Higher Engineering Mathematics, Sixth Edition

248 Higher Engineering Mathematics

Hence

x=

Dx

D

=

70

35

= 2 ,y=

Dy

D

=

− 105

35

=− 3

and z=

Dz

D

=

175

35

= 5

Now try the following exercise

Exercise 100 Further problems on solving

simultaneous equations using Cramersrule

1. Repeat problems 3, 4, 5, 7 and 8 of Exercise
   98 on page 241, using Cramers rule.


2. Repeat problems 3, 4, 8 and 9 of Exercise 99
   on page 244, using Cramers rule.

23.4 Solution of simultaneous

equations using the Gaussian

elimination method

Consider the following simultaneous equations:

x+y+z=4(1)

2 x− 3 y+ 4 z= 33 (2)

3 x− 2 y− 2 z=2(3)

Leaving equation (1) as it is gives:

x+y+z=4(1)

Equation( 2 )− 2 ×equation (1) gives:

0 − 5 y+ 2 z= 25 (2′)

and equation( 3 )− 3 ×equation (1) gives:

0 − 5 y− 5 z=− 10 (3′)

Leaving equations (1) and (2′) as they are gives:

x+y+z=4(1)

0 − 5 y+ 2 z= 25 (2′)

Equation (3′)−equation (2′)gives:

0 + 0 − 7 z=− 35 (3′′)

By appropriately manipulating the three original equa-

tions we have deliberately obtained zeros in the posi-

tions shown in equations (2′)and(3′′).

Working backwards, from equation (3′′),

z=

− 35

− 7

= 5 ,

from equation (2′),

− 5 y+ 2 ( 5 )= 25 ,

from which,

y=

25 − 10

− 5

=− 3

and from equation (1),

x+(− 3 )+ 5 = 4 ,

from which,

x= 4 + 3 − 5 = 2

(This is the same example as Problems 2 and 7, and

a comparison of methods can be made). The above

method is knownas theGaussian eliminationmethod.

We conclude from the above example that if

a 11 x+a 12 y+a 13 z=b 1

a 21 x+a 22 y+a 23 z=b 2

a 31 x+a 32 y+a 33 z=b 3

the three-stepprocedureto solve simultaneous equa-

tions in three unknowns using theGaussian elimina-

tion methodis:

1. Equation( 2 )−

a 21

a 11

×equation (1) to form equa-

tion (2′) and equation (3)−

a 31

a 11

×equation (1) to

form equation (3′).

2. Equation( 3 ′)−

a 32

a 22

×equation (2′) to form equa-

tion (3′′).

3. Determinez from equation (3′′), then y from
   equation (2′) and finally,xfrom equation (1).

Problem 8. 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 (1)

I 1 − 5 I 2 − 3 I 3 =− 87 (2)

− 7 I 1 + 2 I 2 + 6 I 3 = 12 (3)