Question

求下列非齐性线性方程组的解

X1+ X2+ X3+ X4+ X5=3
2X1+ X2+3X3+3X4+4X5=14
3X1+4X2+ X3- 3X4+2X5=-11
X1- X2+4X3+8X4+4X5=31

Answer 1

解: 增广矩阵=
1 1 1 1 1 3
2 1 3 3 4 14
3 4 1 -3 2 -11
1 -1 4 8 4 31

r2-2r1,r3-3r1,r4-r1
1 1 1 1 1 3
0 -1 1 1 2 8
0 1 -2 -6 -1 -20
0 -2 3 7 3 28

r1+r2,r3+r2,r4-2r4
1 0 2 2 3 11
0 -1 1 1 2 8
0 0 -1 -5 1 -12
0 0 1 5 -1 12

r1+2r3,r2+r3,r4+r3
1 0 0 -8 5 -13
0 -1 0 -4 3 -4
0 0 -1 -5 1 -12
0 0 0 0 0 0

r2*(-1),r3*(-1)
1 0 0 -8 5 -13
0 1 0 4 -3 4
0 0 1 5 -1 12
0 0 0 0 0 0

方程组的通解为: (-13,4,12,0,0)^T+c1(8,-4,-5,1,0)^T+c2(-5,3,1,0,1)^T.

Answer 2

记成增广矩阵，化成上三角，再回带就行了

追问

就是不会做，我要全解。详细的谢谢

追答

就是消元法，体力活，用矩阵不过是少写几次x而已