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 )= 113The 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 =− 22Using 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 =− 22Problem 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 exerciseExercise 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]