已知线性方程组{x1+2x2-x3-2x4=02x1-x2-x3+x4=13x1+x2-2x3-x4=1用导出组的基础解系表示方程组的全部解

Question

已知线性方程组{x1+2x2-x3-2x4=0 2x1-x2-x3+x4=1 3x1+x2-2x3-x4=1 用导出组的基础解系表示方程组的全部解

Answer 1

解: 增广矩阵(A,b)=
1 2 -1 -2 0
2 -1 -1 1 1
3 1 -2 -1 1
r3-r1-r2, r2-2r1
1 2 -1 -2 0
0 -5 1 5 1
0 0 0 0 0
r1+r2 (为了避免分数运算,约束变元选x1,x3)
1 -3 0 3 1
0 -5 1 5 1
0 0 0 0 0
方程组的全部解为: (1,0,1,0)^T + k1(3,1,5,0)^T +k2(3,0,5,-1)^T.