Question

1 -1 -1 1 0 1 -1 1 -3 0 1 -1 -2 3 0 的基础解系与其通解怎么求

Answer 1

解: 系数矩阵=

1 -1 -1 1

1 -1 1 -3

1 -1 -2 3

r2-r1,r3-r1

1 -1 -1 1

0 0 2 -4

0 0 -1 2

r2*(1/2),r1+r2,r3+r2

1 -1 0 -1

0 0 1 -2

0 0 0 0

方程组的通解为: c1(1,1,0,0)'+c2(1,2,0,1)'.