系数矩阵 A=
[1 1 0 3 -1]
[1 -1 2 -1 0]
[4 -2 6 3 -4]
[2 4 -2 4 -7]
行初等变换为
[1 1 0 3 -1]
[0 -2 2 -4 1]
[0 -6 6 -9 0]
[0 2 -2 -2 -5]
行初等变换为
[1 1 0 3 -1]
[0 2 -2 4 -1]
[0 0 0 3 -3]
[0 0 0 -6 -4]
行初等变换为
[1 1 0 3 -1]
[0 2 -2 4 -1]
[0 0 0 1 -1]
[0 0 0 0 -10]
行初等变换为
[1 1 0 3 -1]
[0 2 -2 4 -1]
[0 0 0 1 -1]
[0 0 0 0 1]
方程组同解变形为
x1+x2+3x4-x5=0
2x2+4x4-x5=2x3
x4-x5=0
x5=0
取x3=1,得基础解系 (-1,1,1,0,0)^T.