246 Higher Engineering Mathematics
DI 1 =
∣
∣
∣∣
∣
∣
3 − 4 − 26
− 5 − 387
26 − 12
∣
∣
∣∣
∣
∣
=( 3 )
∣
∣
∣
∣
− 387
6 − 12
∣
∣
∣
∣−(−^4 )
∣
∣
∣
∣
− 587
2 − 12
∣
∣
∣
∣
+(− 26 )
∣
∣
∣
∣
− 5 − 3
26
∣∣
∣
∣
= 3 (− 486 )+ 4 (− 114 )− 26 (− 24 )
=− 1290
DI 2 =
∣ ∣ ∣ ∣ ∣ ∣
2 − 4 − 26
1 − 387
− 76 − 12
∣ ∣ ∣ ∣ ∣ ∣
=( 2 )( 36 − 522 )−(− 4 )(− 12 + 609 )
+(− 26 )( 6 − 21 )
=− 972 + 2388 + 390
= 1806
DI 3 =
∣ ∣ ∣ ∣ ∣ ∣
23 − 26
1 − 587
− 72 − 12
∣ ∣ ∣ ∣ ∣ ∣
=( 2 )( 60 − 174 )−( 3 )(− 12 + 609 )
+(− 26 )( 2 − 35 )
=− 228 − 1791 + 858 =− 1161
and D=
∣ ∣ ∣ ∣ ∣ ∣
23 − 4
1 − 5 − 3
− 726
∣ ∣ ∣ ∣ ∣ ∣
=( 2 )(− 30 + 6 )−( 3 )( 6 − 21 )
+(− 4 )( 2 − 35 )
=− 48 + 45 + 132 = 129
Thus
I 1
− 1290
=
−I 2
1806
=
I 3
− 1161
=
− 1
129
giving
I 1 =
− 1290
− 129
=10mA,
I 2 =
1806
129
=14mA
and I 3 =
1161
129
=9mA
Now try the following exercise
Exercise 99 Further problemson solving
simultaneous equations using determinants
In Problems 1 to 5 usedeterminantsto solve the
simultaneous equations given.
- 3x− 5 y=− 17. 6
7 y− 2 x− 22 = 0
[x=− 1. 2 ,y= 2 .8]
- 3 m− 4. 4 n= 6. 84
- 5 n− 6. 7 m= 1. 23
[m=− 6. 4 ,n=− 4 .9]
- 3x+ 4 y+z= 10
2 x− 3 y+ 5 z+ 9 = 0
x+ 2 y−z= 6
[x= 1 ,y= 2 ,z=−1]
- 2 p− 2. 3 q− 3. 1 r+ 10. 1 = 0
- 7 p+ 3. 8 q− 5. 3 r− 21. 5 = 0
- 7 p− 8. 3 q+ 7. 4 r+ 28. 1 = 0
[p= 1. 5 ,q= 4. 5 ,r= 0 .5]
5.
x
2
−
y
3
+
2 z
5
=−
1
20
x
4
+
2 y
3
−
z
2
=
19
40
x+y−z=
59
(^60) [
x=
7
20
,y=
17
40
,z=−
5
24
]
- In a system of forces, the relationshipbetween
two forcesF 1 andF 2 is given by:
5 F 1 + 3 F 2 + 6 = 0
3 F 1 + 5 F 2 + 18 = 0
Use determinants to solve forF 1 andF 2.
[F 1 = 1. 5 ,F 2 =− 4 .5] - Applying mesh-current analysis to an a.c.
circuit results in the following equations:
( 5 −j 4 )I 1 −(−j 4 )I 2 = 100 ∠ 0 ◦
( 4 +j 3 −j 4 )I 2 −(−j 4 )I 1 = 0
Solve the equations forI 1 andI 2.
[
I 1 = 10. 77 ∠ 19. 23 ◦A,
I 2 = 10. 45 ∠− 56. 73 ◦A
]