解下列方程组(1){x+y-z=0,x-3y+2z=1,3x+2y-z=4

x+y-z=0 (1)
x-3y+2z=1 (2)
3x+2y-z=4 (3)
(1)×2+(2)
3x-y=1 (4)
(3)-(1)
2x+y=4 (5)
(4)+(5)
5x=5
所以
x=1
y=3x-1=2
z=x+y=3还有两题（1）{x+y-z=-3,2x+y-3z=-2,3x-y+2z=-11（2）{5x-4y=0，3y-5z=0，2x-y+z=6采纳我，重新问