请问方程组x1+x2-3x4-x5=0,x1-x2+2x3-x4+x5=0,4x1-2x2+6x3-5x4+x5=0的基础解系与通解怎么求

问题描述:

请问方程组x1+x2-3x4-x5=0,x1-x2+2x3-x4+x5=0,4x1-2x2+6x3-5x4+x5=0的基础解系与通解怎么求

系数矩阵 A =1 1 0 -3 -11 -1 2 -1 14 -2 6 -5 1r2-r1,r3-4r11 1 0 -3 -10 -2 2 2 20 -6 6 7 5r2*(-1/2),r1-r2,r3+6r21 0 1 -2 00 1 -1 -1 -10 0 0 1 -1r1+2r3,r2+r21 0 1 0 -20 1 -1 0 -20 0 0 1 -1所以方程组的基...