Question

请教如何用matlab求多元二次方程组？

X1+X2*X3+X3=100
X1-X2+2*X3=90
X1*X3+X2*X3=300
这样的方程如何求解，我不会用matlab .
请大侠仔细地说一下。
谢谢！！
谢谢楼下的朋友解答，关键的是我要解的方程组是23元二次方程组。上面的那个列子是想说明问题用的。这么多的未知数如何手工代入消元啊？
有没有其它的方法啊？
谢谢！！

Answer 1

吾觉得要手工计算代入，求出一元高次方程，再用roots求解，或者直接用solve求解。
[x1,x2,x3]=solve('x1+x2*x3+x3=100','x1-x2+2*x3=90','x1*x3+x2*x3=300')
结果是：
x1 =

-(1/6*(558900+60*i*6884535^(1/2))^(1/3)+1160/(558900+60*i*6884535^(1/2))^(1/3)+15)^2+610+22/3*(558900+60*i*6884535^(1/2))^(1/3)+51040/(558900+60*i*6884535^(1/2))^(1/3)
-(-1/12*(558900+60*i*6884535^(1/2))^(1/3)-580/(558900+60*i*6884535^(1/2))^(1/3)+15+1/2*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3)))^2+610-11/3*(558900+60*i*6884535^(1/2))^(1/3)-25520/(558900+60*i*6884535^(1/2))^(1/3)+22*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3))
-(-1/12*(558900+60*i*6884535^(1/2))^(1/3)-580/(558900+60*i*6884535^(1/2))^(1/3)+15-1/2*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3)))^2+610-11/3*(558900+60*i*6884535^(1/2))^(1/3)-25520/(558900+60*i*6884535^(1/2))^(1/3)-22*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3))

x2 =

-(1/6*(558900+60*i*6884535^(1/2))^(1/3)+1160/(558900+60*i*6884535^(1/2))^(1/3)+15)^2+550+23/3*(558900+60*i*6884535^(1/2))^(1/3)+53360/(558900+60*i*6884535^(1/2))^(1/3)
-(-1/12*(558900+60*i*6884535^(1/2))^(1/3)-580/(558900+60*i*6884535^(1/2))^(1/3)+15+1/2*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3)))^2+550-23/6*(558900+60*i*6884535^(1/2))^(1/3)-26680/(558900+60*i*6884535^(1/2))^(1/3)+23*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3))
-(-1/12*(558900+60*i*6884535^(1/2))^(1/3)-580/(558900+60*i*6884535^(1/2))^(1/3)+15-1/2*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3)))^2+550-23/6*(558900+60*i*6884535^(1/2))^(1/3)-26680/(558900+60*i*6884535^(1/2))^(1/3)-23*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3))

x3 =

1/6*(558900+60*i*6884535^(1/2))^(1/3)+1160/(558900+60*i*6884535^(1/2))^(1/3)+15
-1/12*(558900+60*i*6884535^(1/2))^(1/3)-580/(558900+60*i*6884535^(1/2))^(1/3)+15+1/2*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3))
-1/12*(558900+60*i*6884535^(1/2))^(1/3)-580/(558900+60*i*6884535^(1/2))^(1/3)+15-1/2*i*3^(1/2)*(1/6*(558900+60*i*6884535^(1/2))^(1/3)-1160/(558900+60*i*6884535^(1/2))^(1/3))
即x1,x2,x3分别有3个根。