TEST BANK FOR Elementary Linear Algebra with Applications 9th Edition By Kolman, Hill

Instructor’s Solutions Manual
Elementary Linear
Algebra with
Applications
Ninth Edition
Bernard Kolman
Drexel University
David R. Hill
Temple University
Contents
Preface iii
1 Linear Equations and Matrices 1
1.1 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.4 Algebraic Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.5 Special Types of Matrices and Partitioned Matrices . . . . . . . . . . . . . . . . . . . . . . . . 9
1.6 Matrix Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.7 Computer Graphics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.8 Correlation Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2 Solving Linear Systems 27
2.1 Echelon Form of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.2 Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.3 Elementary Matrices; Finding A−1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.4 Equivalent Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.5 LU-Factorization (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3 Determinants 37
3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.2 Properties of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.3 Cofactor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.4 Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.5 Other Applications of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
4 Real Vector Spaces 45
4.1 Vectors in the Plane and in 3-Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.2 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4.3 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.4 Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.5 Span and Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4.6 Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.7 Homogeneous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.8 Coordinates and Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.9 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
ii CONTENTS
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
5 Inner Product Spaces 71
5.1 Standard Inner Product on R2 and R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
5.2 Cross Product in R3 (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5.3 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5.4 Gram-Schmidt Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
5.5 Orthogonal Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
5.6 Least Squares (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
6 Linear Transformations and Matrices 93
6.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
6.2 Kernel and Range of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
6.3 Matrix of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
6.4 Vector Space of Matrices and Vector Space of Linear Transformations (Optional) . . . . . . . 99
6.5 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
6.6 Introduction to Homogeneous Coordinates (Optional) . . . . . . . . . . . . . . . . . . . . . . 103
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
7 Eigenvalues and Eigenvectors 109
7.1 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
7.2 Diagonalization and Similar Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
7.3 Diagonalization of Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
8 Applications of Eigenvalues and Eigenvectors (Optional) 129
8.1 Stable Age Distribution in a Population; Markov Processes . . . . . . . . . . . . . . . . . . . 129
8.2 Spectral Decomposition and Singular Value Decomposition . . . . . . . . . . . . . . . . . . . 130
8.3 Dominant Eigenvalue and Principal Component Analysis . . . . . . . . . . . . . . . . . . . . 130
8.4 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
8.5 Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
8.6 Real Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
8.7 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
8.8 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
10 MATLAB Exercises 137
Appendix B Complex Numbers 163
B.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
B.2 Complex Numbers in Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
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 = t2
.
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,
ap1s1 + · · · + apnsn = bp
aq1s1 + · · · + aqnsn = 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.
2 Chapter 1
26. (a) A unique point.
(b) There are infinitely many points.
(c) No points simultaneously lie in all three planes.
28. No points of intersection: C1 C2
C2
C1
One point of intersection: C1 C2
Two points of intersection:
C1 C2
Infinitely many points of intersection: C1= C2
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
2. (a) A =
!
"""""#
0 1 0 0 1
1 0 1 1 1
0 1 0 0 0
0 1 0 0 0
1 1 0 0 0
$
%%%%%&
(b) A =
!
"""""#
0 1 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.
6. (a) C + E = E + C =
!
#
5 −5 8
4 2 9
5 3 4
$
&. (b) Impossible. (c)
'
7 −7
0 1
(
.
(d)
!
# −9 3 −9
−12 −3 −15
−6 −3 −9
$
&. (e)
!
#
0 10 −9
8 −1 −2
−5 −4 3
$
&. (f) Impossible.
8. (a) AT =
!
#
1 2
2 1
3 4
$
&, (AT )T =
'
1 2 3
2 1 4
(
. (b)
!
#
5 4 5
−5 2 3
8 9 4
$
&. (c)
'
−6 10
11 17
(
.
Section 1.3 3
(d)
'
0 −4
4 0
(
. (e)
!
#
3 4
6 3
9 10
$
&. (f)
'
17 2
−16 6
(
.
10. Yes: 2
'
1 0
0 1
(
+ 1
'
1 0
0 0
(
=
'
3 0
0 2
(
.
12.
!
#
! − 1 −2 −3
−6 ! + 2 −3
−5 −2 ! − 4
$
&.
14. Because the edges can be traversed in either direction.
16. Let x =
!
"""#
x1
x2
...
xn
$
%%%&
be an n-vector. Then
x + 0 =
!
"""#
x1
x2
...
xn
$
%%%&
+
!
"""#
0
0
...
0
$
%%%&
=
!
"""#
x1 + 0
x2 + 0
...
xn + 0
$
%%%&
=
!
"""#
x1
x2
...
xn
$
%%%&
= x.
18.
)n
i=1
)m
j=1
aij = (a11 + a12 + · · · + a1m) + (a21 + a22 + · · · + a2m) + · · · + (an1 + an2 + · · · + anm)
= (a11 + a21 + · · · + an1) + (a12 + a22 + · · · + an2) + · · · + (a1m + a2m + · · · + anm)
=
)m
j=1
)n
i=1
aij .
19. (a) True.
)n
i=1
(ai + 1) =
)n
i=1
ai +
)n
i=1
1 =
)n
i=1
ai + n.
(b) True.
)n
i=1
*
+
)m
j=1
1
,
- =
)n
i=1
m = mn.
(c) True.
!
#
)n
i=1
ai
$
&
!
#
)m
j=1
bj
$
& = a1
)m
j=1
bj + a2
)m
j=1
bj + · · · + an
)m
j=1
bj
= (a1 + a2 + · · · + an)
)m
j=1
bj
=
)n
i=1
ai
)m
j=1
bj =
)m
j=1
.
)n
i=1
aibj
/
20. “new salaries” = u + .08u = 1.08u.
Section 1.3, p. 30
2. (a) 4. (b) 0. (c) 1. (d) 1.
4. x = 5.
4 Chapter 1
6. x = ±"2, y = ±3.
8. x = ±5.
10. x = 65
, y = 12
5 .
12. (a) Impossible. (b)
!
#
0 −1 1
12 5 17
19 0 22
$
&. (c)
!
#
15 −7 14
23 −5 29
13 −1 17
$
&. (d)
!
#
8 8
14 13
13 9
$
&. (e) Impossible.
14. (a)
'
58 12
66 13
(
. (b) Same as (a). (c)
'
28 8 38
34 4 41
(
.
(d) Same as (c). (e)
'
28 32
16 18
(
; same. (f)
'
−16 −8 −26
−30 0 −31
(
.
16. (a) 1. (b) −6. (c)
0
−3 0 1
1
. (d)
!
#−1 4 2
−2 8 4
3 −12 −6
$
&. (e) 10.
(f)
!
#
9 0 −3
0 0 0
−3 0 1
$
&. (g) Impossible.
18. DI2 = I2D = D.
20.
'
0 0
0 0
(
.
22. (a)
!
""#
1
14
0
13
$
%%&
. (b)
!
""#
0
18
3
13
$
%%&
.
24. col1(AB) = 1
!
#
1
2
3
$
& + 3
!
#−2
4
0
$
& + 2
!
#−1
3
−2
$
&; col2(AB) = −1
!
#
1
2
3
$
& + 2
!
#−2
4
0
$
& + 4
!
#−1
3
−2
$
&.
26. (a) −5. (b) BAT
28. Let A =
0
aij
1
be m × p and B =
0
bij
1
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
k=1
aikbkj = 0 for j = 1, 2, . . . , n.
(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
k=1
bikakj = 0 for i = 1, 2, . . . , m.
30. (a)
!
""#
2 3 −3 1 1
3 0 2 0 3
2 3 0 −4 0
0 0 1 1 1
$
%%&
. (b)
!
""#
2 3 −3 1 1
3 0 2 0 3
2 3 0 −4 0
0 0 1 1 1
$
%%&
!
"""""#
x1
x2
x3
x4
x5
$
%%%%%&
=
!
""#
7
−2
3
5
$
%%&
.
Section 1.3 5
(c)
!
""#
2 3 −3 1 1 7
3 0 2 0 3 −2
2 3 0 −4 0 3
0 0 1 1 1 5
$
%%&
32.
'
−2 3
1 −5
('
x1
x2
(
=
'
5
4
(
.
34. (a)
2x1 + x2 + 3x3 + 4x4 = 0
3x1 − x2 + 2x3 = 3
−2x1 + x2 − 4x3 + 3x4 = 2
(b) same as (a).
36. (a) x1
'
3
1
(
+ x2
'
2
−1
(
+ x3
'
1
4
(
=
'
4
−2
(
. (b) x1
!
#−1
2
3
$
& + x2
!
#
1
−1
1
$
& =
!
#
3
−2
1
$
&.
38. (a)
'
1 2 0
2 5 3
(!
#
x1
x2
x3
$
& =
'
1
1
(
. (b)
!
#
1 2 1
1 1 2
2 0 2
$
&
!
#
x1
x2
x3
$
& =
!
#
0
0
0
$
&.
39. We have
u · v =
)n
i=1
uivi =
0
u1 u2 · · · un
1
!
"""#
v1
v2
...
vn
$
%%%&
= uT v.
40. Possible answer:
!
#
1 0 0
2 0 0
3 0 0
$
&.
42. (a) Can say nothing. (b) Can say nothing.
43. (a) Tr(cA) =
)n
i=1
caii = c
)n
i=1
aii = cTr(A).
(b) Tr(A + B) =
)n
i=1
(aii + bii) =
)n
i=1
aii +
)n
i=1
bii = Tr(A) + Tr(B).
(c) Let AB = C =
0
cij
1
. Then
Tr(AB) = Tr(C) =
)n
i=1
cii =
)n
i=1
)n
k=1
aikbki =
)n
k=1
)n
i=1
bkiaik = Tr(BA).
(d) Since aT
ii = aii, Tr(AT) =
)n
i=1
aT
ii =
)n
i=1
aii = Tr(A).
(e) Let ATA = B =
0
bij
1
. Then
bii =
)n
j=1
aT
ijaji =
)n
j=1
a2
ji =$ Tr(B) = Tr(ATA) =
)n
i=1
bii =
)n
i=1
)n
j=1
a2
ij % 0.
Hence, Tr(ATA) % 0.
6 Chapter 1
44. (a) 4. (b) 1. (c) 3.
45. We have Tr(AB − BA) = Tr(AB) − Tr(BA) = 0, while Tr
2'
1 0
0 1
(3
= 2.
46. (a) Let A =
0
aij
1
and B =
0
bij
1
be m × n and n × p, respectively. Then bj =
!
"""#
b1j
b2j
...
bnj
$
%%%&
and the ith
entry of Abj is
)n
k=1
aikbkj, which is exactly the (i, j) entry of AB.
(b) The ith row of AB is
04
k aikbk1
4
k aikbk2 · · ·
4
k aikbkn
1
. Since ai =
0
ai1 ai2 · · · ain
1
,
we have
aib =
04
k aikbk1
4
k aikbk2 · · ·
4
k aikbkn
1
.
This is the same as the ith row of Ab.
47. Let A =
0
aij
1
and B =
0
bij
1
be m × n and n × p, respectively. Then the jth column of AB is
(AB)j =
!
"#
a11b1j + · · · + a1nbnj
...
am1b1j + · · · + amnbnj
$
%&
= b1j
!
"#
a11
...
am1
$
%&
+ · · · + bnj
!
"#
a1n
... amn
$
%&
= b1jCol1(A) + · · · + bnjColn(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
+ · · · + x2
n % 0.
(b) x = 0.
52. Let a = (a1, a2, . . . , an), b = (b1, b2, . . . , bn), and c = (c1, c2, . . . , cn). Then
(a) a · b =
)n
i=1
aibi and b · a =
)n
i=1
biai, so a · b = b · a.
(b) (a + b) · c =
)n
i=1
(ai + bi)ci =
)n
i=1
aici +
)n
i=1
bici = a · c + b · c.
(c) (ka) · b =
)n
i=1
(kai)bi = k
)n
i=1
aibi = k(a · b).
Section 1.4 7
53. The i, ith element of the matrix AAT is
)n
k=1
aikaT
ki =
)n
k=1
aikaik =
)n
k=1
(aik)2.
Thus if AAT = O, then each sum of squares
)n
k=1
(aik)2 equals zero, which implies aik = 0 for each i
and k. Thus A = O.
54. AC =
'
17 2 22
18 3 23
(
. CA cannot be computed.
55. BTB will be 6 × 6 while BBT is 1 × 1.
Section 1.4, p. 40
1. Let A =
0
aij
1
, B =
0
bij
1
, C =
0
cij
1
. 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 =
0
aij
1
, let B =
0
−aij
1
.
4. Let A =
0
aij
1
, B =
0
bij
1
, C =
0
cij
1
. Then the (i, j) entry of (A+B)C is
)n
k=1
(aik + bik)ckj and that of
AC +BC is
)n
k=1
aikckj +
)n
k=1
bikckj. By the distributive and additive associative laws for real numbers,
these two expressions for the (i, j) entry are equal.
6. Let A =
0
aij
1
, where aii = k and aij = 0 if i &= j, and let B =
0
bij
1
. Then, if i &= j, the (i, j) entry of
AB is
)n
s=1
aisbsj = kbij , while if i = j, the (i, i) entry of AB is
)n
s=1
aisbsi = kbii. Therefore AB = kB.
7. Let A =
0
aij
1
and C =
0
c1 c2 · · · cm
1
. Then CA is a 1 × n matrix whose ith entry is
)n
j=1
cjaij .
Since Aj =
!
"""#
a1j
a2j
...
amj
$
%%%&
, the ith entry of
)n
j=1
cjAj is
)m
j=1
cjaij .
8. (a)
'
cos 2" sin 2"
−sin 2" cos 2"
(
. (b)
'
cos 3" sin 3"
−sin 3" cos 3"
(
. (c)
'
cos k" sin k"
−sin k" cos k"
(
.
(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
Ak+1 = AkA =
'
cos k" sin k"
−sin k" cos k"
('
cos " sin "
−sin " cos "
(
=
'
cos k" cos " − sin k" sin " cos k" sin " + sin k" cos "
−sin k" cos " − cos k" sin " cos k" cos " − sin k" sin "
(
=
'
cos(k + 1)" sin(k + 1)"
−sin(k + 1)" cos(k + 1)"
(
.
Hence, it is true for all positive integers k.
8 Chapter 1
10. Possible answers: A =
'
1 0
0 1
(
; A =
'
0 1
1 0
(
; A =
!
#
1 "2
1 "2
1 "2 − 1 "2
$
&.
12. Possible answers: A =
'
1 1
−1 −1
(
; A =
'
0 0
0 0
(
; A =
'
0 1
0 0
(
.
13. Let A =
0
aij
1
. The (i, j) entry of r(sA) is r(saij ), which equals (rs)aij and s(raij ).
14. Let A =
0
aij
1
. 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 =
0
aij
1
, and B =
0
bij
1
. Then r(aij + bij) = raij + rbij .
18. Let A =
0
aij
1
and B =
0
bij
1
. The (i, j) entry of A(rB) is
)n
k=1
aik(rbkj), which equals r
)n
k=1
aikbkj, the
(i, j) entry of r(AB).
20. 16
A, 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
0
aij + bij
1
, which is to say, aji + bji.
(d) Let A =
0
aij
1
and let bij = aji. Then the (i, j) entry of (cA)T is the (j, i) entry of
0
caij
1
, which
is to say, cbij .
28. (A + B)T =
!
#
5 0
5 2
1 2
$
&, (rA)T =
!
# −4 −8
−12 −4
−8 12
$
&.
30. (a)
!
#−34
17
−51
$
&. (b)
!
#−34
17
−51
$
&. (c) BTC is a real number (a 1 × 1 matrix).
32. Possible answers: A =
'
1 −3
0 0
(
; B =
'
1 2
23
1
(
; C =
'
−1 2
0 1
(
.
A =
'
2 0
3 0
(
; B =
'
0 0
1 0
(
; C =
'
0 0
0 1
(
.
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
c d
(
be such that AB = BA for any 2 × 2 matrix B. Then in particular,
'
a b
c d
('
1 0
0 0
(
=
'
1 0
0 0
('
a b
c d
(
'
a 0
c 0
(
=
'
a b
0 0
(
so b = c = 0, A =
'
a 0
0 d
(
.
Section 1.5 9
Also
'
a 0
0 d
('
1 1
0 0
(
=
'
1 1
0 0
('
a 0
0 d
(
'
a a
0 0
(
=
'
a d
0 0
(
,
which implies that a = d. Thus A =
'
a 0
0 a
(
for some number a.
35. We have
(A − B)T = (A + (−1)B)T
= AT + ((−1)B)T
= AT + (−1)BT = AT − BT 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 =
0
dij
1
so dij = 1 if i = j and 0 otherwise. Then the (i, j) entry of ImA is
)m
k=1
dikakj = diiaij (since all other d’s = 0)
= aij (since dii = 1).
2. We prove that the product of two upper triangular matrices is upper triangular: Let A =
0
aij
1
with
aij = 0 for i > j; let B =
0
bij
1
with bij = 0 for i > j. Then AB =
0
cij
1
where cij =
)n
k=1
aikbkj. For
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
0
cij
1
is upper triangular.
3. Let A =
0
aij
1
and B =
0
bij
1
, where both aij = 0 and bij = 0 if i &= j. Then if AB = C =
0
cij
1
, we
have cij =
)n
k=1
aikbkj = 0 if i &= j.
4. A + B =
!
#
9 −1 1
0 −2 7
0 0 3
$
& and AB =
!
#
18 −5 11
0 −8 −7
0 0 0
$
&.
5. All diagonal matrices.
10 Chapter 1
6. (a)
'
7 −2
−3 10
(
(b)
'
−9 −11
22 13
(
(c)
'
20 −20
4 76
(
8. ApAq = (A · A· · ·A)
5 67 8
p factors
(A · A· · ·A)
5 67 8
q factors 5 67 8
p + q factors
= Ap+q; (Ap)q = ApApAp · · ·Ap
5 67 8
q factors
= A
q summands 7 85 6
p + p + · · · + p = Apq.
9. We are given that AB = BA. For p = 2, (AB)2 = (AB)(AB) = A(BA)B = A(AB)B = A2B2.
Assume that for p = k, (AB)k = AkBk. Then
(AB)k+1 = (AB)k(AB) = AkBk · A · B = Ak(Bk−1AB)B
= Ak(Bk−2AB2)B = · · · = Ak+1Bk+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 = A0B0.
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 = ckAk, then for k + 1:
(cA)k+1 = (cA)k(cA) = ckAk · cA = ck(Akc)A = ck(cAk)A = (ckc)(AkA) = ck+1Ak+1.
11. True for p = 0: (AT )0 = In = IT
n = (A0)T . Assume true for p = n. Then
(AT )n+1 = (AT )nAT = (An)TAT = (AAn)T = (An+1)T .
12. True for p = 0: (A0)−1 = I−1
n = In. Assume true for p = n. Then
(An+1)−1 = (AnA)−1 = A−1(An)−1 = A−1(A−1)n = (A−1)n+1.
13.
91k
A−1
:
(kA) =
91k
· k
:
A−1A = In and (kA)
91k
A−1
:
=
9
k · 1k
:
AA−1 = In. Hence, (kA)−1 = 1k
A−1 for
k &= 0.
14. (a) Let A = kIn. Then AT = (kIn)T = kIT
n = kIn = A.
(b) If k = 0, then A = kIn = 0In = O, which is singular. If k &= 0, then A−1 = (kA)−1 = 1k
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
0 a
(
. Infinitely many.
17. The result is false. Let A =
'
1 2
3 4
(
. Then AAT =
'
5 11
11 25
(
and ATA =
'
10 14
14 20
(
.
18. (a) A is symmetric if and only if AT = A, or if and only if aij = aT
ij = aji.
(b) A is skew symmetric if and only if AT = −A, or if and only if aT
ij = 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 )TAT = AAT .
22. (a) (A + AT )T = AT + (AT )T = AT + A = A + AT .
Section 1.5 11
(b) (A − AT )T = AT − (AT )T = AT − A = −(A − AT ).
23. (Ak)T = (AT )k = Ak.
24. (a) (A + B)T = AT + BT = A + B.
(b) If AB is symmetric, then (AB)T = AB, but (AB)T = BTAT = BA, so AB = BA. Conversely, if
AB = BA, then (AB)T = BTAT = BA = AB, so AB is symmetric.
25. (a) Let A =
0
aij
1
be upper triangular, so that aij = 0 for i > j. Since AT =
0
aT
ij
1
, where aT
ij = aji,
we have aT
ij = 0 for j > i, or aT
ij = 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.
28. Suppose that A is skew symmetric, so AT = −A. Then (Ak)T = (AT )k = (−A)k = −Ak if k is a
positive odd integer, so Ak is skew symmetric.
29. Let S =
912
:
(A + AT ) and K =
912
:
(A − AT ). Then S is symmetric and K is skew symmetric, by
Exercise 18. Thus
S + K =
912
:
(A + AT + A − AT) =
912
:
(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 = ST + KT = S − K
A + AT = (S + K) + (S − K) = 2S, S =
912
:
(A + AT ),
A − AT = (S + K) − (S − K) = 2K, K =
912 :
(A − AT )
30. S =
1
2
!
#
2 7 3
7 12 3
3 3 6
$
& and K =
1
2
!
#
0 −1 −7
1 0 1
7 −1 0
$
&.
31. Form
'
2 3
4 6
('
w x
y z
(
=
'
1 0
0 1
(
. Since the linear systems
2w + 3y = 1
4w + 6y = 0
and
2x + 3z = 0
4x + 6z = 1
have no solutions, we conclude that the given matrix is singular.
32. D−1 =
!
"""#
14
0 0
0 −12
0
0 0 13
$
%%%&.
34. A =
!
#−12
1
2
2 −1
$
&.
36. (a)
'
1 2
1 3
('
4
6
(
=
'
16
22
(
. (b)
'
38
53
(
.
12 Chapter 1
38.
'
−9
−6
(
.
40.
'
8
9
(
.
42. Possible answer:
'
1 0
0 0
(
+
'
0 0
0 1
(
=
'
1 0
0 1
(
.
43. Possible answer:
'
1 2
3 4
(
+
'
−1 −2
3 4
(
=
'
0 0
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
(A1A2 · · ·Ar)−1 = [(A1A2 · · ·Ar−1)Ar]−1 = A−1
r (A1A2 · · ·Ar−1)−1 = A−1
r A−1
r−1 · · ·A−1
2 A−1
1 .
45. We have A−1A = 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
A =
!
"""#
a11 0 0 · · · 0
0 a22 0 · · · 0
...
0 0 · · · · · · ann
$
%%%&
,
where aii &= 0 for i = 1, 2, . . . , n. Then
A−1 =
!
""""""#
1
a11 0 0 · · · 0
0 1
a22 0 · · · 0
...
0 0 · · · · · · 1
ann
$
%%%%%%&
as can be verified by computing AA−1 = A−1A = In.
48. A4 =
!
#
16 0 0
0 81 0
0 0 625
$
&.
49. Ap =
!
"""#
ap
11 0 0 · · · 0
0 ap
22 0 · · · 0
...
0 0 · · · · · · ap
nn
$
%%%&
.
50. Multiply both sides of the equation by A−1.
51. Multiply both sides by A−1.
Section 1.5 13
52. Form
'
a b
c d
('
w x
y z
(
=
'
1 0
0 1
(
. This leads to the linear systems
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 = IT
n = In. Now
(AA−1)T = (A−1)TAT = (A−1)TA, which means that (A−1)T = A−1.
55. A + B =
!
#
4 5 0
0 4 1
6 −2 6
$
& is one possible answer.
56. A =
!
#
2 × 2 2 × 2 2 × 1
2 × 2 2 × 2 2 × 1
2 × 2 2 × 2 2 × 1
$
& and B =
!
#
2 × 2 2 × 3
2 × 2 2 × 3
1 × 2 1 × 3
$
&.
A =
'
3 × 3 3 × 2
3 × 3 3 × 2
(
and B =
'
3 × 3 3 × 2
2 × 3 2 × 2
(
.
AB =
!
"""""""#
21 48 41 48 40
18 26 34 33 5
24 26 42 47 16
28 38 54 70 35
33 33 56 74 42
34 37 58 79 54
$
%%%%%%%&
.
57. A symmetric matrix. To show this, let A1, . . . , An be symmetric matrices and let x1, . . . , xn be scalars.
Then AT1
= A1, . . . , AT
n = An. Therefore
(x1A1 + · · · + xnAn)T = (x1A1)T + · · · + (xnAn)T
= x1AT1
+ · · · + xnAT
n
= x1A1 + · · · + xnAn.
Hence the linear combination x1A1 + · · · + xnAn is symmetric.
58. A scalar matrix. To show this, let A1, . . . , An be scalar matrices and let x1, . . . , xn be scalars. Then
Ai = ciIn for scalars c1, . . . , cn. Therefore
x1A1 + · · · + xnAn = x1(c1I1) + · · · + xn(cnIn) = (x1c1 + · · · + xncn)In
which is the scalar matrix whose diagonal entries are all equal to x1c1 + · · · + xncn.
59. (a) w1 =
'
5
1
(
, w2 =
'
19
5
(
, w3 =
'
65
19
(
, w4 =
'
214
65
(
; u2 = 5, u3 = 19, u4 = 65, u5 = 214.
(b) wn−1 = An−1w0.
60. (a) w1 =
'
4
2
(
, w2 =
'
8
4
(
, w3 =
'
16
8
(
.
(b) wn−1 = An−1w0.
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−1B. 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.