Question

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

先谢谢了

Answer 1

解: 增广矩阵 =
1 1 2 4 3
3 1 6 2 3
-1 2 -2 1 1

r2-3r1,r3+r1
1 1 2 4 3
0 -2 0 -10 -6
0 3 0 5 4

r2*(-1/2), r1-r2,r3-3r2
1 0 2 -1 0
0 1 0 5 3
0 0 0 -10 -5

r3*(-1/10), r1+r3,r2-5r3
1 0 2 0 1/2
0 1 0 0 1/2
0 0 0 1 1/2

得同解方程组:
x1 = -2x3 + 1/2
x2 = 1/2
x4 = 1/2

令x3=k, 即得
x1=-2k+1/2, x2=1/2, x3=k, x4=1/2