Page 235 - Advanced engineering mathematics

7.5 Homogeneous Systems 215

The reduced system A R X = O is
35 13
x 1 − x 4 + x 5 = 0,
16 16
28 20
x 2 + x 4 − x 5 = 0,
16 16
and
7 9
x 3 + x 4 − x 5 = 0.
16 16
This system is easy to solve:
35 13
x 1 = x 4 − x 5 ,
16 16
28 20
x 2 =− x 4 + x 5 ,
16 16
and
7 9
x 3 =− x 4 + x 5
16 16
in which x 4 and x 5 (the free variables) can be given any values and these determine x 1 , x 2 and
x 3 . Again, note that these two free variables are in the two columns of the reduced matrix that
contain no leading element of any row.
We can express this solution more neatly by setting x 4 = 16α and x 5 = 16β with α and β
arbitrary numbers and writing
x 1 = 35α − 13β,
x 2 =−28α + 20β,
x 3 =−7α + 9β,
x 4 = 16α,
and
x 5 = 16β.
Here α and β are any numbers. This is the general solution of the reduced system, and it is routine
to verify that it is also the solution of the original system. As a column matrix, this solution is
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
35α − 13β 35 −13
⎜ −28α + 20β ⎟ ⎜ −28 ⎟ ⎜ 20 ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
X = ⎜ −7α + 9β ⎟ = α ⎜ −7 ⎟ + β ⎜ 9 ⎟ .
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
16α
⎝ ⎠ ⎝ 16 ⎠ ⎝ 0 ⎠
16β 0 16
This way of writing the general solution reveals its structure as being two dimensional,
depending on two arbitrary constants.

These examples illustrate the strategy outlined at the beginning of this section. This will
depend on the crucial fact that the reduced system has the same solutions as the original system,
as we will now verify.

THEOREM 7.11
Let A be n × m. Then the systems AX = O and A R X = O have the same solutions.

Proof First, we know that there is a matrix
= E r E r−1 ···E 2 E 1 ,
a product of elementary matrices, that reduces A:
A = A R .

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October 14, 2010 14:23 THM/NEIL Page-215 27410_07_ch07_p187-246