matrix Ax=0

A=(1 -3 3 -5) = (1) (1 -3 3 -5)
....(2 -6 6 -10)...(2)

If Ax = 0, then (1 -3 3 -5)x = 0, where x is a column matrix 4*1.
let x = (r1)
...........(r2)
...........(r3)
...........(r4)
so, r1 - 3r2 + 3r3 - 5r4 = 0
In order to make Ax = 0, x is a 4*n matrix such that each column of the matrix
satify r1 - 3r2 + 3r3 - 5r4 = 0 where n is any positive integer.

2012-09-21 09:11:57 補充：
Example :
........(5)......(0).....(2)
x=x1(1)+x2(2)+x3(0)
........(1)......(2).....(1)
........(1)......(0).....(1)

A is a 2 x 4 matrix, so X is a 4 x i matrix in order to be well define.
Denote P(i,j) be the element of i th row and j th column in matrix P,

set i = 1, AX is then become 2 x 1 matrix
Let X(1,1) = a, X(2,1) = b, X(3,1) = c, X(4,1) = d,
AX(1,1) = a - 3b + 3c - 5d = 0
AX(2,1) = 2a - 6b + 6c - 10d = 0 => a - 3b + 3c - 5d = 0

set i = 2, AX is then become 2 x 2 matrix
Let X(1,1) = a, X(2,1) = b, X(3,1) = c, X(4,1) = d,
X(1,2) = e, X(2,2) = f, X(3,2) = g, X(4,2) = h,
AX(1,1) = a - 3b + 3c - 5d = 0
AX(2,1) = 2a - 6b + 6c - 10d = 0 => a - 3b + 3c - 5d = 0
AX(1,2) = e - 3f + 3g - 5h = 0
AX(2,1) = 2e - 6f + 6g - 10h = 0 => e - 3f + 3g - 5h = 0

Without loss of generality, X is a 4 x i matrix and the elements from each column satisfy [X(1,i)] - 3[X(2,i)] + 3X[(3,i)] - 5[X(4,i)] = 0