x+2y+3z=12x+3y+z=23x+y+2z=3.

问题描述:

x+2y+3z=1
2x+3y+z=2
3x+y+2z=3

x+2y+3z=1            ①
2x+3y+z=2              ②
3x+y+2z=3              ③

①×2-②,可得:
y+5z=0,④;
②×3-③×2,可得:
7y-z=0,⑤;
由⑤可得z=7y,
把z=7y代入④,可得:y+5×7y=0,
则y=0,
把y=0,代入z=7y,可得z=0;
把y=0,z=0代入①可得:x+0+0=1,
则x=1,
所以这个方程组的解是:
x=1
y=0
z=0