Question

求线性方程组的全部解，并用对应导出组的基础解系表示

X1+3X2+5X3-4X4=1 X1+3X2+2X3-2X4+X5=-1 X1-2X2+X3-X4-X5=3 X1-4X2+X3+X4-X5=3 X1+2X2+X3-X4+X5=-1

Answer 1

解: 增广矩阵 =
1 3 5 -4 0 1
1 3 2 -2 1 -1
1 -2 1 -1 -1 3
1 -4 1 1 -1 3
1 2 1 -1 1 -1

ri-r1, i=2,3,4,5
1 3 5 -4 0 1
0 0 -3 2 1 -2
0 -5 -4 3 -1 2
0 -7 -4 5 -1 2
0 -1 -4 3 1 -2

r1+3r5,r3-5r5,r4-7r5,r5*(-1)
1 0 7 5 3 -5
0 0 -3 2 1 -2
0 0 16 -12 -6 12
0 0 24 -16 -8 16
0 1 4 -3 -1 2

r1-3r2,r3+6r2,r4+8r2,r5+r2 (这样可避免分数运算)
1 0 -2 -1 0 1
0 0 -3 2 1 -2
0 0 -2 0 0 0
0 0 0 0 0 0
0 1 1 -1 0 0

r3*(-1/2),r1+2r3,r2+3r3,r5-r3
1 0 0 -1 0 1
0 0 0 2 1 -2
0 0 1 0 0 0
0 0 0 0 0 0
0 1 0 -1 0 0

所以方程组的全部解为: (1,0,0,0,-2)^T+c(1,1,0,1,-2)^T.