3.1 SOLUTIONS
Notes:
Some exercises in this section provide practice in computing determinants, while others allow the
1. Expanding along the first row:
30 4 32 22 23
2 3 2 3 0 4 3( 13) 4(10) 1
=−+=−+=
2. Expanding along the first row:
051 30 40 4 3
4 3 0 0 5 1 5(4) 1(22) 2
41 21 2 4
241
−−
−= − + =−+=
Expanding along the second column:
3. Expanding along the first row:
243 12 32 31
3 1 2 2 ( 4) 3 2( 9) 4( 5) (3)(11) 5
−
=−−+=−+−+=−
4. Expanding along the first row:
135 11 21 21
2 1 1 1 3 5 1( 2) 3(1) 5(5) 20
42 32 34
342
=−+ =−−+=
Expanding along the second column:
5. Expanding along the first row:
23 4 05 45 40
−
6. Expanding along the first row:
524 35 05 03
−−−
−
7. Expanding along the first row:
430 52 62 65
8. Expanding along the first row:
816 03 43 4 0
−
9. First expand along the third row, then expand along the first row of the remaining matrix:
3.1 • Solutions 169
31 13
600 5 00 5
172 5 72
( 1) 2 7 2 5 2 ( 1) 5 10(1) 10
++
−=− ⋅ − = ⋅− ⋅ = =
10. First expand along the second row, then expand along either the third row or the second column of
the remaining matrix.
23
1252 122
0030
(1) 32 6 5
2675 504
5044
+
−−
=− ⋅ −
−−
11. There are many ways to do this determinant efficiently. One strategy is to always expand along the
first column of each matrix:
3584 23 7
−−−
12. There are many ways to do this determinant efficiently. One strategy is to always expand along the
first row of each matrix:
11 11
4000 10 0
7100 30
(1) 4 6 3 0 4(1) (1)
++
−
−=− ⋅ = ⋅− ⋅− −
= 4(–1)( –9) = 36
13. First expand along either the second row or the second column. Using the second row,
40 7 3 5 40 3 5
00 2 0 0 73 4 8
−− −
Now expand along the second column to find:
23 22
40 3 5 435
73 4 8
(1) 2 2(1) 35 2 3
50 2 3 012
++
−⎛⎞−
−⎜⎟
−⋅ =−−⋅ −
⎜⎟
−⎜⎟
14. First expand along either the fourth row or the fifth column. Using the fifth column,
63240 63 24
90410 90 41
−−
Now expand along the third row to find:
35 31
63 24 324
90 41
++
⎛⎞
−⎜⎟
Now expand along either the first column or second row. The first column is used below.
324
⎛⎞
15.
30 4
23 2
05 1
=
−
(3)(3)(–1) + (0)(2)(0) + (4)(2)(5) – (0)(3)(4) – (5)(2)(3) – (–1)(2)(0)
3.1 • Solutions 171
16.
051
430
241
−=
(0)(–3)(1) + (5)(0)(2) + (1)(4)(4) – (2)(–3)(1) – (4)(0)(0) – (1)(4)(5)
17.
243
312
141
−
=
−
(2)(1)(–1) + (–4)(2)(1) + (3)(3)(4) – (1)(1)(3) – (4)(2)(2) – (–1)(3)(–4)
18.
135
211
342
=
(1)(1)(2) + (3)(1)(3) + (5)(2)(4) – (3)(1)(5) – (4)(1)(1) – (2)(2)(3)
19.
,
ab ad bc
cd
=−
()
cd cb da ad bc
ab
=− =− −
20.
,
ab ad bc
cd
=−
()() ( )
abakd kcb kad kbc kad bc
kc kd =−=−=−
21.
3418 20 2,
56
=− =−
34
3(6 4 ) (5 3 )4 2
53 64 kk
kk
=+−+ =−
++
The row operation replaces row 2 with k times row 1 plus row 2, and the determinant is unchanged.
172 CHAPTER 3 • Determinants
25. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
26. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
27. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
28. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
29. A cofactor expansion along row 1 gives
010 10
30. A cofactor expansion along row 1 gives
001 01
31. A 3 × 3 elementary row replacement matrix looks like one of the six matrices
100 100 100 100 10 1 0
kk
⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤
32. A 3 × 3 elementary scaling matrix with k on the diagonal looks like one of the three matrices
00 100 100
k
⎡⎤⎡⎤⎡⎤
33.
01
,
10
E⎡⎤
=⎢⎥
⎣⎦ ,
ab
Acd
⎡⎤
=⎢⎥
⎣⎦
cd
EA ab
⎡⎤
=⎢⎥
⎣⎦
34.
10
,
0
Ek
⎡⎤
=⎢⎥
⎣⎦ ,
ab
Acd
⎡⎤
=⎢⎥
⎣⎦
ab
EA kc kd
⎡⎤
=⎢⎥
⎣⎦
35.
1,
01
k
E⎡⎤
=⎢⎥
⎣⎦ ,
ab
Acd
⎡⎤
=⎢⎥
⎣⎦
akc bkd
EA cd
++
⎡⎤
=⎢⎥
⎣⎦
36.
10
,
1
Ek
⎡⎤
=⎢⎥
⎣⎦ ,
ab
Acd
⎡⎤
=⎢⎥
⎣⎦
ab
EA ka c kb d
⎡⎤
=⎢⎥
++
⎣⎦
37.
31
,
42
A⎡⎤
=⎢⎥
⎣⎦
15 5
5,
20 10
A⎡⎤
=⎢⎥
⎣⎦
det A = 2, det 5A = 50 ≠ 5det A
38.
,
ab
Acd
⎡⎤
=⎢⎥
⎣⎦
,
ka kb
kA kc kd
⎡⎤
=⎢⎥
⎣⎦
det A = ad – bc,
39. a. True. See the paragraph preceding the definition of the determinant.
174 CHAPTER 3 • Determinants
41. The area of the parallelogram determined by
3,
0
⎡
⎤
=
⎢
⎥
⎣
⎦
u
1,
2
⎡
⎤
=
⎢
⎥
⎣
⎦
v
u + v, and 0 is 6, since the base of
the parallelogram has length 3 and the height of the parallelogram is 2. By the same reasoning, the
area of the parallelogram determined by
3,
0
⎡
⎤
=
⎢
⎥
⎣
⎦
u
,
2
x
⎡
⎤
=
⎢
⎥
⎣
⎦
x
u + x, and 0 is also 6.
42. The area of the parallelogram determined by
a
b
⎤
=
⎥
⎦
u
,
0
c
⎡
⎤
=
⎢
⎥
⎣
⎦
v
, u + v, and 0 is cb, since the base of
the parallelogram has length c and the height of the parallelogram is b.
43. [M] Answers will vary. The conclusion should be that det (A + B) ≠ det A + det B.
X
2
a
b
c
X
1
U
V
3.2 • Solutions 175
45. [M] Answers will vary. For 4 × 4 matrices, the conclusions should be that
det det ,
T
AA
=
det(–A) = det A, det(2A) = 16det A, and
4
det (10 ) 10 det
AA
=
. For 5 × 5 matrices, the conclusions
46. [M] Answers will vary. The conclusion should be that
1
det 1/ det .
AA
−
=
3.2 SOLUTIONS
Notes:
This section presents the main properties of the determinant, including the effects of row
operations on the determinant of a matrix. These properties are first studied by examples in Exercises 1–
20. The properties are treated in a more theoretical manner in later exercises. An efficient method for
1. Rows 1 and 2 are interchanged, so the determinant changes sign (Theorem 3b.).
3. The row replacement operation does not change the determinant (Theorem 3a.).
5.
1440120123
−− = −= −=
6.
3 3 3 0 18 12 6 0 3 2 6 0 3 2 (6)( 3) 18
−=− =−=−=−=−
7.
0
3 5 2 1 0 4 2 5 0 0 30 27 0 0 30 27
−−
8.
0
2543012 50000
===
−−−
176 CHAPTER 3 • Determinants
9.
1130 1130 1130 1130
015401540154 0154 (3) 3
1 2 85 0 1 55 0 0 0 1 0 0 3 5
−− −− −− −−
== =− =−−=
−−−
10.
262390003500035
373870208100477
===
−−
−−− − −−
11. First use a row replacement to create zeros in the second column, and then expand down the second
column:
25 3 1 25 3 1 313
−− −− −
12. First use a row replacement to create zeros in the fourth column, and then expand down the fourth
column:
1230 12 30 12 3
−− −
13. First use a row replacement to create zeros in the fourth column, and then expand down the fourth
column:
3.2 • Solutions 177
2541 2541 032
−−
14. First use a row replacement to create zeros in the third column, and then expand down the third
column:
3214 3214 133
−− − −− − −
15.
55(7)35
abc abc
def def
===
16.
333 3 3(7)21
abc abc
def def
===
17.
7
abc abc
gh i de f
=− =−
18.
(7) 7
abc gh i de f
=− =− − =− − =
⎜⎟
20.
7
defdef
==
21. Since
134 10
=− ≠
, the matrix is invertible.
22. Since
1320
−−=
, the matrix is not invertible.
23. Since
0
, the matrix is not invertible.
24. Since
605110
−= ≠
, the columns of the matrix form a linearly independent set.
25. Since
450 10
−=−≠
, the columns of the matrix form a linearly independent set.
26. Since
0
, the columns of the matrix form a linearly dependent set.
27. a. True. See Theorem 3.
28. a. True. See Theorem 3.
b . False. See the paragraphs following Example 2.
3.2 • Solutions 179
30. Suppose the two rows of a square matrix A are equal. By swapping these two rows, the matrix A is
not changed so its determinant should not change. But since swapping rows changes the sign of the
32. By factoring an r out of each of the n rows,
det ( ) det .
n
rA r A
=
34. By Theorem 6 and Exercise 31,
111
det ( ) (det )(det )(det ) (det )(det )(det )
PAP P A P P P A
−−−
==
35. By Theorem 6 and Theorem 5,
2
det (det )(det ) (det ) .
TT
UU U U U
==
Since ,
T
UU I=
det det 1
T
UU I
==
, so
2
(det ) 1.
U
=
Thus det U = ±1.
37. One may compute using Theorem 2 that det A = 3 and det B = 8, while
60
17 4
AB ⎡⎤
=⎢⎥
⎣⎦
. Thus
det AB = 24 = 3 × 8 = (det A)(det B).
39. a. By Theorem 6, det AB = (det A)(det B) = 4 × –3 = –12.
b . By Exercise 32,
3
det 5 5 det 125 4 500
AA
==×=
.
40. a. By Theorem 6, det AB = (det A)(det B) = –1 × 2 = –2.
b . By Theorem 6,
555
det (det ) 2 32
BB
===
.
180 CHAPTER 3 • Determinants
41. det A = (a + e)d – c(b + f) = ad + ed – bc – cf = (ad – bc) + (ed – cf) = det B + det C.
1
ab
+
43. Compute det A by using a cofactor expansion down the third column:
1 1 13 2 2 23 3 3 33
det ( )det ( )det ( )detAuv A uv A uv A=+ −+ ++
44. By Theorem 5,
det det ( ) .
T
AE AE
=
Since
()
TTT
AE E A
=
,
det det( ).
TT
AE E A
=
Now
T
E
is itself
45. [M] Answers will vary, but will show that
det T
AA
always equals 0 while
det T
AA
should seldom
be zero. To see why
T
AA
– should not be invertible (and thus
det 0
T
AA
=
), let A be a matrix with
46. [M] One may compute for this matrix that det A = –4008 and cond A ≈ 16.3. Note that this is the
2
A
condition number, which is used in Section 2.3. Since det A ≠ 0, it is invertible and
1
837 181 207 297
750 574 30 654
1
−
−−−
⎡⎤
⎢⎥
−−
3.3 • Solutions 181
3.3 SOLUTIONS
Notes:
This section features several independent topics from which to choose. The geometric
interpretation of the determinant (Theorem 10) provides the key to changes of variables in multiple
integrals. Students of economics and engineering are likely to need Cramer’s Rule in later courses.
1. The system is equivalent to Ax = b, where
57
24
A
⎡
⎤
=
⎢
⎥
⎣
⎦
and
3
1
⎡
⎤
=
⎢
⎥
⎣
⎦
b
. We compute
12 12
37 53
() , () ,det 6,det () 5,det () 1,
14 21
AA AAA
⎡⎤ ⎡⎤
=====−
⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
bb bb
2. The system is equivalent to Ax = b, where
41
52
A
⎡
⎤
=
⎢
⎥
⎣
⎦
and
6
7
⎤
=
⎥
⎦
b
. We compute
12 12
61 46
( ) , ( ) , det 3, det ( ) 5, det ( ) 2,
72 57
AA AAA
⎡⎤ ⎡⎤
=====−
⎢⎥ ⎢⎥
bb bb
3. The system is equivalent to Ax = b, where
32
56
A−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
and
7
5
⎡
⎤
=
⎢
⎥
−
⎣
⎦
b
. We compute
4. The system is equivalent to Ax = b, where
53
31
A−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
and
9
5
⎡
⎤
=
⎢
⎥
−
⎣
⎦
b
. We compute