Question

线性方程组λx1+x2+x3=λ-3 x1+λx2+x3=-2 x1+x2+λx3=-2

Answer 1

由题目可知 λx1+x2+x3=λ-3-------------------------(1) x1+λx2+x3=-2--------------------------(2)

x1+x2+λx3=-2--------------------------(3)

(1)-λ*(2),

x2-λ^2 x2+x3-λx3=λ-3-2λ---------------(4)

(1)-λ*(3)

x2-λ x2+x3-λ^2x3=λ-3-2λ---------------(5)

从(4),

x2(1-λ^2)+x3(1-λ)=x2(1-λ)(1+λ)+x3(1-λ)=-3-λ-----------(6)

从(5),

x2(1-λ)+x3(1-λ^2)=-3-λ--------------------------------(7)

从(6),(7),

x3(1-λ)[1+(1+λ)^2]=-(λ+3)(1+λ)

设λ≠1,

x3=-(λ+3)(1+λ)/{(1-λ)[2+2λ+λ^2]}----------------(8)

从(5),(8)

x2=[-3-λ-x3(1-λ^2)]/(1-λ)----------------------------(9)

从(2),(8),(9),

x1=-λx2-x3-2---------------------------------------------(10)

方程组通解:(8),(9),(10)

考虑有唯一解:

系数行列式H=(λ+2)(λ-1)^2.

∴λ≠-2且λ≠1时,方程组有唯一解.

λ=-2时,三个方程相加(1)+(2)+(3)得0=-5-2-2,矛盾,方程组无解.

λ=1时,三个方程相同,(1)=(2)=(3),都是x1+x2+x3=-2,方程组有无穷多解.

（引自 wong6764 ）