182 CHAPTER 3 • Determinants
5. The system is equivalent to Ax = b, where
210
301
012
A
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
⎣
⎦
and
7
8
3
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
−
⎣
⎦
b
. We compute
710 2 70 21 7
⎡⎤⎡ ⎤⎡ ⎤
12 3
det 4,det ( ) 6,det ( ) 16, det ( ) 14,AA A A=== =−bb b
6. The system is equivalent to Ax = b, where
211
102
313
A
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
⎣
⎦
and
4
2
2
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
−
⎣
⎦
b
. We compute
123
det 4, det ( ) 16, det ( ) 52, det ( ) 4,AA A A==− = =−bbb
7. The system is equivalent to Ax = b, where
64
92
s
As
⎡
⎤
=
⎢
⎥
⎣
⎦
and
5
2
⎡
⎤
=
⎢
⎥
−
⎣
⎦
b
. We compute
8. The system is equivalent to Ax = b, where
35
95
s
As
−
⎡
⎤
=
⎢
⎥
⎣
⎦
and
3
2
⎡
⎤
=
⎢
⎥
⎣
⎦
b
. We compute
12 1 2
35 33
( ) , ( ) , det ( ) 15 10, det ( ) 6 27.
25 92
s
AA AsAs
s
−
⎡⎤ ⎡⎤
== =+=−
⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
bb b b
3.3 • Solutions 183
9. The system is equivalent to Ax = b, where
2
36
ss
As
−
⎡
⎤
=
⎢
⎥
⎣
⎦
and
1
4
−
⎡
⎤
=
⎢
⎥
⎣
⎦
b
. We compute
12 12
12 1
() , () ,det () 2,det () 4 3.
46 34
ss
AA AsAs
s
−− −
⎡⎤ ⎡⎤
== ==+
⎢⎥ ⎢⎥
bb bb
10. The system is equivalent to Ax = b, where
21
36
s
Ass
⎡
⎤
=
⎢
⎥
⎣
⎦
and
1
2
⎡
⎤
=
⎢
⎥
⎣
⎦
b
. We compute
12 1 2
11 21
() , () ,det () 6 2,det () .
26 3 2
s
AA AsAs
ss
⎡⎤ ⎡⎤
== =−=
⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
bb b b
11. Since det A = 3 and the cofactors of the given matrix are
11
00 0,
11
C==
12
30 3,
11
C=− =−
−
13
30 3,
11
C==
−
12. Since det A = 5 and the cofactors of the given matrix are
11
21 1,
10
C−
==−
12
210,
00
C=− =
13
22
2,
01
C−
==
13. Since det A = 6 and the cofactors of the given matrix are
11
01 1,
11
C==−
12
11 1,
21
C=− =
13
10 1,
21
C==
21
54 1,
11
C=− =−
22
34 5,
21
C==−
23
35 7,
21
C=− =
14. Since det A = –1 and the cofactors of the given matrix are
11
215,
34
C==
12
012,
24
C=− =
13
02 4,
23
C==−
21
67 3,
34
C=− =−
22
37 2,
24
C==−
23
36 3,
23
C=− =
15. Since det A = 6 and the cofactors of the given matrix are
11
10 2,
32
C==
12
10 2,
22
C−
=− =
−
13
11 1,
23
C−
==−
−
21
00 0,
32
C=− =
22
30 6,
22
C==
−
23
30 9,
23
C=− =−
−
16. Since det A = –9 and the cofactors of the given matrix are
11
31 9,
03
C−
==−
12
010,
03
C=− =
13
03
0,
00
C−
==
21
24 6,
03
C=− =−
22
14 3,
03
C==
23
12 0,
00
C=− =
3.3 • Solutions 185
⎡
⎢
⎢
⎢
⎣
17. Let
ab
Acd
⎡⎤
=
⎢⎥
⎣⎦
. Then the cofactors of A are
11
,Cdd==
12
,Ccc=− =−
21
Cbb=− =−
, and
18. Each cofactor of A is an integer since it is a sum of products of entries in A. Hence all entries in adj A
will be integers. Since det A = 1, the inverse formula in Theorem 8 shows that all the entries in
1
A
−
will be integers.
⎡
⎢
⎣
20. The parallelogram is determined by the columns of
14
35
A−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
, so the area of the parallelogram
is |det A| = |–7| = 7.
21. First translate one vertex to the origin. For example, subtract (–1, 0) from each vertex to get a new
parallelogram with vertices (0, 0),(1, 5),(2, –4), and (3, 1). This parallelogram has the same area as
⎡
⎢
⎣
22. First translate one vertex to the origin. For example, subtract (0, –2) from each vertex to get a new
parallelogram with vertices (0, 0),(6, 1),(–3, 3), and (3, 4). This parallelogram has the same area as
⎡
⎢
⎣
23. The parallelepiped is determined by the columns of
117
021
240
A
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
−
⎣
⎦
, so the volume of the
186 CHAPTER 3 • Determinants
24. The parallelepiped is determined by the columns of
121
452
⎣
A
−−
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
, so the volume of the
25. The Invertible Matrix Theorem says that a 3 × 3 matrix A is not invertible if and only if its columns
are linearly dependent. This will happen if and only if one of the columns is a linear combination of
the others; that is, if one of the vectors is in the plane spanned by the other two vectors. This is
26. By definition, p + S is the set of all vectors of the form p + v, where v is in S. Applying T to a typical
vector in p + S, we have T(p + v) = T(p) + T(v). This vector is in the set denoted by T(p) + T(S). This
27. Since the parallelogram S is determined by the columns of
22
35
−−
⎡
⎤
⎢
⎥
⎣
⎦
, the area of S is
22
det | 4 | 4.
35
−−
⎡⎤
=− =
⎢⎥
⎣⎦
The matrix A has
62
det 6
32
A−
==
−
. By Theorem 10, the area of T(S)
28. Since the parallelogram S is determined by the columns of
40
71
⎡
⎤
⎢
⎥
−
⎣
⎦
, the area of S is
40
det | 4 | 4
71
⎡⎤
==
⎢⎥
−
⎣⎦
. The matrix A has
72
det 5
11
A==
. By Theorem 10, the area of T(S) is
29. The area of the triangle will be one half of the area of the parallelogram determined by
1
v
and
2
.v
By Theorem 9, the area of the triangle will be (1/2)|det A|, where
[]
12
.A=
vv
30. Translate R to a new triangle of equal area by subtracting
33
(, )xy
from each vertex. The new triangle
3.3 • Solutions 187
13 23
13 23
1det .
2
xx x x
yy yy
−−
⎡⎤
⎢⎥
−−
⎣⎦
31. a. To show that T(S) is bounded by the ellipsoid with equation
2
22
312
222
1
xxx
abc
++=
, let
1
2
3
u
u
u
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
u
and
let
1
2
3
x
xA
x
⎡⎤
⎢⎥
==
⎢⎥
⎢⎥
⎣⎦
xu
. Then
11
/
uxa=
,
22
/
uxb=
, and
33
/
uxc=
, and u lies inside S (or
32. a. A linear transformation T that maps S onto S′will map
1
e
to
1
,v
2
e
to
2
,v
and
3
e
to
3
;v
that is,
188 CHAPTER 3 • Determinants
34. [M] Answers will vary, as will the commands which produce the second entry of x. For example, the
35. [M] MATLAB Student Version 4.0 uses 57,771 flops for inv A and 14,269,045 flops for the inverse
Chapter 3 SUPPLEMENTARY EXERCISES
1. a. True. The columns of A are linearly dependent.
b . True. See Exercise 30 in Section 3.2.
c. False. See Theorem 3(c); in this case
3
det 5 5 detAA=
.
d . False. Consider
20
01
A
⎡⎤
=
⎢⎥
⎣⎦
,
10
03
B
⎡
⎤
=
⎢
⎥
⎣
⎦
, and
30
04
AB
⎡
⎤
+=
⎢
⎥
⎣
⎦
.
k . True. See Theorems 6 and 5;
2
det (det )
T
AA A=
.
l. False. The coefficient matrix must be invertible.
m. False. The area of the triangle is 5.
2.
12 13 14 12 13 14
15 16 17 3 3 3 0
18 19 20 6 6 6
==
11 1
abc a bc abc
++ +
Chapter 3 • Supplementary Exercises 189
4.
111 0
abcabc abc
axbxcx xxx xy
+++= = =
5.
91999 9992 405
90992 4050
(1) (1)(2)9 3 9
40050 9390 607
=− =− −
6.
48885 4885 485
01000 6887 45
(1) (1)(2) 6 8 7 (1)(2)( 3) (1)( 2)( 3)( 2) 12
68887 0830 67
= = =− =−−=
7. Expand along the first row to obtain
11 1 1
11
22 2 2
111
11 0.
11
1
xy xy y x
xy x y
xy y x
xy
=−+=
8. Expand along the first row to obtain
11 1 1
111
11 1()()(1)0.
xy xy y x
xy x y mxyxmy
=−+=−−+=
This equation may
9.
22 2
222
11 1
det 1 0 0 ( )( )
aa a a aa
Tbb baba bababa
==−−=−−+
190 CHAPTER 3 • Determinants
10. Expanding along the first row will show that
23
01 2 3
() det .ft V c ct ct ct==+++
By Exercise 9,
2
11
2
3 2 2 2 13 13 2
1
1 ( )( )( ) 0
xx
cxxxxxxxx
=− = − − − ≠
11. To tell if a quadrilateral determined by four points is a parallelogram, first translate one of the
vertices to the origin. If we label the vertices of this new quadrilateral as 0,
1
v
,
2
v
, and
3
v
, then
they will be the vertices of a parallelogram if one of
1
v
,
2
v
, or
3
v
is the sum of the other two. In
12. A 2 × 2 matrix A is invertible if and only if the parallelogram determined by the columns of A has
13. By Theorem 8,
1
1
(adj ) det
AAAAI
A
−
⋅==
. By the Invertible Matrix Theorem, adj A is invertible
14. a. Consider the matrix
k
k
AO
AOI
⎡⎤
=
⎢⎥
⎣⎦
, where 1 ≤ k ≤ n and O is an appropriately sized zero matrix.
We will show that
det det
k
AA=
for all 1 ≤ k ≤ n by mathematical induction.
First let k = 1. Expand along the last row to obtain
(1)(1)
1
det det ( 1) 1 det det .
1
nn
AO
AAA
O
++ +
⎡⎤
==−⋅⋅=
⎢⎥
b. Consider the matrix
k
k
k
IO
ACD
⎡⎤
=
⎢⎥
⎣⎦
, where 1 ≤ k ≤ n,
k
C
is an n × k matrix and O is an
Chapter 3 • Supplementary Exercises 191
First let k = 1. Expand along the first row to obtain
11
1
1
1
det det ( 1) 1 det det .
O
ADD
CD
+
⎡⎤
==−⋅⋅=
⎢⎥
⎣⎦
c. By combining parts a. and b., we have shown that
det det det (det )(det ).
AO AO I O AD
CD OI CD
⎛⎞⎛⎞
⎡⎤ ⎡⎤⎡⎤
==
⎜⎟⎜⎟
⎢⎥ ⎢⎥⎢⎥
⎣⎦ ⎣⎦⎣⎦
⎝⎠⎝⎠
From this result and Theorem 5, we have
15. a. Compute the right side of the equation:
IOAB A B
X I O Y XA XB Y
⎡⎤⎡⎤⎡ ⎤
=
⎢⎥⎢⎥⎢ ⎥
+
⎣⎦⎣⎦⎣ ⎦
Set this equal to the left side of the equation:
b. From part a.,
11
det (det )(det ( )) det[ ( )]
AB ADCAB ADCAB
CD
−−
⎡⎤
=−=−
⎢⎥
⎣⎦
16. a. Doing the given operations does not change the determinant of A since the given operations are
all row replacement operations. The resulting matrix is
00
00
ab ab
ab ab
−−+ …
⎡⎤
⎢⎥
−−+…
192 CHAPTER 3 • Determinants
b. Since column replacement operations are equivalent to row operations on
T
A
and
det det
T
AA=
,
the given operations do not change the determinant of the matrix. The resulting matrix is
00 0
00 0
ab
ab
−…
⎡⎤
⎢⎥
−…
17. First consider the case n = 2. In this case
2
det ( ),det ,
0
ab b b b
BaabCabb
aba
−
==−==−
bb a
…
since the matrix in the above formula is a (k – 1) × (k – 1) matrix. We can perform a series of row
operations on C to “zero out” below the first pivot, and produce the following matrix whose
determinant is det C:
bb b
…
⎡⎤
18. [M] Since the first matrix has a = 3, b = 8, and n = 4, its determinant is
41 3
(3 8) (3 (4 1)8) ( 5) (3 24) ( 125)(27) 3375.
−
−+−=−+=− =−
Since the second matrix has a = 8, b =
19. [M] We find that
11111
1111
111 12222
1222
122 1, 1, 1.
12333
== =
Our conjecture then is that
111 1
122 2
…
…
To show this, consider using row replacement operations to “zero out” below the first pivot. The
resulting matrix is
111 1
011 1
…
⎡⎤
⎢⎥
…
⎢⎥
Now use row replacement operations to “zero out” below the second pivot, and so on. The final
matrix which results from this process is
111 1
011 1
,
001 1
…
⎡⎤
⎢⎥
…
⎢⎥
⎢⎥
…
20. [M] We find that
1111 1
111 1
111 1333 3
194 CHAPTER 3 • Determinants
111 1
133 3
…
…
To show this, consider using row replacement operations to “zero out” below the first pivot. The
resulting matrix is
111 1
022 2
…
⎡⎤
⎢⎥
…
Now use row replacement operations to “zero out” below the second pivot. The matrix which results
from this process is
11111 1 1
02222 2 2
00333 3 3
…
⎡⎤
⎢⎥
…
⎢⎥
⎢⎥
…
This matrix has the same determinant as the original matrix, and is recognizable as a block matrix of
the form
,
AB
OD
⎡⎤
⎢⎥
⎣⎦
where
333 3 3 1111 1
366 6 6 1222 2
11
and 3 .
369 9 9 1233 3
⎢
⎢
⎢
⎣
AD
……
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
……
⎢
⎥⎢ ⎥
⎡⎤
⎢
⎥⎢ ⎥
== =
……
Chapter 3 • Supplementary Exercises 195
1111 1
1222 2
…
…