The solution of simultaneous equations by matrices and determinants 249
Use the Gaussian elimination method to solve for
I 1 , I 2 and I 3 .
(This is the same example as Problem 6 on page 243,
and a comparison of methods may be made)
Following the above procedure:
1. 2I 1 + 3I 2 − 4I 3 = 26
(1)
Equation (2) −
1
2
× equation (1) gives:
0 − 6.5I 2 − I 3 = −100
(2 )
Equation (3) −
−7
2
× equation (1) gives:
0 + 12.5I 2 − 8I 3 = 103
(3 )
2. 2I 1 + 3I 2 − 4I 3 = 26
(1)
0 − 6.5I 2 − I 3 = −100
(2 )
Equation (3 ) −
12.5
−6.5
× equation (2 ) gives:
0 + 0 − 9.923I 3 = −89.308
(3 )
3. From equation (3 ),
I 3 =
−89.308
−9.923
= 9 mA,
from equation (2 ), −6.5I 2 − 9 =−100,
from which, I 2 =
−100 +9
−6.5
= 14 mA
and from equation (1), 2I 1 + 3(14) − 4(9) = 26,
from which, I 1 =
26 − 42 + 36
2
=
20
2
= 10 mA
Now try the following exercise
Exercise 101 Further problems on solving
simultaneous equations using Gaussian
elimination
1. In a mass-spring-damper system, the acceleration ¨
x m/s 2 , velocity ˙
x m/s and displacement
x m are related by the following simultaneous
equations:
6.2 ¨
x + 7.9 ˙
x + 12.6x = 18.0
7.5 ¨
x + 4.8 ˙
x + 4.8x = 6.39
13.0 ¨
x + 3.5 ˙
x − 13.0x = −17.4
By using Gaussian elimination, determine the
acceleration, velocity and displacement for the
system, correct to 2 decimal places.
[ ¨
x = −0.30, ˙
x = 0.60, x = 1.20]
2. The tensions, T 1 , T 2 and T 3 in a simple framework are given by the equations:
5T 1 + 5T 2 + 5T 3 = 7.0
T 1 + 2T 2 + 4T 3 = 2.4
4T 1 + 2T 2
= 4.0
Determine T 1 , T 2 and T 3 using Gaussian elimination.
[T 1 = 0.8, T 2 = 0.4, T 3 = 0.2]
3. Repeat problems 3, 4, 5, 7 and 8 of Exercise 98
on page 241, using the Gaussian elimination
method.
4. Repeat problems 3, 4, 8 and 9 of Exercise 99
on page 244, using the Gaussian elimination
method.