Question

线性代数求教,p=0,q=2(1)求齐次方程组Ax=0的基础解系(2)求方程组Ax=b的通解

Answer 1

增广矩阵 B=(A, b)=

[1 1 1 1 1 1]

[3 2 1 1 -3 0]

[0 1 2 2 6 3]

[5 4 3 3 -1 2]

初等行变换为

[1 1 1 1 1 1]

[0 -1 -2 -2 -6 -3]

[0 1 2 2 6 3]

[0 -1 -2 -2 -6 -3]

初等行变换为

[1 1 1 1 1 1]

[0 1 2 2 6 3]

[0 0 0 0 0 0]

[0 0 0 0 0 0]

方程同解变形为

x1+x2=-x3-x4-x5+1

x2=-2x3-2x4-6x5+3

导出组的基础解系为 (1 -2 1 0 0)^T, (1 -2 0 1 0)^T, (5 -6 0 0 1)^T.

特解为 (-2 3 0 0 0)^T,

方程组的通解为 x=k1(1 -2 1 0 0)^T+k2(1 -2 0 1 0)^T

+k3(5 -6 0 0 1)^T+(-2 3 0 0 0)^T.