238 Higher Engineering Mathematics
the products of the elements and their cofactors is
2 × 11 + 7 ×13, i.e.,
∣ ∣ ∣ ∣ ∣ ∣
34 − 1
207
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 )
∣
∣
∣
∣
20
1 − 3
∣
∣
∣
∣−^7
∣
∣
∣
∣
34
1 − 3
∣
∣
∣
∣+(−^2 )
∣
∣
∣
∣
34
20
∣
∣
∣
∣
= 6 + 91 + 16 = 113 ,as obtained previously.
Problem 15. Evaluate
∣ ∣ ∣ ∣ ∣ ∣
14 − 3
− 526
− 1 − 42
∣ ∣ ∣ ∣ ∣ ∣
Using the first row:
∣∣
∣
∣
∣
∣
14 − 3
− 526
− 1 − 42
∣∣
∣
∣
∣
∣
= 1
∣
∣
∣
∣
26
− 42
∣
∣
∣
∣−^4
∣
∣
∣
∣
− 56
− 12
∣
∣
∣
∣+(−^3 )
∣
∣
∣
∣
− 52
− 1 − 4
∣
∣
∣
∣
=( 4 + 24 )− 4 (− 10 + 6 )− 3 ( 20 + 2 )
= 28 + 16 − 66 =− 22
Using the second column:
∣ ∣ ∣ ∣ ∣ ∣
14 − 3
− 526
− 1 − 42
∣ ∣ ∣ ∣ ∣ ∣
=− 4
∣
∣
∣
∣
− 56
− 12
∣
∣
∣
∣+^2
∣
∣
∣
∣
1 − 3
− 12
∣
∣
∣
∣−(−^4 )
∣
∣
∣
∣
1 − 3
− 56
∣
∣
∣
∣
=− 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 45
∣ ∣ ∣ ∣ ∣ ∣ ∣
Using the first column, the value of the determinant is:
(j 2 )
∣
∣
∣
∣
∣
1 j
j 45
∣
∣
∣
∣
∣
−( 1 −j)
∣
∣
∣
∣
∣
( 1 +j) 3
j 45
∣
∣
∣
∣
∣
+( 0 )
∣
∣
∣
∣
∣
( 1 +j) 3
1 j
∣
∣
∣
∣
∣
=j 2 ( 5 −j^24 )−( 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 −j12]
=j 18 + 2 +j 12 = 2 +j 30 or30.07∠86.19◦
Now try the following exercise
Exercise 96 Further problems on 3 by 3
determinants
- Find the matrix of minors of
⎛
⎝
4 − 76
− 240
57 − 4
⎞
⎠
⎡
⎣
⎛
⎝
− 16 8 − 34
− 14 −46 63
−24 12 2
⎞
⎠
⎤
⎦
- Find the matrix of cofactors of
⎛
⎝
4 − 76
− 240
57 − 4
⎞
⎠
⎡
⎣
⎛
⎝
− 16 − 8 − 34
14 − 46 − 63
− 24 − 12 2
⎞
⎠
⎤
⎦
- Calculate the determinant of
⎛
⎝
4 − 76
− 240
57 − 4
⎞
⎠ [−212]
- Evaluate
∣ ∣ ∣ ∣ ∣ ∣
8 − 2 − 10
2 − 3 − 2
63 8
∣ ∣ ∣ ∣ ∣ ∣
[−328]
- Calculate the determinant of
⎛
⎝
3. 12. 46. 4
− 1. 63. 8 − 1. 9
5. 33. 4 − 4. 8
⎞
⎠ [−242.83]
- Evaluate
∣ ∣ ∣ ∣ ∣ ∣
j 22 j
( 1 +j) 1 − 3
5 −j 40
∣ ∣ ∣ ∣ ∣ ∣
[− 2 −j]