238 Higher Engineering Mathematics
the products of the elements and their cofactors is
2 × 11 +7 × 13, i.e.,
3
4 −1
2
0
7
1 −3 −2
= 2(11) + 0 + 7(13) = 113
The same result will be obtained whichever row or
column is selected. For example, the third column
expansion is
(−1)
2
0
1 −3
− 7
3
4
1 −3
+ (−2)
3 4
2 0
= 6 + 91 + 16 = 113, as obtained previously.
Problem 15. Evaluate
1
4 −3
−5
2
6
−1 −4
2
Using the first row:
1
4 −3
−5
2
6
−1 −4
2
= 1
2 6
−4 2
− 4
−5 6
−1 2
+ (−3)
−5
2
−1 −4
= (4 + 24) − 4(−10 + 6) − 3(20 + 2)
= 28 + 16 − 66 = −22
Using the second column:
1
4 −3
−5
2
6
−1 −4
2
= −4
−5 6
−1 2
+ 2
1 −3
−1
2
−(−4)
1 −3
−5
6
= −4(−10 + 6) + 2(2 − 3) + 4(6 − 15)
= 16 − 2 − 36 = −22
Problem 16. Determine the value of
j 2
(1 + j ) 3
(1 − j )
1
j
0
j 4
5
Using the first column, the value of the determinant is:
( j 2)
1 j
j 4 5
− (1 − j )
(1 + j ) 3
j 4
5
+ (0)
(1 + j ) 3
1
j
= j 2(5 − j
2 4) − (1 − j )(5 + j 5 − j 12) + 0
= j 2(9) − (1 − j )(5 − j 7)
= j 18 − [5 − j 7 − j 5 + j
2
7]
= j 18 − [−2 − j 12]
= j 18 + 2 + j 12 = 2 + j 30 or 30.07∠86.19
◦
Now try the following exercise
Exercise 96 Further problems on 3 by 3
determinants
1. Find the matrix of minors of
⎛
⎝
4 −7
6
−2
4
0
5
7 −4
⎞
⎠
⎡
⎣
⎛
⎝
−16
8 −34
−14 −46
63
−24
12
2
⎞
⎠
⎤
⎦
2. Find the matrix of cofactors of
⎛
⎝
4 −7
6
−2
4
0
5
7 −4
⎞
⎠
⎡
⎣
⎛
⎝
−16 −8 −34
14 −46 −63
−24 −12
2
⎞
⎠
⎤
⎦
3. Calculate the determinant of
⎛
⎝
4 −7
6
−2
4
0
5
7 −4
⎞
⎠
[−212]
4. Evaluate
8 −2 −10
2 −3 −2
6
3
8
[−328]
5. Calculate the determinant of
⎛
⎝
3.1 2.4
6.4
−1.6 3.8 −1.9
5.3 3.4 −4.8
⎞
⎠
[−242.83]
6. Evaluate
j 2
2
j
(1 + j ) 1 −3
5
− j 4 0
[−2 − j ]
the products of the elements and their cofactors is
2 × 11 +7 × 13, i.e.,
3
4 −1
2
0
7
1 −3 −2
= 2(11) + 0 + 7(13) = 113
The same result will be obtained whichever row or
column is selected. For example, the third column
expansion is
(−1)
2
0
1 −3
− 7
3
4
1 −3
+ (−2)
3 4
2 0
= 6 + 91 + 16 = 113, as obtained previously.
Problem 15. Evaluate
1
4 −3
−5
2
6
−1 −4
2
Using the first row:
1
4 −3
−5
2
6
−1 −4
2
= 1
2 6
−4 2
− 4
−5 6
−1 2
+ (−3)
−5
2
−1 −4
= (4 + 24) − 4(−10 + 6) − 3(20 + 2)
= 28 + 16 − 66 = −22
Using the second column:
1
4 −3
−5
2
6
−1 −4
2
= −4
−5 6
−1 2
+ 2
1 −3
−1
2
−(−4)
1 −3
−5
6
= −4(−10 + 6) + 2(2 − 3) + 4(6 − 15)
= 16 − 2 − 36 = −22
Problem 16. Determine the value of
j 2
(1 + j ) 3
(1 − j )
1
j
0
j 4
5
Using the first column, the value of the determinant is:
( j 2)
1 j
j 4 5
− (1 − j )
(1 + j ) 3
j 4
5
+ (0)
(1 + j ) 3
1
j
= j 2(5 − j
2 4) − (1 − j )(5 + j 5 − j 12) + 0
= j 2(9) − (1 − j )(5 − j 7)
= j 18 − [5 − j 7 − j 5 + j
2
7]
= j 18 − [−2 − j 12]
= j 18 + 2 + j 12 = 2 + j 30 or 30.07∠86.19
◦
Now try the following exercise
Exercise 96 Further problems on 3 by 3
determinants
1. Find the matrix of minors of
⎛
⎝
4 −7
6
−2
4
0
5
7 −4
⎞
⎠
⎡
⎣
⎛
⎝
−16
8 −34
−14 −46
63
−24
12
2
⎞
⎠
⎤
⎦
2. Find the matrix of cofactors of
⎛
⎝
4 −7
6
−2
4
0
5
7 −4
⎞
⎠
⎡
⎣
⎛
⎝
−16 −8 −34
14 −46 −63
−24 −12
2
⎞
⎠
⎤
⎦
3. Calculate the determinant of
⎛
⎝
4 −7
6
−2
4
0
5
7 −4
⎞
⎠
[−212]
4. Evaluate
8 −2 −10
2 −3 −2
6
3
8
[−328]
5. Calculate the determinant of
⎛
⎝
3.1 2.4
6.4
−1.6 3.8 −1.9
5.3 3.4 −4.8
⎞
⎠
[−242.83]
6. Evaluate
j 2
2
j
(1 + j ) 1 −3
5
− j 4 0
[−2 − j ]
