The following linear system of three unknowns has a non-zero solution if and only if λ=\lambda = λ= ___:
{x1+2x2−2x3=0,2x1−x2+λx3=0,3x1+x2+x3=0.\begin{cases} x_1 + 2x_2 - 2x_3 = 0, \\ 2x_1 - x_2 + \lambda x_3 = 0, \\ 3x_1 + x_2 + x_3 = 0. \end{cases}⎩⎨⎧x1+2x2−2x3=0,2x1−x2+λx3=0,3x1+x2+x3=0.
參考答案與解析
λ=3\lambda=3λ=3。非零解存在若且唯若係數矩陣行列式為0：det[12−22−1λ311]=1(−1−λ)−2(2−3λ)−2(2+3)=5λ−15=0\det\begin{bmatrix}1&2&-2\\2&-1&\lambda\\3&1&1\end{bmatrix}=1(-1-\lambda)-2(2-3\lambda)-2(2+3)=5\lambda-15=0det1232−11−2λ1=1(−1−λ)−2(2−3λ)−2(2+3)=5λ−15=0，解得λ=3\lambda=3λ=3。
