Find a basis for the null space of the matrix

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A can be transformed to the reduced row echelon form:
1 -2 0 -1 3
0 0 1 2 -2
0 0 0 0 0
It corresponds to (in x1 - x5, the numbers are subscript):
x1 - 2x2 -x4 + 3x5 = 0
x3 + 2x4 -2x5 = 0
0 = 0
The system has infinitely many solutions:
x1 = 2x2 + x4 -3x5
x2 = arbitrary
x3 = -2x4 + 2x5
x4 = arbitrary
x5 = arbitrary
The solution can be written in vector form:
2
1
0 c2
0
0
+
1
0
-2 c4
1
0
+
-3
0
2
0
1
c5
Therefore the null space has a basis formed by the set { \x09
2
1
0
0
0
\x09, \x09
1
0
-2
1
0
\x09, \x09
-3
0
2
0
1
\x09}.

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