用消元法解线性方程组x1+x2+2x3+4x4=33x1+x2+6x3+2x4=3-x1+2x2-2x3+x4=1麻烦用矩阵做,做到最简形以后,再解出方程.答案是x1=-2k+1/2,x2=1/2,x3=k,x4=1/2
问题描述：
用消元法解线性方程组
x1+x2+2x3+4x4=3
3x1+x2+6x3+2x4=3
-x1+2x2-2x3+x4=1
麻烦用矩阵做,做到最简形以后,再解出方程.
答案是x1=-2k+1/2,x2=1/2,x3=k,x4=1/2
答
增广矩阵 = 1 1 2 4 33 1 6 2 3-1 2 -2 1 1r2-3r1,r3+r11 1 2 4 30 -2 0 -10 -60 3 0 5 4r2*(-1/2),r1-r2,r3-3r21 0 2 -1 00 1 0 5 30 0 0 -10 -5r3*(-1/10),r1+r3,r2-5r31 0 2 0 1/20 1 0 0 1/20 0 0 1 1/2得同解方...