Higher Engineering Mathematics

THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 281

F

I 1

(− 20 +j40)+(40+j15)

=

−I 2

(30−j60)−(30+j40)

=

1

(36+j123)−(− 28 +j96)

I 1

20 +j 55

=

−I 2

−j 100

=

1

64 +j 27

HenceI 1 =

20 +j 55

64 +j 27

=

58. 52 ∠ 70. 02 ◦

69. 46 ∠ 22. 87 ◦

= 0. 84 ∠ 47. 15 ◦A

and I 2 =

100 ∠ 90 ◦

69. 46 ∠ 22. 87 ◦

= 1. 44 ∠ 67. 13 ◦A

(b) When solving simultaneous equations inthree
unknowns using determinants:
(i) Write the equations in the form
a 1 x+b 1 y+c 1 z+d 1 = 0
a 2 x+b 2 y+c 2 z+d 2 = 0
a 3 x+b 3 y+c 3 z+d 3 = 0
and then
(ii) the solution is given by
x
Dx

=

−y

Dy

=

z

Dz

=

− 1

D

whereDxis

∣

∣

∣

∣

∣

b 1 c 1 d 1

b 2 c 2 d 2

b 3 c 3 d 3

∣

∣

∣

∣

∣

i.e. the determinant of the coefficients

obtained by covering up thexcolumn.

Dyis

∣

∣

∣

∣

∣

a 1 c 1 d 1

a 2 c 2 d 2

a 3 c 3 d 3

∣

∣

∣

∣

∣

i.e., the determinant of the coefficients

obtained by covering up theycolumn.

Dzis

∣

∣

∣

∣

∣

a 1 b 1 d 1

a 2 b 2 d 2

a 3 b 3 d 3

∣

∣

∣

∣

∣

i.e. the determinant of the coefficients

obtained by covering up thezcolumn.

andDis

∣

∣

∣

∣

∣

a 1 b 1 c 1

a 2 b 2 c 2

a 3 b 3 c 3

∣

∣

∣

∣

∣

i.e. the determinant of the coefficients

obtained by covering up the constants

column.

Problem 6. A d.c. circuit comprises three

closed loops. Applying Kirchhoff’s laws to the

closed loops gives the following equations for

current flow in milliamperes:

2 I 1 + 3 I 2 − 4 I 3 = 26

I 1 − 5 I 2 − 3 I 3 =− 87

− 7 I 1 + 2 I 2 + 6 I 3 = 12

Use determinants to solve forI 1 ,I 2 andI 3

(i) Writing the equations in the

a 1 x+b 1 y+c 1 z+d 1 =0 form gives:

2 I 1 + 3 I 2 − 4 I 3 − 26 = 0

I 1 − 5 I 2 − 3 I 3 + 87 = 0

− 7 I 1 + 2 I 2 + 6 I 3 − 12 = 0

(ii) the solution is given by

I 1

DI 1

=

−I 2

DI 2

=

I 3

DI 3

=

− 1

D

whereDI 1 is the determinant of coefficients

obtained by covering up theI 1 column, i.e.,

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