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

zheyizhang2022-10-04 11:39:541条回答

求方程组
X1+X2+X3+4xX4-3xX5=0
2xX1+X2+3xX3+5xX4-5xX5=0
X1-X2+3xX3-2xX4-X5=0
3xX1+X2+5xX3+6xX4-8xX5=0

已提交，审核后显示！提交回复

休仙人共回答了19个问题 | 采纳率84.2%: 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; 1年前