1 sin θ2 − sin θ1 cos θ2
=
,
sin θ1
cos θ1
sin θ1 cos θ2 − cos θ1 sin θ2 cos θ1 cos θ2 + sin θ1 sin θ2
cos(θ1 − θ2 ) = cos θ1 cos θ2 + sin θ1 sin θ2

sin(θ1 − θ2 ) = sin θ1 cos θ2 − cos θ1 sin θ2 .

Section 1.8, p. 79
2. Correlation coefficient = 0.9981. Quite highly correlated.
10
8
6
4
2
0

0

5

10

4. Correlation coefficient = 0.8774. Moderately positively correlated.
100
80
60
40
20
0

0

50

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100


19

Supplementary Exercises

Supplementary Exercises for Chapter 1, p. 80
b1

.
0
b
b
k=2
B = 11 12 .
0
0
b
b
b
k=3
B = 11 12 13 .
0
0
0
b
b
b
b
k=4
B = 11 12 13 14 .
0
0
0
0
(b) The answers are not unique. The only requirement is that row 2 of B have all zero entries.


1 0 0

1 12
4. (a)
.
(b)  0 0 0 .
(c) I4 .
0 1
0 0 0

2. (a) k = 1

(d) Let A =

B=

a b
0 1
a2 + bc ab + bd
. Then A2 =
=
= B implies
c d
0 0
ac + dc bc + d2
b(a + d) = 1
c(a + d) = 0.

It follows that a + d = 0 and c = 0. Thus
A2 =

a2

b

0
b(a + d)
=
2
0
d

1
.
0

Hence, a = d = 0, which is a contradiction; thus, B has no square root.
5. (a) (AT A)ii = (rowi AT ) × (coli A) = (coli A)T × (coli A)
(b) From part (a)
(AT A)ii = a1i a2i · · · ani


a1i
 a2i 


× . =
 .. 
ani


n
j=1

a2ji ≥ 0.

(c) AT A = On if and only if (AT A)ii = 0 for i = 1, . . . , n. But this is possible if and only if aij = 0
for i = 1, . . . , n and j = 1, . . . , n
6. (Ak )T = (A · A · · · A)T = AT AT · · · AT = (AT )k .
k times

k times

7. Let A be a symmetric upper (lower) triangular matrix. Then aij = aji and aij = 0 for j > i (j < i).
Thus, aij = 0 whenever i = j, so A is diagonal.
8. If A is skew symmetric then AT = −A. Note that xT Ax is a scalar, thus (xT Ax)T = xT Ax. That is,
xT Ax = (xT Ax)T = xT AT x = −(xT Ax).

The only scalar equal to its negative is zero. Hence xT Ax = 0 for all x.

9. We are asked to prove an “if and only if” statement. Hence two things must be proved.
(a) If A is nonsingular, then aii = 0 for i = 1, . . . , n.
Proof: If A is nonsingular then A is row equivalent to In . Since A is upper triangular, this can
occur only if we can multiply row i by 1/aii for each i. Hence aii = 0 for i = 1, . . . , n. (Other
row operations will then be needed to get In .)

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