急 设a1,a2,a3是互不相等的常数,求方程组(x1)+(a1)(x2)+(a1)^2(x3)=1,(x1)+(a2)
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急 设a1,a2,a3是互不相等的常数,求方程组(x1)+(a1)(x2)+(a1)^2(x3)=1,(x1)+(a2)(x
求方程组(x1)+(a1)(x2)+(a1)^2(x3)=1,(x1)+(a2)(x2)+(a2)^2(x3)=1,(x1)+(a3)(x2)+(a3)^2(x3)=1,
求方程组(x1)+(a1)(x2)+(a1)^2(x3)=1,(x1)+(a2)(x2)+(a2)^2(x3)=1,(x1)+(a3)(x2)+(a3)^2(x3)=1,
(x1)+(a1)(x2)+(a1)^2(x3)=1 (1)
(x1)+(a2)(x2)+(a2)^2(x3)=1 (2)
(x1)+(a3)(x2)+(a3)^2(x3)=1 (3)
(2)-(1)得
(a2-a1)[x2+(a2+a1)X3]=0
同理(3)-(2),(3)-(1)
(a3-a2)[x2+(a2+a3)X3]=0
(a3-a1)[x2+(a3+a1)X3]=0
因为a1,a2,a3是互不相等的常数
所以
x2+(a2+a1)X3= x2+(a2+a3)X3= x2+(a3+a1)X3=0
得出X3=0;X2=0;X1=1
(x1)+(a2)(x2)+(a2)^2(x3)=1 (2)
(x1)+(a3)(x2)+(a3)^2(x3)=1 (3)
(2)-(1)得
(a2-a1)[x2+(a2+a1)X3]=0
同理(3)-(2),(3)-(1)
(a3-a2)[x2+(a2+a3)X3]=0
(a3-a1)[x2+(a3+a1)X3]=0
因为a1,a2,a3是互不相等的常数
所以
x2+(a2+a1)X3= x2+(a2+a3)X3= x2+(a3+a1)X3=0
得出X3=0;X2=0;X1=1
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