Question

用矩阵的初等变换解线性方程组 {X1+2X2+X3-X4=0 3X1+6X2-X3-3X4=0 5X1+10X2+X3-5X4=0}

老师这道题怎么做，谢谢哦

Answer 1

解: 系数矩阵A=
1 2 1 -1
3 6 -1 -3
5 10 1 -5

r2-3r1, r3-5r1
1 2 1 -1
0 0 -4 0
0 0 -4 0

r3-r2, r2*(-1/4),r1-r2
1 2 0 -1
0 0 1 0
0 0 0 0

所以方程组的通解为 k1(-2,1,0,0)^T + k2(1,0,0,1)^T

Answer 2

A =
1 2 1 -1 0

3 6 -1 -3 0
5 10 1 -5 0
化为标准型：
1 2 0 -1 0
0 0 1 0 0
0 0 0 0 0
x1=-2C1+C2
x2=C1
x3=0
x4=C2