Higher Engineering Mathematics, Sixth Edition

The solution of simultaneous equations by matrices and determinants 245

Following the procedure:

(i) ( 9 +j 12 )I 1 −( 6 +j 8 )I 2 − 5 = 0

−( 6 +j 8 )I 1 +( 8 +j 3 )I 2 −( 2 +j 4 )= 0

(ii)

I 1

∣

∣

∣

∣

−( 6 +j 8 ) − 5

( 8 +j 3 ) −( 2 +j 4 )

∣

∣

∣

∣

=

−I 2

∣

∣

∣

∣

( 9 +j 12 ) − 5

−( 6 +j 8 ) −( 2 +j 4 )

∣

∣

∣

∣

=

1

∣

∣

∣

∣

( 9 +j 12 ) −( 6 +j 8 )

−( 6 +j 8 )( 8 +j 3 )

∣

∣

∣

∣

I 1

(− 20 +j 40 )+( 40 +j 15 )

=

−I 2

( 30 −j 60 )−( 30 +j 40 )

=

1

( 36 +j 123 )−(− 28 +j 96 )

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 =0formgives:

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

where DI 1 is the determinant of coefficients

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