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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
…
…
