Instructor’s Solutions Manual for Elementary Linear Algebra: A Matrix Approach (Classic Version) by Lawrence E. Spence, Arnold J. Insel, and Stephen H. Friedberg, 2nd Edition. All chapters are covered.

I n s t r u c t o r ’s Manual



Elementary Linear Algebra
A Matrix Approach
Second Edition
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Spence Insel Friedberg
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Upper Saddle River, NJ 07458
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,Chapter 1

Matrices, Vectors, and
Systems of Linear
Equations

2 3 2 3
1.1 MATRICES AND VECTORS 12 4 −7 −1
6−4 207 6 3 07
22. 6
4 8 −245
7 23. 6
4−3
7
»
8 −4 20
–
−2
»
1 −5
– 35
1. 2. 16 −8 −4 4
12 16 4 −3 −4 −1
2 3
»
6 −4 24
– »
8 −3 11
– −7 −1
3. 4.
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6 3 07
8 10 −4 13 18 11 24. 6 7 25. −2 26. 0
2 3 2 3 4−3 35
2 4 4 7 −4 4
5. 4 0 65 6. 4−1 105 2 3 2 3
−4 8 1 9 3 −2 » –
405 2
27. 28. 41.65 29.
2
4 7
3 2e
»
3 −1 3
– 2π 5
A

7. 8. 4−1 105 » –
5 7 5 0.4
1 9 30. 31. [2 − 3 0.4] 32. [2e 12 0]
2 3 0
2 3 » –
−1
2 3
1 7 150
9. 4−1 45 10. √
1 1 −3 33. 4150 35 mph
5 1
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2 3 2 3 10
−1 −2 −1 −2 »√ –
−35 −35 2
11. 4 0 12. 4 0 34. (a) The swimmer’s velocity is u = √ mph.
2 −4 2 −4 2
2 3
−12 0
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» –
−3 1 −2 −4 6 6 157
13. 14. 6 7
−1 −5 6 2 4 −3 −95 North
0 6 y
» –
−6 2 −4 −8
15. 
−2 −10 12 4 swimmer in
still water
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» –
−8 4 −2 −0
16. 17. not possible
0 10 −6 4 .... ◦
...45
2
7 1
3 ...
.
x
» – East
7 −3 3 4 6−3 07
18. 19. 4
6 7
1 0 −3 −4 3 −35 Figure for Exercise 34(a)
4 −4
» –
1 1 4 12
20. 21. not possible
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3 25 −24 −2


1
a

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,2 Chapter 1 Matrices, Vectors, and Systems of Linear Equations

» –
0 A+B, which is aij +bij . By definition, the
(b) The water’s velocity is v = mph. So
1 ith components of aj and bj are aij and
the new velocity
» √ of – the swimmer is bij , respectively. So the ith component of
2 aj + bj is also aij + bij . Thus the jth
u+v = √ mph. The correspond-
2+1 columns of A + B and aj + bj are equal.
p √
ing speed is 5 + 2 2 ≈ 2.798 mph. (b) The proof is similar to the proof of (a).
58. Since A is an m × n matrix, 0A is also an m × n
North
y matrix. Because the (i, j)-entry of 0A is 0aij =
combined 7 6 0, we see that 0A equals the m × n zero matrix.
velocity 
59. Since A is an m × n matrix, 1A is also an m × n
 water current
 matrix. Because the (i, j)-entry of 1A is 1aij =
 aij , we see that 1A equals A.

 60. Because both A and B are m × n matrices, both
 A + B and B + A are m × n matrices. The (i, j)-
 entry of A + B is aij + bij , and the (i, j)-entry
.
.... ◦ of B + A is bij + aij . Since aij + bij = bij + aij
 ..45
 .... x by the commutative property of addition of real
East
numbers, the (i, j)-entries of A+B and B+A are
equal for all i and j. Thus, since the matrices
Figure for Exercise 34(b)
A + B and B + A have the same size and all
» √ – pairs of corresponding entries are equal, A+B =
150 2√+ 50
35. (a) mph B + A.
150 2
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p √ 61. If O is the m × n zero matrix, then both A
(b) 50 37 + 6 2 ≈ 337.21 mph and A + O are m × n matrices; so we need
36. The three components of the vector represent, only show they have equal corresponding en-
respectively, the average blood pressure, average tries. The (i, j)-entry of A + O is aij + 0 = aij ,
pulse rate, and the average cholesterol reading which is the (i, j)-entry of A.
of the 20 people. 62. The proof is similar to the proof of Exercise 61.
A

37. True 38. True 39. True 63. The matrices (st)A, tA, and s(tA) are all m × n
matrices; so we need only show that the corre-
40. False, a scalar multiple of the zero matrix is the
sponding entries of (st)A and s(tA) are equal.
zero matrix.
The (i, j)-entry of s(tA) is s times the (i, j)-
41. False, the transpose of an m × n matrix is an entry of tA, and so it equals s(taij ) = st(aij ),
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n × m matrix. which is the (i, j)-entry of (st)A. Therefore
42. True (st)A = s(tA).
43. False, the rows of B are 1 × 4 vectors. 64. The matrices (s+t)A, sA, and tA are m×n ma-
trices. Hence the matrices (s + t)A and sA + tA
44. False, the (3, 4)-entry of a matrix lies in row 3 are m × n matrices; so we need only show they
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and column 4. have equal corresponding entries. The (i, j)-
45. True entry of sA + tA is the sum of the (i, j)-entries
of sA and tA, that is, saij + taij . And the (i, j)-
46. False, an m × n matrix has mn entries.
entry of (s + t)A is (s + t)aij = saij + taij .
47. True 48. True 49. True
65. The matrices (sA)T and sAT are n × m matri-
50. False, matrices must have the same size to be ces; so we need only show they have equal corre-
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equal. sponding entries. The (i, j)-entry of (sA)T is the
51. True 52. True 53. True (j, i)-entry of sA, which is saji . The (i, j)-entry
of sAT is the product of s and the (i, j)-entry of
54. True 55. True 56. True AT , which is also saji .
57. Suppose that A and B are m × n matrices. 66. The matrix AT is an n × m matrix; so the ma-
(a) The jth column of A + B and aj + bj are trix (AT )T is an m × n matrix. Thus we need
m × 1 vectors. The ith component of the only show that (AT )T and A have equal corre-
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jth column of A + B is the (i, j)-entry of sponding entries. The (i, j)-entry of (AT )T is
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, 1.2 Linear Combinations, Matrix-Vector Products, and Special Matrices 3


the (j, i)-entry of AT , which in turn is the (i, j)- 81. Let A1 = 12 (A + AT ) and A2 = 21 (A − AT ). It
entry of A. is easy to show that A = A1 + A2 . By Exercises
67. If i 6= j, then the (i, j)-entry of a square zero 75 and 74, A1 is symmetric. Also, by Theorem
matrix is 0. Because such a matrix is square, it 1.2(b), (a), and (c), we have
is a diagonal matrix. 1 1
68. If B is a diagonal matrix, then B is square. AT2 = (A − AT )T = [AT − (AT )T ]
2 2
Hence cB is square, and the (i, j)-entry of cB 1 T 1
is cbij = c · 0 = 0 if i 6= j. Thus cB is a diagonal = (A − A) = − (A − AT ) = −A2 .
2 2
matrix.
69. If B is a diagonal matrix, then B is square. Since 82. (a) Because the (i, i)-entry of A+B is aii +bii ,
B T is the same size as B in this case, B T is we have
square. If i 6= j, then the (i, j)-entry of B T is trace(A + B)
bji = 0. So B T is a diagonal matrix. = (a11 + b11 ) + · · · + (ann + bnn )
70. Suppose that B and C are n × n diagonal ma- = (a11 + · · · + ann ) + (b11 + · · · + bnn )
trices. Then B + C is also an n × n matrix.
= trace(A) + trace(B).
Moreover, if i 6= j, the (i, j)-entry of B + C is
bij + cij = 0 + 0 = 0. So B + C is a diagonal (b) The proof is similar to the proof of (a).
matrix. 2 3 (c) The proof is similar to the proof of (a).
» – 2 5 6
2 5 83. The ith component of ap + bq is api + bqi , which
71. and 45 7 85
5 8 is nonnegative. Also, the sum of the components
6 8 4
of ap + bq is
72. Let A be a symmetric matrix. Then A = AT .
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So the (i, j)-entry of A equals the (i, j)-entry of (ap1 + bq1 ) + · · · + (apn + bqn )
AT , which is the (j, i)-entry of A. = a(p1 + · · · + pn ) + b(q1 + · · · + qn )
73. Let O be a square zero matrix. The (i, j)-entry = a(1) + b(1) = a + b = 1.
of O is zero, whereas the (i, j)-entry of OT is the
(j, i)-entry of O, which is also zero. So O = OT ,
2 3
6.5 −0.5 −1.9 −2.8
and hence O is a symmetric matrix. 6 9.6 −2.9 1.5 −3.07
A

6 7
74. By Theorem 1.2(b), (cB)T = cB T = cB. 84. 6 17.4
(a) 6 0.4 −15.5 5.27
7
75. By Theorem 1.1(a) and Theorem 1.2(a) and (c), 4−1.0 −3.7 −7.3 17.55
we have 5.2 1.4 3.5 16.8
2 3
(B + B T )T = B T + (B T )T = B T + B = B + B T . −1.3 3.4 −4.0 10.4
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6 3.0 4.9 −2.4 6.67
76. By Theorem 1.2(a), (B + C)T = B T + C T =
6 7
6−3.9
(b) 6 −4.1 9.4 −8.67
7
B + C. 4 1.7 −0.1 −14.5 −0.25
2 3
2 5 6 » – −4.7 4.1 −0.7 −1.8
2 6
77. No. Consider 5 7 8 and
4 5 . 2 3
5 8 3.9 7.4 10.3 −0.1 1.9
6 8 4
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6 0.8 −0.3 −1.1 −2.5 2.37
78. Let A be a diagonal matrix. If i 6= j, then aij = (c) 4
6 7
−2.6 0.2 −7.2 −9.7 2.15
0 and aji = 0 by definition. Also, aij = aji if i = 1.6 0.2 0.6 11.6 10.6
j. So every entry of A equals the corresponding
entry of AT . Therefore A = AT .
1.2 LINEAR COMBINATIONS,
79. The (i, i)-entries must all equal zero. By equat-
MATRIX-VECTOR PRODUCTS,
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ing the (i, i)-entries of AT and −A, we obtain
aii = −aii , and so aii = 0. AND SPECIAL MATRICES
» –
0 1 2 3 23
80. Take B = . If C is any 2 × 2 skew- » – −5 9 » –
−1 0 12 22
1. 2. 4 45 3. 4 05 4.
symmetric matrix, then C T = −C. Therefore 14
7 10
32
c12 = −c21 . By Exercise 79, c11 = c22 = 0. So 2 3
» – » – » – »– a
0 −c21 0 1 a 22
C= = −c21 = −c21 B. 5. 6. [18] 7. 8. 4 b 5
c21 0 −1 0 b 5
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