.lambda = __ 时,齐次线性方程组 ) lambda (x)_(1)+2(x)_(2)+2(x)_(4)+2(x)_(4)=0 2(x)_(1)+lambda (x)_(2)+2(x)_(3)+2(x)_(4)=0 2(x)_(1)+2(x)_(2)+lambda (x) . 有非

步骤 1：整理方程组
将方程组整理为标准形式：
\[
\left \{ \begin{matrix}
(\lambda +2) {x}_{1}+2{x}_{2}+2{x}_{3}+0{x}_{4}=0\\
2{x}_{1}+(\lambda +2) {x}_{2}+2{x}_{3}+2{x}_{4}=0\\
2{x}_{1}+2{x}_{2}+(\lambda +2) {x}_{3}+2{x}_{4}=0\\
2{x}_{1}+2{x}_{2}+2{x}_{3}+(\lambda +2) {x}_{4}=0
\end{matrix} \right.
\]

步骤 2：写出系数矩阵
系数矩阵为：
\[
A = \begin{pmatrix}
\lambda +2 & 2 & 2 & 0 \\
2 & \lambda +2 & 2 & 2 \\
2 & 2 & \lambda +2 & 2 \\
2 & 2 & 2 & \lambda +2
\end{pmatrix}
\]

步骤 3：求行列式
方程组有非零解的条件是系数矩阵的行列式为0，即：
\[
\det(A) = 0
\]
计算行列式：
\[
\det(A) = (\lambda +2)^4 - 4(\lambda +2)^3 + 6(\lambda +2)^2 - 4(\lambda +2) = (\lambda +2)^2((\lambda +2)^2 - 4(\lambda +2) + 4) = (\lambda +2)^2(\lambda^2 + 4\lambda + 4 - 4\lambda - 8 + 4) = (\lambda +2)^2(\lambda^2 - 4)
\]
\[
\det(A) = (\lambda +2)^2(\lambda -2)(\lambda +2) = (\lambda +2)^3(\lambda -2)
\]
令行列式为0，得到：
\[
(\lambda +2)^3(\lambda -2) = 0
\]
解得：
\[
\lambda = -2 \quad \text{或} \quad \lambda = 2
\]