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素材/整合素材/线代素材/一个计算题的解答.md

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 ## 解答
 
 其特征多项式为  
-$$\det(A - \lambda E) = \begin{vmatrix} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 2 \\ 0 & 2 & 3-\lambda \end{vmatrix} = (2-\lambda)\left[(3-\lambda)^2 - 4\right] = (2-\lambda)(\lambda^2 - 6\lambda + 5) = (2-\lambda)(\lambda-1)(\lambda-5).$$  
+$$\begin{aligned}det(A - \lambda E) &= \begin{vmatrix} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 2 \\ 0 & 2 & 3-\lambda \end{vmatrix} \\
+&= (2-\lambda)\left[(3-\lambda)^2 - 4\right] \\
+&= (2-\lambda)(\lambda^2 - 6\lambda + 5)\\&=(2-\lambda)(\lambda-1)(\lambda-5).\end{aligned}$$  
 特征值为  $\lambda_1 = 1,\quad \lambda_2 = 2,\quad \lambda_3 = 5.$
 
 - 对于 $\lambda_1 = 1$：解 $(A - E)\mathbf{X} = \mathbf{0}$， $A - E = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix},$ 得基础解系 $\mathbf{x}_1 = (0, -1, 1)^\mathrm{T}$。