Question

求齐次线性方程组x1+2x2+x3+x4+x5=1 2x1+4x2+3x3+x4+x5=2 -x1-2x2+x3+3x4-3x5=5 2x3+4x-2x5=6的一般解

Answer 1

增广矩阵=
1 2 1 1 1 1
2 4 3 1 1 2
-1 -2 1 3 -3 5
0 0 2 4 -2 6
用初等行变换化为行最简形
1 2 0 0 2 -1
0 0 1 0 -1 1
0 0 0 1 0 1
0 0 0 0 0 0

一般解为: (-1,0,1,1,0)^T+k1(-2,1,0,0,0)^T+k2(-2,0,1,0,1)^T.

追问

可以写一下变化过程吗？

追答

增广矩阵=
1 2 1 1 1 1
2 4 3 1 1 2
-1 -2 1 3 -3 5
0 0 2 4 -2 6

r2-2r1,r3+r1
1 2 1 1 1 1
0 0 1 -1 -1 0
0 0 2 4 -2 6
0 0 2 4 -2 6

r4-r3lr1-r2,r3-2r2
1 2 0 2 2 1
0 0 1 -1 -1 0
0 0 0 6 0 6
0 0 0 0 0 0

r3*(1/6),r1-2r3,r2+r3
1 2 0 0 2 -1
0 0 1 0 -1 1
0 0 0 1 0 1
0 0 0 0 0 0

一般解为: (-1,0,1,1,0)^T+k1(-2,1,0,0,0)^T+k2(-2,0,1,0,1)^T.