Question

急！！线性代数作业，求详细步骤，在线等

如有解答，感激不尽
求线性方程组基础解系
x1+x2+3x4-x5=0
x1-x2+2x3-x4=0
4x1-2x2+6x3+3x4-4x5=0
2x1+4x2-2x3+4x4-7x5=0

Answer 1

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.

Answer 2

X1x2x3x4x5 全部为0不就行了？