求解三元一次方程,{x+2y+3z=1,y+2z+3x=2,z+2x+3y=3 问x,y,z个是多少?
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求解三元一次方程,{x+2y+3z=1,y+2z+3x=2,z+2x+3y=3 问x,y,z个是多少?
求解三元一次方程,{x+2y+3z=1,y+2z+3x=2,z+2x+3y=3 问x,y,z个是多少?
求解三元一次方程,{x+2y+3z=1,y+2z+3x=2,z+2x+3y=3 问x,y,z个是多少?
x+2y+3z=1(1) y+2z+3x=2(2) z+2x+3y=3(3)
由(1)得:x=1-2y-3z(4)
将(4)带入(2)和(3)
得到方程组:5y+7z=1 -5z-y=1
解方程组可以得到:y=2/3 z=-1/3
将y=2/3 z=-1/3带入(4)得到:
x=2/3
得出原三元一次方程组得解是:
x=2/3 y=2/3 z=-1/3
x+2y+3z=1,(1)
y+2z+3x=2,(2)
z+2x+3y=3 (3)
(3)*2-(2)
x+5y=4 (4)
(3)*3-(1)
5x+7y=8 (5)
(4)*5-(5)
y=2/3
x=2/3
z=-1/3
x=2/3
y=2/3
z=-1/3
x+2y+3z=1①
3x+y+2z=2②
2x+3y+z=3③
①+②+③
6x+6y+6z=6
x+y+z=1
①-④
y+2z=0⑤
③-2×④
y-z=1...
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x+2y+3z=1①
3x+y+2z=2②
2x+3y+z=3③
①+②+③
6x+6y+6z=6
x+y+z=1
①-④
y+2z=0⑤
③-2×④
y-z=1⑥
⑤-⑥
3z=-1
z=-1/3
把z=-1/3代入⑥
y-(-1/3)=1
y+1/3=1
y=2/3
把y=2/3,z=-1/3代入①
x+2×2/3+3×-1/3=1
x+4/3-1=1
x+1/3=1
x=1/3
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