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

A=(1 1 1 1 1) b=(1)
(3 2 1 1 -3) (p)
(0 1 2 2 6) (3)
(5 4 3 3 -1) (q)

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.