解下列方程组(1){x+y-z=0,x-3y+2z=1,3x+2y-z=4
人气:483 ℃ 时间:2019-12-17 03:01:16
解答
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
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