Higher Engineering Mathematics

284 MATRICES AND DETERMINANTS

26.4 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

equations we have deliberately obtained zeros in the

positions 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 known as theGaussian elimination

method.

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-step procedure to solve simultaneous

equations in three unknowns using theGaussian

elimination 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. Determinezfrom equation (3′′), thenyfrom
   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)

Use the Gaussian elimination method to solve

forI 1 ,I 2 andI 3.

(This is the same example as Problem 6 on page 281,

and a comparison of methods may be made)

Following the above procedure:

1. 2I 1 + 3 I 2 − 4 I 3 = 26 (1)

Equation (2)−

1

2

×equation (1) gives:

0 − 6. 5 I 2 −I 3 =− 100 (2′)

Equation (3)−

− 7

2

×equation (1) gives:

0 + 12. 5 I 2 − 8 I 3 = 103 (3′)

2. 2I 1 + 3 I 2 − 4 I 3 = 26 (1)

0 − 6. 5 I 2 −I 3 =− 100 (2′)

Equation (3′)−

12. 5

− 6. 5

×equation (2′) gives:

0 + 0 − 9. 923 I 3 =− 89. 308 (3′′)