问λ取何值时,齐次线性方程组{(5-λ)x1+2x2+2x3=0;2x1+(6-λ)x2=0;问λ取何值时,齐次线性方程组{(5-λ)x1+2x2+2x3=0;2x1+(6-λ)x2=0;2x1+(4-λ)x3=0.

Question

问λ取何值时,齐次线性方程组{ (5-λ)x1+2x2+2x3=0; 2x1+(6-λ)x2=0;
问λ取何值时,齐次线性方程组{
(5-λ)x1+2x2+2x3=0;
2x1+(6-λ)x2=0;
2x1+(4-λ)x3=0.

Answer 1

系数行列式
5-λ 2 2
2 6-λ 0
2 0 4-λ
= (4-λ)(5-λ)(6-λ)-4(6-λ)-4(4-λ) --直接对角线法则得
= (4-λ)(5-λ)(6-λ)-4(10-2λ) --后两项合并得因子(5-λ)
= (4-λ)(5-λ)(6-λ)-8(5-λ)
= (5-λ)[(4-λ)(6-λ)-8]
= (5-λ)[λ^2-10λ+16] -- 十字相乘法分解
= (5-λ)(2-λ)(8-λ)
注:这个计算比较特殊,不能用行列式的性质化简提出λ的因子
所以 λ 等于2 或 5 或 8 时,方程组有非零解