Page 234 - Advanced engineering mathematics

214 CHAPTER 7 Matrices and Linear Systems

Of course, we do not need matrices to solve this system, but we want to illustrate a point. The
matrix of coefficients is

1 −3 2
A = .
−2 1 −3
It is routine to find

10 7/5
A R = .
01 −1/5
The reduced system A R X = O is
7
x 1 + x 3 = 0
5
1
x 2 − x 3 = 0.
5
This reduced system can be solved by inspection:
7 1
x 1 =− x 3 , x 2 = x 3 , x 3 is arbitrary.
5 5
We can give x 3 any numerical value, and this determines x 1 and x 2 to yield a solution.
It will be useful to write this solution as a column matrix:
−7/5 −7/5
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x 1
= α ⎝ 1/5 ⎠ ,
⎝ ⎠
⎝ 1/5 ⎠
X = x 2 = x 3
1 1
x 3
in which we have written x 3 = α because it looks neater. Here α can be any number.
This general solution of the reduced system is also the solution of the original system. In
this example the general solution depends on one arbitrary constant, hence is, in a sense to be
discussed, a one-dimensional solution.
In Example 7.18, x 3 is called a free variable, since it can assume any value. This example
had one free variable, but the general solution of a system AX = O might have any number.

Free variables occur in columns of A R that contain no leading entry of a row.

EXAMPLE 7.19
Consider the 3 × 5system
x 1 − 3x 2 + x 3 − 7x 4 + 4x 5 = 0
x 1 + 2x 2 − 3x 3 = 0
x 2 − 4x 3 + x 5 = 0.
The matrix of coefficients is
⎛ ⎞
1 −3 1 −7 4
A = 1 2 −3 0 0 ⎠ .
⎝
0 1 −4 0 1
A routine calculation yields
⎛ ⎞
1 0 0 −35/16 13/16
A R = 0 1 0 28/16 −20/16 ⎠ .
⎝
0 0 1 7/16 −9/16

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October 14, 2010 14:23 THM/NEIL Page-214 27410_07_ch07_p187-246