Question

求方程组X1+X2+X3+4xX4-3xX5=02xX1+X2+3xX3+5xX4-5xX5=0X1-X2+3xX3-2

Answer 1

系数矩阵A=

1 1 1 4 -3

2 1 3 5 -5

1 -1 3 -2 -1

3 1 5 6 -8

r2-2r1,r3-r1,r4-3r1

1 1 1 4 -3

0 -1 1 -3 1

0 -2 2 -6 2

0 -2 2 -6 1

r1+r2,r3-2r2,r4-2r2,r2*(-1)

1 0 2 1 -2

0 1 -1 3 -1

0 0 0 0 0

0 0 0 0 -1

r4*(-1),r1+2r4,r2+r4

1 0 2 1 0

0 1 -1 3 0

0 0 0 0 0

0 0 0 0 1

通解为 k1(-2,1,1,0,0)^T + k2(-1,-3,0,1,0)^T