Higher Engineering Mathematics, Sixth Edition

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.

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
   
   
   2. 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 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]


7. 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

]