2.1 Deterrminants by C
Cofactor Expaansion
CHA
APTER 2: DETE
ERMINA
ANTS
2.1 Deterrminants by
b Cofacto
or Expansion
1 2
1.
6
M11
1 
3
7 1
7 1 
 29
1 4
1 4
1 2
6
M12
1 
3
6
M 21
2 
3
C13   1
M13  M13  27
1 3
C22   1
22
M 22  M 22  13
C23   1
2 3
M 23   M 23  5
3 1
M 31  M 31  19
2 1
M 21   M 21  11
3
3
1 2
 5
7 1 
3
1
1 4
3
2 3
7 1 
 19
7 1
1 4
C31   1
3
1 3
 19
7 1 
6 1
1 4
1 2
6
M33
3 
3
M12   M12  21
1 2
1 3
7 1 
 13
3 4
1 4
1 2
M32
6
3 
3
C12   1
C21   1
1 2
6
M31
3 
3
M11  M11  29
2 3
 11
1 4
1 2
M 23
6
2 
3
3
7 1 
1 4
1 2
6
M 22
2 
3
11
3
6 7
7 1 
 27
3 1
1 4
1 2
C11   1
3
6 1
7 1 
 21
3 4
1 4
1 2
6
M13
1 
3
3
C32   1
3 2
M 32   M 32  19
3 3
M 33  M 33  19
3
1 2
7 1 
 19
6
7
1 4
C33   1
1
Cofactor Expaansion
2.1 Deterrminants by C
1 1 2
3 6
M11  3 3 6 
6
1 4
0 1 4
2.
C11   1
11
M11  M 11  6
1 1 2
3 6
M12  3 3 6 
 12
1
0 4
0 1 4
C12   1
M12   M12  12
C13   1
M13  M13  3
1 2
1 1 2
3 3
M13  3 3 6 
3
0 1
0 1 4
1 3
1 1 2
1 2
M 21  3 3 6 
2
1 4
0 1 4
C21   1
2 1
M 21   M 21  2
C22   1
22
M 22  M 22  4
1 1 2
1 2
M 22  3 3 6 
4
0 4
0 1 4
1 1 2
1 1
M 23  3 3 6 
1
0 1
0 1 4
C23   1
23
M 23   M 23  1
3 1
M 31  M 31  0
1 1 2
1 2
M 31  3 3 6 
0
3 6
0 1 4
C31   1
1 1 2
1 2
M 32  3 3 6 
0
3 6
0 1 4
C32   1
3 2
M 32   M 32  0
3 3
M 33  M 33  0
1 1 2
1 1
M 33  3 3 6 
0
3 3
0 1 4
0 0
3.
(a)
M13
 4
4
3
1 14
4 14
4 1
1 14  0
3
0
4 2
1 2
4 1
1 2
 0  0  3 0  0
C13
   1
1 3
C33   1
M133  M13  0
cofactor expanssion
along the first roow
2
2.1 Determinants by Cofactor Expansion
(b)
M 23
4 1 6
1 14
4 14
4 1
4
1 14  4
  1
6
1 2
4 2
4 1
4
1 2
cofactor expansion
along the first row
 4  12   1 48   6  0   96
(c)
C23
  1
M22
4 1 6
1 6
4 6
4 1
 4 0 14  4
0
 14
3 2
4 2
4 3
4 3 2
2 3
M 23   M 23  96
cofactor expansion
along the second row
 4  16   0  14  8   48
(d)
C22
  1
M21
1 1 6
1 6
1 6
1 1
 1 0 14  1
0
 14
3 2
1 2
1 3
1 3 2
22
M 22  M 22  48
cofactor expansion
along the second row
 1 16   0  14  4   72
4.
(a)
C21
   1
M 32
2 1 1
0 3
3 3
3 0
 3 0 3  2
  1
1
1 4
3 4
3 1
3 1 4
2 1
M 21   M 21  72
cofactor expansion
along the first row
 2  3  1 21  1 3  30
(b)
C32
  1
M44
2
3 1
2 0
3 0
3 2
 3 2 0  2
3
  1
2 1
3 1
3 2
3 2
1
3 2
M 32   M 32  30
 2  2   3  3  1 0   13
C 44
   1
44
M 44  M 44  13
cofactor expansion
along the first row
3
2.1 Deterrminants by C
Cofactor Expaansion
3 1 1
(c)
0 3
2 3
2 0
  1
1
0 3 3
1 0
2 0
2 1
1 0
M41  2
2
coofactor expanssion
allong the first roow
 3  3  1 6   1 2   1
  1
C41
4 1
3 1
2
(d)
M 41   M 41  1
3 1
2 1
3 2
1 2
  1
3
3 1
2 1
3 2
1
M24  3 2
3 2
coofactor expanssion
allong the first roow
 2  0   3  0   1 0   0
   1
C24
2 4
M 24  M 24  0
5.
3 5
 4 5   112
  3 4    5  2   12  10  22  0 . Inverrse: 221 
1
2 4
 2 3   11
6.
4 1
  4  2   1 8   0 ; The matriix is not inverrtible.
8 2
7.
5 7
 2 7   592
1
4  59  0 . Inverse:
  5  2    7  7   10  49
I
7
59 
7 2
 7 5  59
8.
9.

a3
5
3 a  2

2
5 1 2
3 8 4
 5
3
7
1
4
3 5 7
1 6 2
1
3 6



 3

 4
7
59
5
59



1
 6 3 2
 4
2   3 6
5
a3
  a  3 a  2   5  3  a 2  5a  6  15  a 2  5a  21
3 a  2
6
2
11.
6
4
2
10.
 2  3    6   4   6  4 6  3 6  0 . Inverrse:
3
2
5
22
3
22
2
7
6 2
1 2 5
8 4 3
1
4 2
 3 5 7
1 6 2
7
1   8  42  24
40   18  322  140   0
8
1
3 5   20  7  72    20  844  6   65
1 6


1
3 3

1
3
4
2.1 Deterrminants by C
Cofactor Expaansion
1
12.
0
1 1
2 1 1
3
0
 3 0 5 3 0   0  5  42    0  35  6   4
1 7 2 1 7
0
0
 2 1 5
1 9 4
2 1 5
1 9 4
c 4
14.
2
3 0 5
1 7 2
3
13.
1
3
0
2 1  12  0  0    0  135  0   123
1 9
3
c 4
3 c 4
c2  2
2
1
1
c2 2
1  2c  16c 2  6  c  1   12   c  1 c 3  16 
4 c 1 2
4 c 1 2 4 c 1
 2c  16c 2  6c  6  122  c 4  c 3  16
1  c 4  c 3  16c 2  8c  2
15.
det  A  
 2
5
1
  λ  2  λ  4   1 5   λ 2  2λ  3   λ  3  λ  1
4
The determinant is zero iff   3 or   1 .
16.
Calculate thee determinantt by a cofacto
or expansion along
a
the firstt row:
4 0
det  A   0

0
0
2
3  1
   4

2
00
3  1
    4      1  6      4   2    6      4    3   2 
The determinant is zero iff   2 ,   3 , or   4 .
17.
det  A  
 1
2
0
  λ  1 λ  1
 1
The determinant is zero iff   1 or   1 .
18.
or expansion along
a
the thirdd row:
Calculate thee determinantt by a cofacto
4 4
det  A   1 
0
0
0
0
 5
 0  0     5
 4 4
1 
    5    4    4      5   2  4  4      5   2 
2
5
2.1 Determinants by Cofactor Expansion
The determinant is zero if   2 or   5 .
19.
20.
21.
(a)
3
1 5
 0  0  3  41  123
9 4
(b)
3
1 5
0 0
0 0
2
1
 3  41  2  0   1 0   123
9 4
9 4
1 5
(c)
2
(d)
0   1
(e)
1
(f)
05
3 0
3 0
  4 
 5  27   4  3  123
1 9
2 1
(a)
 1
0 5
3 5
3 0
1
2
  1 35  111  2  21  4
7 2
1 2
1 7
(b)
 1
0 5
1 2
1 2
3
1
  1 35   3  12   1 5   4
7 2
7 2
0 5
(c)
3
1 2
1 1
 0   5
 3  12   0  5  8   4
7 2
1 7
(d)
1
3 5
1 2
07
 111  0  7  1  4
1 2
3 5
(e)
1
1 2
1 1
1 2
7
2
 1 5  7  1  2  3   4
0 5
3 5
3 0
(f)
2
3 0
1 1
1 1
  5 
2
 2  21  5  8   2  3   4
1 7
1 7
3 0
0 0
3 0
3 0
  1
5
 2  0   1 12   5  27   123
9 4
1 4
1 9
3 0
3 0
9
 1 12   9 15  123
1 4
2 5
0 0
3 0
3 0
9
  4 
 1 0   9 15  4  3  123
1 5
2 5
2 1
Calculate the determinant by a cofactor expansion along the second column:
0  5
22.
Calculate the determinant by a cofactor expansion along the second row:
1
23.
3 7
 0  5  8   40
1 5
3 1
3 3
 0   4 
 118   0  4  12   66
3 5
1 3
Calculate the determinant by a cofactor expansion along the first column:
6
2.1 Determinants by Cofactor Expansion
1
24.
k k2
k k2
k k2

1

1
 1  0   1 0   1  0   0
k k2
k k2
k k2
Calculate the determinant by a cofactor expansion along the second column:
  k  1
k 1 7
k 1 7
2 4
  k  3
  k  1
k
5 k
5
2
4
   k  1 2 k  20    k  3    k  1 k  35    k  1  4  k  1  14 
 k 3  8k 2  10 k  95
25.
Calculate the determinant by a cofactor expansion along the third column:
3 3
5
3 3
5
det  A   0  0   3  2 2 2  3 2 2 2
2 10
2
4 1 0
Calculate the determinants in the third and fourth terms by a cofactor expansion along the first row:
3 3
5
2 2
2 2
2 2
2 2 2  3
3
5
 3  24   3  8   5 16   128
10
2
2
2
2 10
2 10
2
3 3
5
2 2
2 2
2 2
2 2 2  3
3
5
 3  2   3  8   5  6   48
1 0
4
0
4 1
4 1 0
Therefore det  A   0  0  3 128   3  48   240.
26.
Calculate the determinant by a cofactor expansion along the first row:
3 3 1 0
det  A   4
3 3 3 0
2 4
4 6
2 3
1 2 4 3
 0  0 1
0
2 3
9 4 6 3
2 4
2 3
2 2 4 3
Calculate each of the two determinants by a cofactor expansion along its first row:
3 3 1 0
2 4
4 6
2 4
4 2 3
2 2 3
2 4 3
2 3
 3 6 2 3  3 4 2 3   1 4 6 3  0  3  0   3  0   1 0   0  0
2 3
4 2 3
2 2 3
2 4 3
2 3
3 3 3 0
2 4 3
1 4 3
1 2 3
1 2 4 3
 3 4 6 3  3 9 6 3  3 9 4 3  0  3  0   3  6   3  6   0  0
9 4 6 3
2 4 3
2 4 3
2 2 3
2 2 4 3
Therefore det  A   4  0   0  0  1 0   0.
7
2.1 Determinants by Cofactor Expansion
27.
By Theorem 2.1.2, determinant of a diagonal matrix is the product of the entries on the main diagonal:
det  A   1 11  1.
28.
By Theorem 2.1.2, determinant of a diagonal matrix is the product of the entries on the main diagonal:
det  A    2  2  2   8.
29.
By Theorem 2.1.2, determinant of a lower triangular matrix is the product of the entries on the main diagonal:
det  A    0  2  3 8   0 .
30.
By Theorem 2.1.2, determinant of an upper triangular matrix is the product of the entries on the main diagonal:
det  A   1 2  3 4   24.
31.
By Theorem 2.1.2, determinant of an upper triangular matrix is the product of the entries on the main diagonal:
det  A   11 2  3  6.
32.
By Theorem 2.1.2, determinant of a lower triangular matrix is the product of the entries on the main diagonal:
det  A    3 2  1 3  18 .
33.
(a)
sin 
 cos
cos
  sin   sin     cos   cos   sin 2   cos2   1
sin 
(b)
Calculate the determinant by a cofactor expansion along the third column:
0  0 1
35.
sin 
 cos
cos
 0  0  11  1 (we used the result of part (a))
sin 
The minor M11 in both determinants is
1 f
 1 . Expanding both determinants along the first row yields
0 1
d1    d2 .
37.
If n 1 then the determinant is 1 .
If n  2 then
1 1
0.
1 1
If n  3 then a cofactor expansion will involves minors
1 1
 0 . Therefore the determinant is 0 .
1 1
By induction, we can show that the determinant will be 0 for all n  3 as well.
43.
Calculate the determinant by a cofactor expansion along the first column:
1 x1
x12
1 x2
x22 
1 x3
2
3
x
x2
x22
x1
x12
x1
x12
x3
x3
x3
x3
x2
x22

2

2


 ( x2 x32  x3 x22 )  ( x1 x32  x3 x12 )  ( x1 x22  x2 x12 )   x32  x2  x1   x3 x22  x12   x1 x22  x2 x12 .
8
2.1 Determinants by Cofactor Expansion
Factor out  x2  x1  to get  x2  x1   x32  x2 x3  x1 x3  x1 x2    x2  x1   x32   x2  x1  x3  x1 x2  .
Since x32   x2  x1  x3  x1 x2   x3  x1  x3  x2  , the determinant is  x2  x1  x3  x1  x3  x2  .
True-False Exercises
(a)
False. The determinant is ad  bc .
(b)
False. E.g., det  I 2   det  I 3   1 .
(c)
True. If i  j is even then  1
(d)
a b
True. Let A   b d
 c e
Then C12   1
1 2
b
c
i j
 1 therefore Cij   1
i j
Mij  Mij .
c
e  .
f 
e
c
2 1 b
   bf  ec  and C21   1
   bf  ce  therefore C12  C21 . In the same way,
f
e f
one can show C13  C 31 and C 23  C 32 .
(e)
True. This follows from Theorem 2.1.1.
(f)
True. In formulas (7) and (8), each cofactor Cij is zero.
(g)
False. The determinant of a lower triangular matrix is the product of the entries along the main diagonal.
(h)
False. E.g. det  2 I 2   4  2  2det  I 2  .
(i)
False. E.g., det  I 2  I 2   4  2  det  I 2   det  I 2  .
(j)
  a b  2  a 2  bc ab  bd

True. det  
 a 2  bc bc  d 2   ab  bd  ac  cd 
  c d   ac  cd bc  d 2





 a 2 bc  a 2 d 2  b2 c 2  bcd 2  a 2 bc  abcd  abcd  bcd 2  a 2 d 2  b 2 c 2  2abcd .
2
 a b 2    a b   
   det
  ad  bc   a d  2 adbc  b c therefore det  
 .
  c d      c d   
c d




a b
2
2
2
2
2 2
2.2 Evaluating Determinants by Row Reduction
1.
det  A  
2 3
2 1
  2  4    31  11 ; det AT 
  2  4   1 3  11
1 4
3 4
2.
det  A  
1
6
6 2
  6  2   1 2   10 ; det AT 
  6  2    2 1  10
2 2
1 2
 
 
9
2.2 Evaluating Determinants by Row Reduction
3.
2 1 3
det  A   1 2 4   24  20  9  30  24  6   5 ;
5 3 6
 
det A
T
4.
2 1 5
 1 2 3   24  9  20   30  24  6   5 (we used the arrow technique)
3 4 6
4 2 1
det  A   0 2 3   40  6  0    2  12  0   56 ;
1 1 5
 
det A
T
4 0 1
 2 2
1   40  0  6    2  12  0   56 (we used the arrow technique)
1 3 5
5.
The third row of I 4 was multiplied by 5 . By Theorem 2.2.4, the determinant equals 5.
6.
5 times the first row of I 3 was added to the third row. By Theorem 2.2.4, the determinant equals 1.
7.
The second and the third rows of I 4 were interchanged. By Theorem 2.2.4, the determinant equals 1.
8.
The second row of I 4 was multiplied by  13 . By Theorem 2.2.4, the determinant equals  13 .
9.
3 6
9
1 2
3
2
7 2  3 2
7 2
0
1 5
0
1 5
1 2 3
30
3 4
0
1 5
1 2 3
 3  1 0
1 5
0
3 4
A common factor of 3 from the first row
was taken through the determinant sign.
2 times the first row was added to the
second row.
The second and third rows were
interchanged.
1 2
3
  3  1 0
1
5
0
0 11
3 times the second row was added to the
third row.
1 2 3
  3  1 11 0
1 5
0
0 1
A common factor of 11 from the last row
was taken through the determinant sign.
  3 1 111  33
10
2.2 Evaluating Determinants by Row Reduction
11
Another way to evaluate the determinant would be to use cofactor expansion along the first column after the second
step above:
3 6
9
1 2 3
 3 4

2
 0  0   3 111   33 .
7 2  3 0
3 4  3 1
 1 5

0
1 5
0
1 5
10.
3 6 9
1 2 3
0 0 2  3 0 0 2
2 1 5
2 1 5
1 2 3
 3 0 0 2
0 5 1
1 2 3
 3  1 0 5 1
0 0 2
1 2 3
  3  1 5  0 1  15
0 0 2
1 2 3
  3  1 5  2  0 1  15
0 0
1
A common factor of 3 from the first row
was taken through the determinant sign.
2 times the first row was added to the third
row.
The second and third rows were
interchanged.
A common factor of 5 from the second row
was taken through the determinant sign.
A common factor of 2 from the last row
was taken through the determinant sign.
  3 1 5 2 1  30
Another way to evaluate the determinant would be to use cofactor expansion along the first column after the
second step above:
3 6 9
1 2 3
 0 2

0 0 2  3 0 0 2  3  1
 0  0   3 110    30 .
 5 1

0 5 1
2 1 5
11.
2 1 3 1
1 0 1 1
0 2 1 0
0 1 2 3
  1
1 0 1 1
2 1 3 1
0 2 1 0
0 1 2 3
The first and second rows
were interchanged.
2.2 Evaluating Determinants by Row Reduction
1 0 1 1
0 1 1 1
  1
  1
0 2 1
0 1 2
0
3
1 0
1
0 1 1 1
0 0 1 2
0
  1
1
1
1 0
0 1
2
3
1 1
1 1
0 0 1
0 0
1
2
4
1 0 1
1
  1 1
  1 1
0 1 1 1
0 0 1 2
0 0 1
4
1 0
1
1
0 1 1 1
0 0 1 2
0 0 0
  1 1 6 
6
1 0 1 1
0 1 1 1
0 0 1 2
0 0 0
1
2 times the first row was
added to the second row.
2 times the second row was
added to the third row.
1 times the second row was
added to the fourth row.
A common factor of 1 from
the third row was taken
through the determinant sign.
1 times the third row was
added to the fourth row.
A common factor of 6 from
the third row was taken
through the determinant sign.
  1 1 6 1  6
Another way to evaluate the determinant would be to use cofactor expansions along the first column after the
fourth step above:
2 1 3 1
1 0 1 1
0 2 1 0
0 1 2 3
  1
1 0
0 1
1 1
1 1
0 0 1
0 0 1
1 1 1
1 2
  11 0 1 2   111
2
1 4
0 1 4
4
  111 6   6 .
12.
1 3 0 1 3 0
2
4 1  0 2 1
5 2 2 5 2 2
2 times the first row was
added to the second row.
12
2.2 Evaluating Determinants by Row Reduction
1 3 0
 0 2 1
0 13 2
1 3
0
 2 0
1  12
0 13
2
1 3
0
 2 0
1  12
0
  2  
17
2
0
17
2
1 3
0
 0 1  12
0 0
1
5 times the first row was
added to the third row.
A common factor of 2 from the second
row was taken through the determinant
sign.
13 times the second row was
added to the third row.
A common factor of 172 from the last row
was taken through the determinant sign.
  2   172  1  17
Another way to evaluate the determinant would be to use cofactor expansion along the first column after the
second step above:
1 3 0 1 3 0
2 1
2
 1 17   17 .
4 1  0 2 1  1
13 2
5 2 2 0 13 2
13.
1 3
2 7
0 0
0 0
0
1 5
0 4
1 0
2
1
0 0
1 3
0 1
0 0
0 0
0
1
0
  1 0
0
3
2
1
1
1 1
1
2
1
2
0 0
5
6
0
1
3
8
1
1
1 1
3
1 5 3
1 2 6 8
0
1 0
1
0 2
1 1
0 0
0
2 times the first row was
added to the second row.
1
1
A common factor of 1 from the second row
was taken through the determinant sign.
13
2.2 Evaluating Determinants by Row Reduction
1
0
  1 0
0
3
1 5 3
1 2 6 8
0
1 0
1
0 0
1 1
0 0
1
0
  1 0
0
1
1
3
1 5 3
1 2 6 8
0
1 0
1
0 0
1 1
0 0
1
0
  1 2  0
0
0
0
0
0
0
1 times the fourth row was
added to the fifth row.
2
3
1 5 3
1 2 6 8
0
1 0
1
0
0
1 1
0 0
2 times the third row was
added to the fourth row.
A common factor of 2 from the fifth row
was taken through the determinant sign.
1
  1 2 1  2
Another way to evaluate the determinant would be to use cofactor expansions along the first column after the
third step above:
1 3
2 7
0 0
0 0
0
1 5
0 4
1 0
2
1
0 0
3
1
2
0
1   1 0
1
0
1 1
3
1 5 3
1 2 6 8
1 2 6 8
0
1 0
1
0
1 0
1   11
0
0
1 1
0
0
1 1
0
0
1 1
0 0
0
1 1
1 0
1
1 1
  111 0 1 1   1111
  1111 2   2 .
1 1
0 1 1
14.
1 2
3
1
1 2
3
1
5 9 6
3
0
1 9 2

1 2 6 2 1 2 6 2
2 8 6
1
2 8 6
1
1 2 3
1
0
1 9 2

0 0 3 1
2
8 6
1
5 times the first row was added to the
second row.
The first row was added to the third row.
14
2.2 Evaluating Determinants by Row Reduction
1 2 3
1
0
1 9 2

0 0 3 1
0 12 0 1
1 2
3
1
0
1 9 2

0
0 3 1
0
0 108 23
1 2
3
1
0
1 9 2
 3
1
0 0
1
3
0 0 108 23
1 2
3
1
0
1 9 2
 3
1
0 0
1
3
0 0 0 13
1 2
3
1
0
1 9 2
  3  13 
1
0 0
1
3
0 0 0
1
2 times the first row was added to the
fourth row.
12 times the second row was added to the
fourth row.
A common factor of 3 from the third row
was taken through the determinant sign.
108 times the third row was added to the
fourth row.
A common factor of 13 from the third row
was taken through the determinant sign.
  3 131  39
Another way to evaluate the determinant would be to use cofactor expansions along the first column after the
fourth step above:
1 2
3
1 1 2
3
1
1 9 2
3 1
5 9 6
3 0
1 9 2

 1 0 3 1  11
1 2 6 2 0 0 3 1
108 23
0 108 23
2
8 6
1 0 0 108 23
 111 39   39 .
15.
d e
g h
a b
f
a b
i   1 g h
c
d e
c
i
f
The first and third rows were interchanged.
a b
  1 1 d e
g h
c
f
i
The second and third rows were interchanged.
15
2.2 Evaluating Determinants by Row Reduction
  1 1 6   6
16.
17.
g h
The first and the third rows were interchanged, therefore d e
a b
3a 3b 3c
a b
c
 d e  f  3  d e  f
4 g 4 h 4i
4 g 4 h 4i
c
f    6   6.
i
A common factor of 3 from the first row
was taken through the determinant sign.
a b c
 3  1 d
e f
4 g 4 h 4i
a b
 3  1 4  d e
g h
i
a b
f  d e
c
g h
A common factor of 1 from the second
row was taken through the determinant
sign.
c
f
i
A common factor of 4 from the third row
was taken through the determinant sign.
 3  1 4  6   72
18.
ad be c f
a b
c
d
e
 f   d e  f
g
h
i
g h
i
a b
 1 d e
g h
The second row was added to the first row.
c
f
i
A common factor of 1 from the second
row was taken through the determinant
sign.
  1 6   6
19.
ag bh ci a b
d
e
f d e
g
h
i
g h
c
f
i
1 times the third row was added to the first
row.
 6
20.
a
b
c
a b
c
2d
2e
2 f  2 d 2e 2 f
g  3a h  3b i  3c
g h
i
3 times the first row was added to the last
row.
16
2.2 Evaluating Determinants by Row Reduction
a b
2 d e
g h
c
f
i
A common factor of 2 from the second row
was taken through the determinant sign.
  2  6   12
21.
3a
3b
3c
d
e
f
g  4 d h  4e i  4 f
a
b
c
 3 d
e
f
g  4 d h  4e i  4 f
a b
 3 d e
g h
c
f
i
A common factor of 3 from the first row
was taken through the determinant sign.
4 times the second row was
added to the last row.
  3 6   18
22.
a b c
The third row is proportional to the first row, therefore by Theorem 2.2.5 d
e
f  0.
2 a 2b 2c
(This can also be shown by adding 2 times the first row to the third, then performing a cofactor expansion of the
a b c
resulting determinant d e f along the third row.)
0 0 0
23.
1 1 1
1
a b c  0
a 2 b2 c2 a 2
1
1
ba ca
b2
c2
1
1
 0 ba
0 b2  a 2
1
ca
c2  a2
1
1
1
 0 ba
ca
2
2
0
0
c  a   c  a  b  a 
 1 b  a  c  a  c  a  b  a 
  b  a  c  a  c  b 
a times the first row was added to the
second row.
 a 2 times the first row was added to the third
row.
  b  a  times the second row was added to
the third row.
17
2.2 Evaluating Determinants by Row Reduction
24.
(a)
Interchanging the first row and the third row and applying Theorem 2.1.2 yields
0 a13 
 0
 a31 a32 a33 


det  0 a22 a23    1 det  0 a22 a23   a13 a22 a31
 a31 a32 a33 
 0
0 a13 
(b)
We interchange the first and the fourth row, as well as the second and the third row. Then we use
Theorem 2.1.2 to obtain
0
0 a14 
 0
 a41 a42 a43 a44 
 0

 0 a
0 a23 a24 
a33 a34 
32


  1 1 det
a a a a
det
 0 a32 a33 a34 
 0
0 a23 a24  14 23 32 41




0
0 a14 
 0
 a41 a42 a43 a44 
Generally for any n  n matrix A such that aij  0 if i  j  n we have
det  A    1 a1n a2,n 1  an1 .
n
a1
a2
a3
25.
a1  b1  c1
a2  b2  c2
a3  b3  c3
b1
b2
b3
a1
 a2
a3
b1
b2
b3
b1  c1
b2  c2
b3  c3
1 times the first column was
added to the third column.
b1
b2
b3
1 times the second column was
added to the third column.
a1
 a2
a3
26.

a1  b1t
a1t  b1
a2  b2 t
a2 t  b2
a3  b3 t
a3 t  b3
c1
c2
c3
a1  b1t

 1  t b1
2
a2  b2 t

 1 t

a3  b3 t
1  t  b 1  t  b
2
2
2
c1
2
c1
c2
c3
a1  b1t
b1
c1

3
c2
c3
a2  b2 t
b2
c2
a3  b3 t
b3
c3
 1 t
2

a1
b1
c1
a2
b2
c2
a3
b3
c3
t times the first row was
added to the second row.
A common factor of 1  t 2 from the second
row was taken through the determinant sign.
t times the second row was
added to the first row.
18
2.2 Evaluating Determinants by Row Reduction
27.
28.
a1
a2
a3
a1  b1
a2  b2
a3  b3
a1  b1
a2  b2
a3  b3
c1
c2
c3
a1  b1
 a2  b2
a3  b3
2b1
2b2
2b3
c1
c2
c3
a1  b1
 2 a2  b2
a3  b3
b1
b2
b3
c1
c2
c3
a1
 2 a 2
a3
b1
b2
b3
c1
c2
c3
1 times the first column was added to the
second column.
A common factor of 2 from the second
column was taken through the determinant
sign.
1 times the second column was added to
the first column.
b1  ta1
b2  ta2
b3  ta3
c1  rb1  sa1
c2  rb2  sa2
c3  rb3  sa3
a1
 a2
a3
b1
b2
b3
c1  rb1  sa1
c2  rb2  sa2
c3  rb3  sa3
t times the first column was
added to the second column.
a1
 a2
a3
b1
b2
b3
c1  rb1
c2  rb2
c3  rb3
s times the first column was
added to the third column.
a1
 a2
a3
b1
b2
b3
c1
c2
c3
a1
 b1
c1
a2
b2
c2
a3
b3
c3
r times the second column was
added to the third column.
The matrix was transposed.
(Theorem 2.2.2)
29.
The second column vector is a scalar multiple of the fourth. By Theorem 2.2.5, the determinant is 0.
30.
Adding the second, third, fourth, and fifth rows to the first results in the first row made up of zeros.
31.
1 2 0 3 0
0

1 2 
3 0
det  M   2 5 0 2 1 0   0  0  2
 0  0   4 
   2  12   24
2 5 
2 1

 1 3 2  3 8 4
19
2.2 Evaluating Determinants by Row Reduction
32.
1 2 0
1 2
  111   11   1
det  M   0 1 2
0 1
0 0 1
33.
In order to reverse the order of rows in 2  2 and 3  3 matrix, the first and the last rows can be interchanged, so
det  B   det  A  .
For 4  4 and 5  5 matrices, two such interchanges are needed: the first and last rows can be swapped, then the
second and the penultimate one can follow.
Thus, det  B    1 1 det  A   det  A  in this case.
Generally, to rows in an n  n matrix can be reversed by

interchanging row 1 with row n ,

interchanging row 2 with row n 1 ,



interchanging row n / 2 with row n  n / 2
where x is the greatest integer less than or equal to x (also known as the "floor" of x ).
We conclude that det  B    1
34.
a b b b
b a b b
b b a b
b b b a



det  A  .
a
b
ba ab
ba
ba
0
0
ab
b
b
0
0
a  2b
b
ba ab
b
0
ab
0
ab
0
b
ba ab
0
ba
ab
0
0

n /2
b
0
0
0
ab
b
0
b
0
0
0
0
0
0
ab
0
ab
a  3b
b
b
0
0
0
ab
0
0
ab
0
0
(a)
True. det  B    1 1 det  A   det  A  .
The last column was
added to the first column.
The third column was
added to the first column.
b
0
0
ab
  a  3b  a  b 
True-False Exercises
1 times the first row was added
to each of the remaining rows.
3
The second column was
added to the first column.
20
2.2 Evaluating Determinants by Row Reduction
21
(b)
True. det  B    4   34  det  A   3det  A  .
(c)
False. det  B   det  A  .
(d)
False. det  B   n  n  13  2  1  det  A    n! det  A  .
(e)
True. This follows from Theorem 2.2.5.
(f)
True. Let B be obtained from A by adding the second row to the fourth row, so det  A   det  B  . Since the fourth
row and the sixth row of B are identical, by Theorem 2.2.5 det  B   0 .
2.3 Properties of Determinants; Cramer’s Rule
1.
det  2 A  
2 4
  2  8    4  6   40
6 8
 2  det  A  4
2
2.
det  4 A  
1 2
 4   1 4    2  3    4  10   40
3 4
8 8
  8  8    8  20   224
20 8
 4  det  A   16
2
3.
2 2
 16   2  2    2  5   16  14   224
5 2
We are using the arrow technique to evaluate both determinants.
4
2 6
det  2 A   6 4 2   160  8  288    48  64  120   448
2 8 10
2 1 3
 2  det  A   8 3 2 1   8    20  1  36    6  8  15    8  56   448
1 4 5
3
4.
We are using the cofactor expansion along the first column to evaluate both determinants.
3 3
3
6 9
 3   6  6    9  3     3  63   189
det  3 A   0 6
9 3
3 6
0 3 6
1 1
1
2 3
 27   2  2    3 1    27  7   189
3 det  A   27 0 2
3   27 1
1 2
0 1 2
3
5.
We are using the arrow technique to evaluate the determinants in this problem.
2.3 Properties of Determinants; Cramer’s Rule
9 1 8
det  AB   31 1 17  18  170  0    80  0  62   170 ;
10 0 2
 1 3 6
det  BA   17 11 4   22  120  510    660  20  102   170 ;
10 5 2
3 0 3
det  A  B   10 5 2   45  0  0    75  0  0   30 ;
5 0 3
det  A   16  0  0    0  0  6   10 ;
det  B   1  10  0   15  0  7   17 ;
det  A  B   det  A  det  B 
6.
We are using the arrow technique to evaluate the determinants in this problem.
6 15 26
det  AB   2 4 3   288  90  520    208  180  360   66 ;
2 10 12
5 8 3
det  BA   6 14 7   350  280  36    210  70  240   66 ;
5 2 5
1 7 2
det  A  B   2 1 2  1  28  20    4  10  14   75 ;
1
2 5
det  A    0  16  4    0  2  16   2 ;
det  B    2  0  12    0  18  1  33 ;
det  A  B   det  A  det  B  ;
7.
det  A    6  0  20    10  0  15  1  0 therefore A is invertible by Theorem 2.3.3
8.
det  A    24  0  0    18  0  0   6  0 therefore A is invertible by Theorem 2.3.3
9.
det  A    2 1 2   4  0 therefore A is invertible by Theorem 2.3.3
10.
det  A   0 (second column contains only zeros) therefore A is not invertible by Theorem 2.3.3
11.
det  A    24  24  16    24  16  24   0 therefore A is not invertible by Theorem 2.3.3
12.
det  A   1  0  81   8  36  0   124  0 therefore A is invertible by Theorem 2.3.3
22
2.3 Properties of Determinants; Cramer’s Rule
13.
det  A    2 1 6   12  0 therefore A is invertible by Theorem 2.3.3
14.
det  A   0 (third column contains only zeros) therefore A is not invertible by Theorem 2.3.3
15.
det  A    k  3  k  2    2  2   k 2  5k  2  k  5 2 17

 k 
5  17
2
 . By Theorem 2.3.3, A is invertible if
k  52 17 and k  52 17 .
16.
det  A   k 2  4   k  2  k  2  . By Theorem 2.3.3, A is invertible if k  2 and k  2 .
17.
det  A    2  12k  36    4k  18  12   8  8k  8 1  k  .
By Theorem 2.3.3, A is invertible if k  1 .
18.
det  A   1  0  0    0  2k  2k   1  4k . By Theorem 2.3.3, A is invertible if k  14 .
19.
det  A    6  0  20    10  0  15  1  0 therefore A is invertible by Theorem 2.3.3.
The cofactors of A are:
C11 =
–1 0
4 3
C21 = –
C31 =
= –3
C12 = –
=5
C22 =
=5
C32 = –
5 5
4 3
5 5
–1 0
–1 0
=3
2 3
2 5
2 3
=–4
2 5
–1 0
=–5
C13 =
–1 –1
2
4
2
5
C23 = –
C33 =
2 4
2
=–2
=2
5
=3
–1 –1
5 5
 3 3 2 
 3



The matrix of cofactors is  5 4 2  and the adjoint matrix is adj  A    3 4 5 .
 5 5 3
 2
2
3
5 5   3  5 5 
 3

From Theorem 2.3.6, we have A  det1 A adj  A   11  3 4 5   3 4
5  .
 2
2
3  2 2 3 
1
20.
det  A    24  0  0    18  0  0   6  0 therefore A is invertible by Theorem 2.3.3.
The cofactors of A are:
C11 =
3
0 –4
C21 = –
C31 =
2
0
= –12 C12 = –
3
0 –4
0 3
3 2
=0
= –9
C22 =
0
2
–2 –4
2
3
–2 –4
C32 = –
2 3
0 2
= – 4 C13 =
=–2
=–4
0 3
–2 0
C23 = –
C33 =
=6
2 0
–2 0
2 0
0 3
=6
=0
23
2.3 Properties of Determinants; Cramer’s Rule
0 9 
 12 4 6 
 12



The matrix of cofactors is  0 2 0  and the adjoint matrix is adj  A    4 2 4  .
 9 4 6 
 6
0
6 
3
 12 0 9   2 0
2
 2 1

1
1
1 
2
.
From Theorem 2.3.6, we have A  det  A adj( A)  6  4 2 4    3 3
3


 6 0 6   1 0 1
21.
det  A    2 1 2   4  0 therefore A is invertible by Theorem 2.3.3.
The cofactors of A are:
C11 =
1 3
0
C21 = 
C31 =
2
=2
3 5
0 2
3
5
1 3
C12 = 
0 3
0
2
2 5
=6
C22 =
=4
C32 = 
0 2
2
=0
=4
5
0 3
=6
C13 =
0
1
2 3
C23 = 
C33 =
=0
0 0
0
0
2 3
0
1
=0
=2
2 0 0
2 6 4 


The matrix of cofactors is  6 4 0  and the adjoint matrix is adj  A   0 4 6  .
 4 6 2 
0 0 2 
2 6 4   12 32 1 


 
From Theorem 2.3.6, we have A1  det1 A adj( A)  14 0 4 6   0 1 23  .
0 0 2  0 0 12 
22.
det  A    2 1 6   12 is nonzero, therefore by Theorem 2.3.3, A is invertible.
The cofactors of A are:
C11 
1 0
3 6
C21  
C31 
6
0 0
1 0
8 0
C22 
0
C32  
5 6
2 0
8 0
8 1
 48
C13 
 12
C23  
0
C33 
5 6
2 0
0
3 6
0 0
C12  
5 3
2 0
5 3
2 0
8
 29
1
 6
2
0 0
6 48 29 
 6



The matrix of cofactors is 0 12 6  and the adjoint matrix is adj  A    48 12 0  .
0
 29 6 2 
0 2 
0 0
 6 0 0   12




1
1 0 .
From Theorem 2.3.6, we have A  det1 A adj  A   121  48 12 0    4
29
 29 6 2   12
 12 61 
24
2.3 Properties of Determinants; Cramer’s Rule
23.
1
2
1
1
3
5
3
3
1
2
8
2
1 1 3
2 0 1

9 0 0
2 0 0
1
0
7
1
1
0
8
1
1 3
0 1

0 0
0 0
1
0
1
7
1
0
1
8
The third row and the fourth row were interchanged.
1 3
0 1

0 0
0 0
1
0
1
0
1
0
1
1
7 times the third row was added to the fourth row
2 times the first row was added to the second row; 1
times the first row was added to the third and fourth
rows.
  1 111  1
The determinant of A is nonzero therefore by Theorem 2.3.3, A is invertible.
The cofactors of A are:
5 2 2
C11  3 8 9   80  54  12    48  90  12   4
3 2 2
2 2 2
C12   1 8 9    32  18  4   16  36  4    2
1 2 2
2 5 2
C13  1 3 9  12  45  6    6  54  10   7
1 3 2
2 5 2
C14   1 3 8   12  40  6    6  48  10    6
1 3 2
3 1 1
C21   3 8 9    48  27  6    24  54  6    3
3 2 2
1 1 1
C22  1 8 9  16  9  2    8  18  2   1
1 2 2
1 3 1
C23   1 3 9    6  27  3    3  27  6    0
1 3 2
25
2.3 Properties of Determinants; Cramer’s Rule
1 3 1
C24  1 3 8   6  24  3    3  24  6   0
1 3 2
3 1 1
C31  5 2 2  12  6  10    6  12  10   0
3 2 2
1 1 1
C32   2 2 2    4  2  4    2  4  4    0
1 2 2
1 3 1
C33  2 5 2  10  6  6    5  6  12   1
1 3 2
1 3 1
C34   2 5 2   10  6  6    5  6  12    1
1 3 2
3 1 1
C41   5 2 2    54  6  40    6  48  45    1
3 8 9
1 1 1
C42  2 2 2  18  2  16    2  16  18   0
1 8 9
1 3 1
C43   2 5 2    45  6  6    5  6  54    8
1 3 9
1 3 1
C44  2 5 2   40  6  6    5  6  48   7
1 3 8
 4 2 7 6 
 4 3 0 1
 3 1 0 0 
 2 1 0 0 
.


and the adjoint matrix is adj( A)  
The matrix of cofactors is
 0 0 1 1
 7 0 1 8 




 1 0 8 7 
 6 0 1 7 
 4 3 0 1  4 3 0 1
 2 1 0 0   2 1 0 0 
1
1
1

.
From Theorem 2.3.6, we have A  det  A adj  A   1
 7 0 1 8   7 0 1 8 

 

 6 0 1 7   6 0 1 7 
24.
7 3
7 2 
3 2 
det A
det  A 
26
A
, A1  
, A2  
; x1  det A   13
 1 , x2  det  A   13
2



13
3
1
5
1
3
5






1
2
26
2.3 Properties of Determinants; Cramer’s Rule
25.
4 5 0
det  A   11 1 2   8  10  0    0  40  110   132 ,
1 5 2
2 5 0
det  A1   3 1 2   4  10  0    0  20  30   36 ,
1 5 2
4 2 0
det  A2   11 3 2   24  4  0    0  8  44   24 ,
1 1 2
4 5 2
det  A3   11 1 3   4  15  110    2  60  55   12 ;
1 5 1
det  A 
det  A 
det  A 
36
24
x  det  A1  132
 113 , y  det  A2  132
 112 , z  det  A3  12
  111 .
132
26.
1 4
1
det  A   4 1 2   3  16  8    2  4  48   55 ,
2
2 3
6 4
1
det  A1   1 1 2  18  160  2    20  24  12   144 ,
2 3
20
1
6
1
det  A2   4
1 2   3  24  80    2  40  72   61 ,
2 20 3
1 4
6
det  A3   4 1 1   20  8  48    12  2  320   230 ;
2
2 20
det  A 
det  A 
det  A 
61
46
, y  det  A2  6155   55
, z  det  A3  230
.
x  det  A1  144
  144
 11
55
55
55
27.
1 3
1
det  A   2 1 0   3  0  0    4  0  18   11 ,
4 0 3
4 3
1
4 3
  3  4  6   30 ,
det  A1   2 1 0  3
2 1
0 0 3
27
2.3 Properties of Determinants; Cramer’s Rule
1 4
1
det  A2   2 2 0   6  0  0    8  0  24   38 ,
4
0 3
1 3 4
3 4
  4  6  4   40 ;
det  A3   2 1 2  4
 1 2
4 0
0
det  A 
det  A 
det  A 
38
, x2  det  A2  3811   11
, x3  det  A3  4011   40
.
x1  det  A1  3011   30
11
11
28.
1 4 2
1
2 1 7 9
det  A  
1 1 3
1
1 2 1 4
1 7
9
2 7
9
2 1 9
2 1 7
1  4 1 3
1  2 1
1
1  1 1
1 3
 1 1 3
1 1 4
1 2 4
1 2 1
2 1 4
  12  14  9    54  1  28    4  24  7  9    27  2  28  
2  8  1  18    9  4  4     2  3  14    7  12  1 
= 90  332 +16  17 = 423
32 4 2
1
14 1 7 9
det  A1  
11 1 3
1
4 2 1 4
1 7
9
14 7
9
14 1 9
14 1 7
1  4 11 3
1  2 11
1
1  1 11
1 3
 32 1 3
 2 1 4
4 1 4
4 2 4
4 2 1
 32 12  14  9    54  1  28    4  168  28  99    108  14  308  
2  56  4  198    36  28  44    14  12  154    28  84  11 
= 2880 +1220  460 + 5 = 2115
1 32 2
1
2 14 7 9
det  A2  
1 11 3
1
1 4 1 4
14 7
9
2 7
9
2 14
9
2 14 7
1  32 1 3
1  2 1 11
1  1 1 11 3
 1 11 3
1 1 4
1 4 4
1 4 1
4 1 4
   168  28  99    108  14  308    32  24  7  9    27  2  28  
28
2.3 Properties of Determinants; Cramer’s Rule
2  88  14  36    99  8  56     22  42  28    77  24  14  
= 305  2656  370  53 = 3384
1 4 32
1
2 1 14
9
det  A3  
1 1 11 1
1 2 4 4
1 14
9
2 14
9
2 1 9
2 1 14
1  4 1 11
1  32 1
1
1  1 1
1 11
 1 1 11
1 4 4
1 2 4
1 2 4
2 4 4
   44  28  36    198  4  56    4  88  14  36    99  8  56  
32  8  1  18    9  4  4     8  11  28   14  44  4  
= 230  740  256  43 = 1269
1 4 2 32
2 1 7 14
det  A4  
1 1 3
11
1 2 1 4
1 7 14
2 7 14
2 1 14
2 1 7
1 11  32 1
1 3
 1 1 3 11  4 1 3 11  2 1
1 1 4
1 2 4
1 2 1
 2 1 4
  12  154  14    84  11  28    4  24  77  14    42  22  28  
2  8  11  28   14  44  4    32  2  3  14    7  12  1 
= 5  212 +86 + 544 = 423
det  A 
x1  det  A1  2115
 5,
423
det  A 
x3  det  A3  1269
 3,
423
det  A 
x2  det  A2  3384
8,
423
det  A 
x 4  det  A4  423
 1
423
29.
det  A   0 therefore Cramer’s rule does not apply.
30.
det  A   cos2   sin2   1 is nonzero for all values of  , therefore by Theorem 2.3.3, A is invertible.
The cofactors of A are:
C11  cos
C12  sin 
C13  0
C21   sin 
C31  0
C22  cos
C32  0
C23  0
C33  cos2   sin 2   1
The matrix of cofactors is
 cos sin  0 
  sin  cos 0 



0
0 1
29
2.3 Properties of Determinants; Cramer’s Rule
and the adjoint matrix is
 cos  sin  0 
adj  A    sin 
cos 0 

0
0 1
From Theorem 2.3.6, we have
1
A 
1
det  A 
cos  sin  0   cos  sin  0 
1
adj  A    sin 
cos 0    sin 
cos 0  .
1

0
0 1 
0
0 1
31.
4
3
det  A  
7
1
1 1
7 1
3 5
1 1
4 6
1
1
3 1 1
1
 424 ; det  A2  
7 3 5
8
1 3 1
2
32.
4
3
A
7

1
1 1
7 1
3 5
1 1
1
1
,
8

2
(a)
 6
 1
A1  
 3

 3
1 1
7 1
3 5
1 1
det  A 
1
1
4 6


1
3 1 1
, A2  
 7 3 5
8


2
1 3 1
det  A 
1
1
det A
0
 0 ; y  det A2  424
0
8
2
1
4

3
1
, A3  
7
8


2
1
1 6
7
1
3 3
1 3
det  A 
1
4

3
1
, A4  
7
8


2
1
1 1 6
7 1 1
;
3 5 3

1 1 3
det  A 
0
848
0
x  det  A1  424
 1 , y  det  A2  424
 0 , z  det  A3  424
 2 , w  det  A4  424
0
424
(b)
4
3
The augmented matrix of the system 
7

1
1
0

0

0
33.
0
1
0
0
0
0
1
0
0
0
0
1
1 1
7 1
3 5
1 1
1 6
1 1
has the reduced row echelon form
8 3

2 3
1
0 
therefore the system has only one solution: x 1 , y  0 , z  2 , and w  0 .
2

0
(c)
The method in part (b) requires fewer computations.
(a)
det  3 A  33 det  A    27  7  189 (using Formula (1))
(b)
det A 1  det1 A  17   17 (using Theorem 2.3.5)
 
30
2.3 Properties of Determinants; Cramer’s Rule



1
 
(c)
det 2 A1  23 det A 1  det8 A  87   87 (using Formula (1) and Theorem 2.3.5)
(d)
det  2 A 
(e)
a g
b h
c i

1
det  2 A 
d
a
e b
f
c
d
e
f
 23 det1  A  817   561 (using Theorem 2.3.5 and Formula (1))
g
a b
h d e
i
g h
c
f    7   7 (in the first step we interchanged the last two columns
i
applying Theorem 2.2.3(b); in the second step we transposed the matrix applying Theorem 2.2.2)
34.
35.
(a)
det   A   det   1 A    1 det  A   det  A   2 (using Formula (1))
(b)
det A 1  det1 A  12   12 (using Theorem 2.3.5)
(c)
det 2 AT  2 4 det AT  16det  A   32 (using Formula (1) and Theorem 2.2.2)
(d)
det A3  det  AAA   det  A  det  A  det  A    2   8 (using Theorem 2.3.4)
(a)
det  3 A   33 det  A    27  7   189 (using Formula (1))
(b)
det A1  det1 A  17 (using Theorem 2.3.5)
(c)
det 2 A1  23 det A1  det8 A  87 (using Formula (1) and Theorem 2.3.5)
(d)
det  2 A 
4
 


 
 
3
 



1
 

1
det  2 A 
 23 det1  A  81 7  561 (using Theorem 2.3.5 and Formula (1))
True-False Exercises
(a)
False. By Formula (1), det  2 A   23 det  A   8det  A  .
(b)
1 0 
0 0 
False. E.g. A  
and B  

 have det  A   det  B   0 but det  A  B   1  2 det  A  .
0 0 
0 1 
(c)
True. By Theorems 2.3.4 and 2.3.5,


 
det A 1 BA  det A1 det  B  det  A   det1 A det  B  det  A   det  B  .
(d)
False. A square matrix A is invertible if and only if det  A   0 .
(e)
True. This follows from Definition 1.
(f)
True. This is Formula (8).
(g)
True. If det  A   0 then by Theorem 2.3.8 Ax  0 must have only the trivial solution, which contradicts our
assumption. Consequently, det  A   0 .
31
2.3 Properties of Determinants; Cramer’s Rule
(h)
True. If the reduced row echelon form of A is I n then by Theorem 2.3.8 Ax  b is consistent for every b , which
contradicts our assumption. Consequently, the reduced row echelon form of A cannot be I n .
(i)
True. Since the reduced row echelon form of E is I then by Theorem 2.3.8 Ex  0 must have only the
trivial solution.
(j)
True. If A is invertible, so is A 1 . By Theorem 2.3.8, each system has only the trivial solution.
(k)
True. From Theorem 2.3.6, A1  det1 A adj  A  therefore adj  A   det  A  A1 . Consequently,

(l)
1
det  A 


A adj  A   det1 A A
  det  A A  
1
det  A 
det  A 
 AA   I so  adj A     A .
1
1
n
1
det A
False. If the k th row of A contains only zeros then all cofactors C jk where j  i are zero (since each of them
involves a determinant of a matrix with a zero row). This means the matrix of cofactors contains at least one zero
row, therefore adj  A  has a column of zeros.
Chapter 2 Supplementary Exercises
1.
(a)
(b)
Cofactor expansion along the first row:
4 2
3 3

3 3
4 2
   3
1 1
4 2
   3
1 1
0 6
   31 6   18
2.
(a)
(b)
Cofactor expansion along the first row:
7 1
2 6

2 6
7 1
   2 
   2 
1
1
3
7 1
3
0 22
4 2
  4  3    2  3  12  6  18
3 3
The first and second rows were interchanged.
A common factor of 3 from the first row
was taken through the determinant sign.
4 times the first row was added to the second row
Use Theorem 2.1.2.
7 1
2 6
  7  6    1 2   42  2  44
The first and second rows were interchanged.
A common factor of 2 from the first row
was taken through the determinant sign.
7 times the first row was added to the second row
32
Supplementary Exercises
   2 1 22 
Use Theorem 2.1.2.
 44
3.
(a)
Cofactor expansion along the second row:
1 5 2
1 2
1 5
0 2 1  0  2
  1
3 1
3 1
3 1 1
 0  2  11   2  3     1  11   5  3  
 0   2  5   114   0  10  14  24
(b)
1 5 2
1 5 2
0 2 1   1 0 2 1
3 1 1
3
1 1
1 5 2
  1 0
2 1
0 14 5
1 5 2
  1 0 2
1
0 0 12
  11 2  12   24
4.
(a)
A common factor of 1 from the first row
was taken through the determinant sign.
3 times the first row was added to the third
row.
7 times the second row was added to the third.
Use Theorem 2.1.2.
Cofactor expansion along the first row:
1 2 3
5 6
4 6
4 5
4 5 6   1
  2 
  3 
8 9
7 9
7 8
7 8 9
  1  5  9    6  8     2   4  9    6  7  
  3   4  8    5  7  
  1 3   2  6    3 3  3  12  9  0
(b)
1 2 3
1 2
3
4 5 6   1 4 5 6
7 8 9
7 8 9
1 2 3
  1 0 3 6
0 6 12
A common factor of 1 from the first row
was taken through the determinant sign.
4 times the first row was added to the
second row and 7 times the first row was
added to the third row
33
Supplementary Exercises
1 2 3
  1 0 3 6
0 0 0
  1 0   0
5.
(a)
2 times the second row was added to the
third row
Use Theorem 2.2.1.
Cofactor expansion along the first row:
3 0 1
1 1
1 1
1 1 1  3
 0   1
4 2
0 4
0 4 2
  3  1 2   1 4    0   1 1 4   1 0  
  3 2   0   1 4   6  0  4  10
(b)
3 0 1
1 1 1
1 1 1   1 3 0 1
0 4 2
0 4 2
1 1
1
  1 0 3 4
0 4
2
3 times the first row was added to the second.
1 1
1
  1 0 3 4
0
1 2
The second row was added to the third row
1 1
1
  1 1 0
1 2
0 3 4
The second and third rows were interchanged.
1 1
1
  1 1 0 1 2
0 0 10
3 times the second row was added to the third.
  1 111 10   10
6.
(a)
The first and second rows were interchanged.
Use Theorem 2.1.2.
Cofactor expansion along the second row:
5
1 4
1 4
5
1
3 0 2  3
02
2 2
1 2
1 2 2
 3 1 2    4  2    2  5  2   11    3 10   2  9   30  18  48
34
Supplementary Exercises
5
1 4
1 2 2
3 0 2  3 0 2
1 2 2
5
1 4
(b)
The first and third rows were interchanged.
1 2
2
0
6 4
0 9 14
3 times the first row was added to the second
row and 5 times the first row was added to the
third row
1 2
2
 6 0
1  23
0 9 14
A common factor of 6 from the second row
was taken through the determinant sign.
1 2
2
 6 0
1  23
0
0
8
9 times the second row was added to the third
row.
 6 11 8   48
7.
(a)
Use Theorem 2.1.2.
We perform cofactor expansions along the first row in the 4x4 determinant. In each of the 3x3 determinants,
we expand along the second row:
3
2
1
9
6 0
3
1
0 1
2 2
1
3
1 4
1 4
1
2
2 3
4
 3 0 1 1  6 1 1 1  0  1 1 0 1
1
2 2 2
9 2 2
9 2 2
2

3 4
3
1 
1 4
2 4
2
1
 3  0   1
1
  1
1
  6  1

2 2
2 2   2 2
9 2
9 2 


 0 1 1

3
1
2 3 
 0   1

2 2
9 2 
 3  0  1 2   1 8    6  110   1 32   113    0  1 1 8   0  1 23  
 3 10   6  55  0  1 31
 329
(b)
3
2
1
9
6 0
3
1
0 1
2 2
1
1
4
2
  1
1
3
2
9
0 1
3
1
6 0
2 2
1
4
1
2
The first and third rows were
interchanged.
35
Supplementary Exercises
1
0
  1
0
0
0 1 1
3 1 6
6
3 2
2 11 11
1
0
   1
0
0
0 1
1
3 1
6
0
5 14
31
0 3
7
  1
1 0 1
0 3 1
0 0
0 0
2 times the first row was added to
the second, 3 times the first row
was added to the third and 9 times
the first row was added to the
fourth.
2 times the second row was
added to the third and  23 times
the second row was added to the
fourth.
1
6
31
5 14
0  329
15
15
  11 3  5    329
 329
15 
8.
(a)
times the third row was added
to the fourth.
Use Theorem 2.1.2.
We perform cofactor expansions along the first row in the 4x4 determinant, as well as in each of the
3x3 determinants:
1 2 3 4
4
3 2
1
1 2
3 4
4 3 2 1
3
 1 2
2
3
1
4
4   2  1
3 2 1
  4 
2
3
1
4
4   3  1
4 2 1
4
1
3
2
3
2
1
4
4 3 1
2
3
4 3 2
 3 4
2 4
2
3
 1 3
2
1

3 1
3 2 
 2 1

3 4
1 4
1 3
 2 4
2
1

4 1
4 2 
 2 1
 2 4
1 4
1 2
3  4
3
1

4 1
4 3 
 3 1
 2
3
1 3
1 2
 4 4
3
2

4 2
4 3 
 3 2
36
Supplem
mentary Exerrcises
0   5    2  4  5   2   4  5    2 15   10 
    3  5    2 10
3   4 
 10    3 155  5  4   4  5   3100    2  5 
0000
0
1 2 3 4
4
1
(b)
3
2
4
1

0
4
2
3
4 3 2
1 2 3 4
1
3
0
2
0
1
0
4 3 2
1
0
1 5
9.
1 5
2
 1 2  3
 4 5 6 4 5   45
4  84  96    105  48  72   0
 7  8 9  7  8
3 0 1
3 0 1
1 1
0 4
1
2
1 1
0 4
5
1 4
5
e.g.
3
5
6
0
3 0
1 1  6  0  4    0  122  0   10
0 4
1
2
4 5
1
 3 0
1 2
3 0 2
1 2 2
(a)
2 1 5
 1  2 3  1  2
4  5 6
7 8 9
10.
Theorem 2.2.1.
Use T
 0 2 1 0 2   2  15
1  0    122  1  0   24
3 1 1 3 1
0 2 1
3 1 1
4 0
8 0
The ffirst row was aadded to the
third row.
2
2
4
1
3 0   0  2  24    0  20  6   48
1 2
3 6
8 5
 3 8 5 0  18
  188 15   2700 was easy too calculate byy cofactor
13 10
7 3 7 10
0
13 10
0 0
13 0 10 0
expansions (first, we
w expanded along
a
the seco
ond column, tthen along thee third colum
mn), but wouldd be more
difficu
ult to calculate using elemeentary row op
perations.
37
Supplementary Exercises
(b)
1 2 3 4
4
3 2
1
e.g.,
of Exercise 8 was easy to calculate using elementary row operations, but more
1 2
3 4
4 3 2 1
difficult using cofactor expansion.
11.
In Exercise 1:
4 2
 18  0 therefore the matrix is invertible.
3 3
In Exercise 2:
7 1
 44  0 therefore the matrix is invertible.
2 6
1 5 2
In Exercise 3: 0 2 1  24  0 therefore the matrix is invertible.
3 1 1
1 2 3
In Exercise 4: 4 5 6  0 therefore the matrix is not invertible.
7 8 9
12.
3 0 1
In Exercise 5: 1 1 1  10  0 therefore the matrix is invertible.
0 4 2
5
1 4
In Exercise 6: 3 0 2  48  0 therefore the matrix is invertible.
1 2 2
3
2
In Exercise 7:
1
9
6 0
3
1
0 1
2 2
1
4
 329  0 therefore the matrix is invertible.
1
2
1 2 3 4
4
3 2
1
 0 therefore the matrix is not invertible.
In Exercise 8:
1 2
3 4
4 3 2 1
13.
14.
5
b3
  5 3   b  3 b  2   15  b2  2b  3b  6  b2  5b  21
b  2 3
3
4
2
1
2 
a
1
2
3
2
a  1 4  a  a  2 0 2 a  6
a
2
a
4a2  3
2
0
8a
4 times the second row was added
to the first row and 1  a times the
second row was added to the last
row.
38
Supplementary Exercises
4a 2  3
8a
 0  1 3
0
2
a  a  2 2 a  6



 4a 2  3  2a  6    8  a  a3  a 2  2
Cofactor expansion along
the second column.

 a 4  a 3  16a 2  8a  2
0
0
0
0
15.
0 0
0 3
0 0 4 0
0 1 0 0
2 0
0 0
5 0
5
0
  1 0
0
0
0
0 0 0
0 0 4
0 1 0
2 0 0
0
0
0
0
0 0
5
0
  1 1 0
0
0
0
0 3
0 0 0
2 0 0
0 1 0
0 0 4
0 0
0
The first row and the fifth row were
interchanged.
0
0
0
0
The second row and the fourth row
were interchanged.
0 3
  1 1 5 2  1 4  3  120
16.
x
1
3 1 x
 x 1  x    1 3   x 2  x  3 ;
Adding 2 times the first row to the second row, then performing cofactor expansion along the second row yields
1 0 3
1 0 3
1 3
2 x 6  0 x
0 x
 x  x  5  3  x 2  2 x
1 x5
1 3 x5 1 3 x5
Solve the equation
 x2  x  3  x2  2 x
2 x 2  3 x  3  0
From quadratic formula x  3 4924  3 4 33 or x  3 49 24  3 4 33 .
39
Supplementary Exercises
17.
It was shown in the solution of Exercise 1 that
4 2
 18 . The determinant is nonzero, therefore by
3 3
 4 2 
Theorem 2.3.3, the matrix A  
 is invertible.
 3 3
The cofactors are:
C11  3
C12  3
C21  2
C22  4
 3 2 
 3 3
The matrix of cofactors is 
and the adjoint matrix is adj  A   

.
2 4 
 3 4 
 3 2    61
From Theorem 2.3.6, we have A1  det1 A adj  A   118 
 1
 3 4   6
18.
It was shown in the solution of Exercise 2 that
1
9
2
9

.

7 1
 44 . The determinant is nonzero, therefore by
2 6
 7 1
Theorem 2.3.3, the matrix A  
 is invertible.
 2 6 
The cofactors are:
C11  6
C12  2
C21  1
C22  7
 6 1
 6 2 
The matrix of cofactors is 
and the adjoint matrix is adj  A   
.

 2 7
 1 7
 6 1  223
From Theorem 2.3.6, we have A1  det1 A adj  A   144 
 1
 2 7    22
19.
 441 
.
 447 
1 5 2
It was shown in the solution of Exercise 3 that 0 2 1  24 . The determinant is nonzero, therefore by
3 1 1
 1 5 2 
Theorem 2.3.3, A   0 2 1 is invertible.
 3 1 1
The cofactors of A are:
40
Supplementary Exercises
C11 =
2 –1
1
C21 = –
C31 =
1
5 2
1 1
5
2
2 –1
=3
C12 = –
= –3 C22 =
= –9
0 –1
–3
1
–1 2
–3 1
C32 = –
=3
=5
–1
2
C13 =
–3 1
C23 = –
= –1 C33 =
0 –1
0 2
=6
–1 5
= –14
–3 1
–1 5
0 2
= –2
6
 3 3
 3 3 9 


The matrix of cofactors is  3 5 14  and the adjoint matrix is adj  A    3
5 1 .
6 14 2 
 9 1 2 
 3 3 9   81

From Theorem 2.3.6, we have A1  det1 A adj  A   241  3
5 1   81
6 14 2   14
20.
 81

 83 

 241  .
 121 
5
24
7
12
1 2 3
It was shown in the solution of Exercise 4 that 4 5 6  0 therefore by Theorem 2.3.3, the matrix is
7 8 9
not invertible.
21.
3 0 1
It was shown in the solution of Exercise 5 that 1 1 1  10 . The determinant is nonzero, therefore by
0 4 2
 3 0 1
Theorem 2.3.3, A   1 1 1  is invertible.
 0 4 2 
The cofactors of A are:
C11 
1 1
4 2
C21  
C31 
 2
0 1
4
1
1
1 1
0 2
3 1
 4
C22 
1
C32  
2
0 1
C12  
0
2
C13 
6
C23  
3 1
1
1
1 1
 2
 4
C33 
0 4
4
3 0
0 4
3 0
1 1
 12
3
4
1
 2 2
 2 4



The matrix of cofactors is  4
6 12  and the adjoint matrix is adj  A    2
6 4  .
 1 4
 4 12
3 
3
1  15
 2 4

From Theorem 2.3.6, we have A1  det1 A adj  A   110  2
6 4    15
 4 12
3   25

2
5
3
5
6
5
 101 
2
.
5
3 
 10 
41
Supplementary Exercises
22.
42
5
1 4
It was shown in the solution of Exercise 6 that 3 0 2  48 . The determinant is nonzero, therefore by
1 2 2
1 4
 5

Theorem 2.3.3, A   3 0 2  is invertible.
 1 2 2 
The cofactors of A are:
C11 =
0 2
–2 2
C21 = –
C31 =
=4
1 4
–2 2
1 4
0 2
=2
C12 = –
= –10 C22 =
3 2
1 2
–5 4
1 2
C32 = –
3
= –4
C13 =
= –14
C23 = –
–5 4
3 2
= 22 C33 =
0
1 –2
= –6
–5
1
1 –2
–5 1
3 0
= –9
= –3
 4 4 6 
 4 10 2 


The matrix of cofactors is  10 14 9  and the adjoint matrix is adj  A    4 14 22  .
 2 22 3
 6 9 3
 4 10 2    121

 
From Theorem 2.3.6, we have A1  det1 A adj  A   148  4 14 22    121
 6 9 3  81
23.
3
2
It was shown in the solution of Exercise 7 that
1
9
 3
 2
Theorem 2.3.3, A  
 1

 9
6
0
3
1
0 1
2 2
6
0
3
1
0 1
2 2
1
4 
is invertible.
1

2
The cofactors of A are:
3
1 4
C11  0 1 1   6  2  0    8  6  0   10
2 2 2
2
1 4
C12   1 1 1    4  9  8    36  4  2    55
9 2 2
2 3 4
C13  1 0 1   0  27  8    0  4  6   21
9 2 2
5
24
7
24
3
16
 241 

.
 11
24 
1 
16 
1
4
 329 . The determinant of A is nonzero therefore by
1
2
Supplementary Exercises
2 3
1
C14   1 0 1    0  27  2    0  4  6    31
9 2 2
6
0 1
C21   0 1 1    12  0  0    2  12  0    2
2 2 2
3 0 1
C22  1 1 1   6  0  2    9  6  0   11
9 2 2
3 6 1
C23   1 0 1    0  54  2    0  6  12    70
9 2 2
3 6
0
C24  1 0 1   0  54  0    0  6  12   72
9 2 2
6
0 1
C31  3
1 4  12  0  6    2  48  0   52
2 2 2
3 0 1
C32   2
1 4    6  0  4    9  24  0    43
9 2 2
3 6 1
C33  2 3 4  18  216  4    27  24  24   175
9 2 2
3 6
0
C34   2 3
1    18  54  0    0  6  24    102
9 2 2
6 0 1
C41   3 1 4    6  0  3    0  24  0    27
0 1 1
3 0 1
C42  2
1 4   3  0  2   1  12  0   16
1 1 1
3 6 1
C43   2 3 4    9  24  0    3  0  12    42
1 0 1
43
Supplementary Exercises
44
3 6 0
C44  2 3 1   9  6  0    0  0  12   15
1 0 1
52 27 
 10 55 21 31
 10 2
 2 11


55 11 43 16 
70 72 
.
The matrix of cofactors is 
and adj  A   
 52 43 175 102 
 21 70 175 42 




 31 72 102 15
 27 16 42 15
2
 329
10
52 27   329
 10 2
 55 11 43 16   55
1 
   329
From Theorem 2.3.6, we have A1  det1 A adj  A   329
 21 70 175 42    473

  31
 31 72 102 15   329
24.
11
 329
10
47
72
329
52
329
43
329
25
47
102
329


27
 329

16 
329 
.
 476 

15
 329

1 2 3 4
4
3 2
1
 0 therefore by Theorem 2.3.3, the matrix
It was shown in the solution of Exercise 8 that
1 2
3 4
4 3 2 1
is not invertible.
25.
3
A   45
5
 45 
 x  45 
 35
3
3
9
16
4
4
A


A
,
det
A






1
;
,


 5  5   5  5  25 25
4

1
2
3
3
5
5
y
5
det  A 
x
;
y
det  A 
x   det  A1  35 x  45 y , y'  det  A2  35 y  45 x
26.
cos
A
 sin 
 sin  
cos
 x  sin  
, A1  
, A2  


cos 
 y cos 
 sin 
det  A 
x
;
y 
det  A 
y sin 
x sin 
x   det  A1  xcoscos2   sin
 x cos   y sin  , y  det  A2  ycoscos2   sin
 y cos   x sin 
2
2


27.
 1 1 
The coefficient matrix of the given system is A   1 1   . Coefficient expansion along the first row yields
  1
det  A   1
1 
1 
1 1
1

 1  1
 
 1   2  1             2  2   2      
2
By Theorem 2.3.8, the given system has a nontrivial solution if and only if det  A   0 , i.e.,    .
28.
According to the arrow technique (see Example 7 in Section 2.1), the determinant of a 3  3 matrix can be expressed
as a sum of six terms:
Supplementary Exercises
a11
a21
a31
a12
a22
a32
45
a13
a23  a11a22 a33  a12 a23 a31  a13 a21a32  a13 a22 a31  a11a23 a32  a12 a21a33
a33
If each entry of A is either 0 or 1 , then each of the terms must be either 0 or 1 . The largest value 3 would result
from the terms 1  1  1  0  0  0 , however, this is not possible since the first three terms all equal 1 would require
that all nine matrix entries be equal 1, making the determinant 0 .
0 1 1 
The largest value of the determinant that is actually attainable is 2 , e.g., let A   1 0 1  .
 1 1 0 
29.
(a)
We will justify the third equality, a cos   b cos   c by considering three cases:
CASE I:   2 and   2
Referring to the figure on the right side, we have
x  b cos and y  a cos  .
Since x  y  c we obtain, a cos   b cos   c .
CASE II:   2 and   2
Referring to the picture on the right side, we can write
x  b cos      b cos and y  a cos 
This time we can write c  y  x  a cos    b cos 
therefore once again a cos   b cos   c .
CASE III:   2 and   2 (similarly to case II, c  b cos  a cos      b cos  a cos  )
The first two equations can be justified in the same manner.
Denoting X  cos , Y  cos  , and Z  cos  we can rewrite the linear system as
cY
cX
bX
 bZ
 aZ
 aY
 a
 b
 c
0 c b
We have det  A   c 0 a   0  abc  abc    0  0  0   2 abc and
b a 0
a c b
det  A1   b 0 a   0  ac 2  ab 2   0  a 3  0   a b 2  c 2  a 2 therefore by Cramer's rule
c a 0

cos   X 
det  A1 
det  A 


a b2  c2  a2
2 abc
  b c a .
2
2
2bc
2

Supplementary Exercises
(b)
Using the results obtained in part (a) along with
0 a b
det  A2   c b a   0  a 2 b  bc 2   b3  0  0   b  a 2  c 2  b 2  and
b c 0
0 c a
det  A3   c 0 b   0  b 2 c  a 2 c   0  0  c 3   c a 2  b 2  c 2 therefore by Cramer's rule
b a c

cos   Y 
31.
det( A3 ) a 2  b 2  c 2
det( A2 ) a 2  c 2  b 2
and cos   Z 
.


det( A)
2 ac
det( A)
2 ab
From Theorem 2.3.6, A1  det1 A adj  A  therefore adj  A   det  A  A1 . Consequently,

1
det  A 


A adj  A   det1 A A
  det  A A  
1
 
det  A 
det  A 
 AA   I so  adj A     A .
1
1
n
1
det A
  A     A.
Using Theorem 2.3.5, we can also write adj A1  det A1
33.

1
1
1
det A
1  0 
1




The equality A       means that the homogeneous system Ax  0 has a nontrivial solution x    .
1  0 
1
Consequently, it follows from Theorem 2.3.8 that det  A   0 .
34.
(b)
1
2
3 3 1
4 0 1   192 is the negative of the area of the triangle because it is being traced clockwise; (reversing
2 1 1
the order of the points would change the orientation to counterclockwise, and thereby result in the positive area:
 2 1 1
1
4 0 1  192 .
2
3 3 1
37.
1 x1

In the special case that n  3, the augmented matrix for the system (13) of Section 1.10 is 1 x2
1 x3

1 x1

We apply Cramer’s Rule to the coefficient matrix A  1 x2
1 x3

 y1

A1   y2
 y3

x1
x2
x3
1 y1
x12 

2
x2  , A2  1 y2
1 y3
x32 

x12 
1 x1
2
x2  , and A3  1 x2
1 x3
x32 
x12
x
x
2
2
2
3
y1 

y2  .
y3 
x12 

x22  .
x32 
y1 
y2  so the coefficients of the desired interpolating
y3 
polynomial y  a0  a1 x  a2 x 2 are: a0  det A1 , a1  det A2 , and a2  det A3 . From the result of Exercise 43 of
det A
Section 2.1, det  A   x2  x1  x3  x1  x3  x2  .
det A
det A
46
Supplementary Exercises
  y1

det   y2
 y
 3
x12  

x22    y3 x1 x2  x2  x1   y2 x1 x3  x3  x1   y1 x2 x3  x3  x2  ,
x32  
x1
x2
x3
 1 y1

det  1 y2
 1 y
3

x12  

x22     y3 x22  x12  y2 x32  x12  y1 x32  x22 ,
x32  

 1 x1

and det  1 x2
 1 x
3






y1  

y2    y3  x2  x1   y2  x3  x1   y1  x3  x2  .
y3  
Therefore,
a0 
a1 
y3 x1 x 2 ( x 2 – x1 ) – y2 x1 x3 ( x3 – x1 )  y1 x 2 x3 ( x3 – x2 )
( x 2 – x1 )( x3 – x1 )( x3 – x 2 )




 x – x  x – x  x – x 
2
and a2 
1
3
1
3

2
 x  x  x  x  x  x 
1
3
1
No. For instance, T 1,0,0,1 
3
2
y3 x1 x2
( x3 – x1 )( x3 – x 2 )
y2  x3  x1 
–
y2 x1 x3
( x 2 – x1 )( x3 – x2 )
y3  x2  x1 
–

( x3 – x1 )( x2 – x1 )
y1  x3  x2 
–

2
2
1
3
y3
1
3
2
3
y2

1
2
1
y1

 x  x  x  x   x  x  x  x   x  x  x  x 
3
1
3
2
2
1
3
2
1 0
 1 so that T 1,0,0,1  T 1,0,0,1  2
0 1
but T  1,0,0,1  1,0,0,1   T  2,0,0,2  
y1 x2 x3
 x – x  x – x   x – x  x – x   x – x  x – x 
3
y3  x2  x1   y2  x3  x1   y1  x3  x2 
2
38.

– y3 x22 – x12  y2 x32 – x12 – y1 x32 – x22

2 0
0 2
 4 which shows that additivity fails.
3
1
2
1
.
,
47