用基础解系表示下列线性方程组得全部解:(x1)- 5(x2)+ 2(x3)- 3(x4)=11 5(x1)+3(x2)+6(x3)-(x4)=-1 2(x1)+4(X2)+2(X3)+(x4)=-6
网友回答
增广矩阵 =1 -5 2 -3 11
5 3 6 -1 -1
2 4 2 1 -6
r2-5r1,r3-2r1
1 -5 2 -3 11
0 28 -4 14 -56
0 14 -2 7 -28
r2-2r3,r3*(1/14)
1 -5 2 -3 11
0 0 0 0 00 1 -1/7 1/2 -2
r1+5r3
1 0 9/7 -1/2 1
0 0 0 0 00 1 -1/7 1/2 -2
所以通解为:(1,-2,0,0)'+c1(-9,1,7,0)'+c2(1,-1,0,2),c1,c2 为任意常数