这道题中的矩阵虽然很“宽”，但其实是一个单列矩阵

一、题目

[1]. 求 $A$;

[2]. 求可逆矩阵 $\boldsymbol{P}$ 与对角矩阵 $\boldsymbol{\Lambda}$, 使得 $\boldsymbol{P}^{-1} \boldsymbol{A P}=\boldsymbol{\Lambda}$.

二、解析

[1]

$$A\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]=\left[\begin{array}{l} x_{1}+x_{2}+x_{3} \\ 2 x_{1}-x_{2}+x_{3} \\ x_{2}-x_{3} \end{array}\right] \Rightarrow$$

$$A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 0 & 1 & -1 \end{array}\right]$$

[2]

$$|\lambda E – A|=0 \Rightarrow\left|\begin{array}{ccc} \lambda-1 & -1 & -1 \\ -2 & \lambda+1 & -1 \\ 0 & -1 & \lambda+1 \end{array}\right|=0 \Rightarrow$$

$$(\lambda+1)^{2}(\lambda-1)-2-2(\lambda+1)-(r-1)=0 \Rightarrow$$

$$(\lambda+1)[(\lambda+1)(\lambda-1)-2]-(\lambda-1)-2=0$$

$$(\lambda+1)\left(\lambda^{2}-3\right)-(r+1)=0 \Rightarrow$$

$$(\lambda+1)\left(\lambda^{2}-4\right)=0 \Rightarrow$$

$$(\lambda+1)(\lambda+2)(\lambda-2)=0 \Rightarrow$$

$$\textcolor{orange}{ \lambda_{1}=-1, \quad \lambda_{2}=-2, \quad \lambda_{3}=2 }$$

$$\left(\lambda_{1} E-A\right) x=0 \Rightarrow\left(\lambda_{1} E-A\right)=$$

$$\left[\begin{array}{ccc}-2 & -1 & -1 \\ -2 & 0 & -1 \\ 0 & -1 & 0\end{array}\right] \Rightarrow$$

$$\left[\begin{array}{ccc}0 & -1 & 0 \\ -2 & 0 & -1 \\ 0 & 0 & 0\end{array}\right] \Rightarrow$$

$$\left[\begin{array}{lll}2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right] \Rightarrow$$

$$\textcolor{orange}{ \alpha_{1}=\left(\begin{array}{c}\frac{-1}{2} \\ 0 \\ 1\end{array}\right) }$$

$$\left(\lambda_{2} E-A\right)=\left[\begin{array}{ccc}-3 & -1 & -1 \\ -2 & -1 & -1 \\ 0 & -1 & -1\end{array}\right] \Rightarrow$$

$$\left[\begin{array}{ccc}-1 & 0 & 0 \\ -2 & 0 & 0 \\ 0 & 1 & 1\end{array}\right] \Rightarrow$$

$$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{array}\right] \Rightarrow$$

$$\textcolor{orange}{ \alpha_{2}=\left(\begin{array}{cc}0 \\ -1 \\ 1\end{array}\right) }$$

$$\left(\lambda_{3} E – A \right)=\left[\begin{array}{ccc}1 & -1 & -1 \\ -2 & 3 & -1 \\ 0 & -1 & 3\end{array}\right] \Rightarrow$$

$$\left[\begin{array}{ccc}1 & -1 & -1 \\ 0 & 1 & -3 \\ 0 & 0 & 0\end{array}\right] \Rightarrow$$

$$\left[\begin{array}{ccc}1 & 0 & -4 \\ 0 & 1 & -3 \\ 0 & 0 & 0\end{array}\right] \Rightarrow$$

$$\textcolor{orange}{ \alpha_{3}=\left(\begin{array}{l}4 \\ 3 \\ 1\end{array}\right) }$$

$$\textcolor{springgreen}{P} = \left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right) = \textcolor{springgreen}{ \left[\begin{array}{ccc}\frac{-1}{2} & 0 & 4 \\ 0 & -1 & 3 \\ 1 & 1 & 1\end{array}\right] } \Rightarrow$$

$$\textcolor{springgreen}{ P^{-1} A P} = \textcolor{springgreen}{ \left[\begin{array}{cc}-1 & & \\ & -2 & \\ & & 2 \end{array}\right] }$$