Higher Engineering Mathematics

282 MATRICES AND DETERMINANTS

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

=10 mA,

I 2 =

1806

129

=14 mA

and I 3 =

1161

129

=9mA

Now try the following exercise.

Exercise 114 Further problems on solving

simultaneous equations using determinants

In Problems 1 to 5 usedeterminantsto solve

the simultaneous equations given.

1. 3x− 5 y=− 17. 6
   7 y− 2 x− 22 = 0
   [x=− 1 .2,y= 2 .8]


2. 
   
   
   
   2. 3 m− 4. 4 n= 6. 84
   
   
   3. 5 n− 6. 7 m= 1. 23
      [m=− 6 .4,n=− 4 .9]
   
   
   
   


3. 3x+ 4 y+z= 10
   2 x− 3 y+ 5 z+ 9 = 0
   x+ 2 y−z= 6
   [x=1,y=2,z=−1]
   4. 1. 2 p− 2. 3 q− 3. 1 r+ 10. 1 = 0
   4. 7 p+ 3. 8 q− 5. 3 r− 21. 5 = 0
   3. 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
]

6. In a system of forces, the relationship
   between 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]

7. Applying mesh-current analysis to an a.c.
   circuit results in the following equations:

(5−j4)I 1 −(−j4)I 2 = 100 ∠ 0 ◦

(4+j 3 −j4)I 2 −(−j4)I 1 = 0

Solve the equations forI 1 andI 2.

[

I 1 = 10. 77 ∠ 19. 23 ◦A,

I 2 = 10. 45 ∠− 56. 73 ◦A

]

8. Kirchhoff’s laws are used to determine the
   current equations in an electrical network
   and show that

i 1 + 8 i 2 + 3 i 3 =− 31

3 i 1 − 2 i 2 +i 3 =− 5

2 i 1 − 3 i 2 + 2 i 3 = 6

Use determinants to find the values ofi 1 ,i 2

andi 3. [i 1 =−5,i 2 =−4,i 3 =2]

9. The forces in three members of a framework
   areF 1 ,F 2 andF 3. They are related by the
   simultaneous equations shown below.

1. 4 F 1 + 2. 8 F 2 + 2. 8 F 3 = 5. 6

4. 2 F 1 − 1. 4 F 2 + 5. 6 F 3 = 35. 0

4. 2 F 1 + 2. 8 F 2 − 1. 4 F 3 =− 5. 6