Page 237 - Advanced engineering mathematics

7.5 Homogeneous Systems 217

Since the number of nonzero rows of the reduced matrix is the rank of A R , which is also the
rank of A, then the dimension of the solution space can also be computed as
m − rank(A).

EXAMPLE 7.20
Solve the system

−x 1 + x 3 + x 4 + 2x 5 = 0
x 2 + 3x 3 + 4x 5 = 0
x 1 + 2x 2 + x 3 + x 4 + x 5 = 0
−3x 1 + x 2 + 4x 5 = 0.
The matrix of coefficients is
⎛ ⎞
−10112
0 1304
⎜ ⎟
A = ⎜ ⎟ .
⎝ 1
2111 ⎠
−31004
Routine manipulations yield the reduced form
⎛ ⎞
1000 −9/8
⎜ 0100 5/8 ⎟
A R = ⎜ ⎟ .
⎝ 0010
9/8 ⎠
0001 −1/4
In this example A has m = 5 columns, and the rank of A is 4 because A R has four nonzero rows.
The solution space will have dimension 5 − 4 = 1.
A R is the coefficient matrix of the reduced system
9
x 1 − x 5 = 0,
8
5
x 2 + x 5 = 0,
8
9
x 3 + x 5 = 0,
8
1
x 4 − x 5 = 0.
4
Notice that x 1 through x 4 depend on the single free variable x 5 , which can be chosen arbitrarily.
Set x 5 = α to write the general solution
9 5 9 1
x 1 = α, x 2 =− α, x 3 =− α, x 4 = α, x 5 = α.
8 8 8 4
If we let β = α/8(β is still any number), then
x 1 = 9β, x 2 =−5β, x 3 =−9β, x 4 = 2β, x 5 = 8β.
As a column matrix, this solution is
⎛ ⎞
9
⎜ −5 ⎟
⎜ ⎟
X = β ⎜ −9 .
⎟
⎜ ⎟
⎝ 2 ⎠
8
This gives the general solution as the set of all multiples of one solution, which forms a basis
for the one-dimensional solution space (a subspace of R ).
5

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October 14, 2010 14:23 THM/NEIL Page-217 27410_07_ch07_p187-246