# 2015年考研数二第22题解析：矩阵、逆矩阵

## 题目

$(Ⅰ)$ 求 $a$ 的值；

$(Ⅱ)$ 若矩阵 $X$ 满足 $X -$ $XA^{2} -$ $AX +$ $AXA^{2} =$ $E$, 其中 $E$ 为 $3$ 阶单位矩阵，求 $X$.

## 解析

### 第 $(Ⅰ)$ 问

$$A^{3} = O \Rightarrow$$

$$A \cdot A \cdot A = O \Rightarrow$$

$$|A \cdot A \cdot A| = 0 \Rightarrow$$

$$|A| |A| |A| = 0 \Rightarrow$$

$$|A| = 0 \Rightarrow$$

$$a^{3} – a + a = 0\Rightarrow$$

$$a = 0.$$

### 第 $(Ⅱ)$ 问

$$A = \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{bmatrix}.$$

$$X – XA^{2} – AX + AXA^{2} = E \Rightarrow$$

$$X(E-A^{2}) – AX(E – A^{2}) = E \Rightarrow$$

$$(X-AX)(E – A^{2}) = E \Rightarrow$$

$$(EX-AX)(E – A^{2}) =E \Rightarrow$$

$$(E-A)X(E – A^{2})=E \Rightarrow$$

$$X = (E-A)^{-1}E(E – A^{2})^{-1} \Rightarrow$$

$$X = (E-A)^{-1}(E – A^{2})^{-1}.$$

$$E-A \Rightarrow$$

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}- \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{bmatrix} \Rightarrow$$

$$E-A = \begin{bmatrix} 1 & -1 & 0\\ -1 & 1 & 1\\ 0 & -1 & 1 \end{bmatrix}.$$

$$A^{2} \Rightarrow$$

$$\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{bmatrix} \Rightarrow$$

$$A^{2} = \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & -1 \end{bmatrix}.$$

$$E-A^{2} \Rightarrow$$

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}- \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & -1 \end{bmatrix} \Rightarrow$$

$$E-A^{2} = \begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ -1 & 0 & 2 \end{bmatrix}.$$

$$((E-A) | E) \Rightarrow$$

$$\begin{bmatrix} 1 & -1 & 0 & \vdots & 1 & 0 & 0\\ -1 & 1 & 1 & \vdots & 0 & 1 & 0\\ 0 & -1 & 1 & \vdots & 0 & 0 & 1 \end{bmatrix} \Rightarrow 初等行变换 \Rightarrow$$

$$(E | (E-A)^{-1}) = \begin{bmatrix} 1 & 0 & 0 & \vdots & 2 & 1 & -1\\ 0 & 1 & 0 & \vdots & 1 & 1 & -1\\ 0 & 0 & 1 & \vdots & 1 & 1 & 0 \end{bmatrix} \Rightarrow$$

$$(E-A)^{-1} = \begin{bmatrix} 2 & 1 & -1\\ 1 & 1 & -1\\ 1 & 1 & 0 \end{bmatrix}.$$

$$((E-A^{2}) | E) \Rightarrow$$

$$\begin{bmatrix} 0 & 0 & 1 & \vdots & 1 & 0 & 0\\ 0 & 1 & 0 & \vdots & 0 & 1 & 0\\ -1 & 0 & 2 & \vdots & 0 & 0 & 1 \end{bmatrix} \Rightarrow 初等行变换 \Rightarrow$$

$$(E | (E-A^{2})^{-1}) = \begin{bmatrix} 1 & 0 & 0 & \vdots & 2 & 0 & -1\\ 0 & 1 & 0 & \vdots & 0 & 1 & 0\\ 0 & 0 & 1 & \vdots & 1 & 0 & 0 \end{bmatrix} \Rightarrow$$

$$(E-A^{2})^{-1} = \begin{bmatrix} 2 & 0 & -1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{bmatrix}$$

$$X = (E-A)^{-1} (E-A^{2})^{-1} \Rightarrow$$

$$X = \begin{bmatrix} 2 & 1 & -1\\ 1 & 1 & -1\\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & -1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{bmatrix} \Rightarrow$$

$$X = \begin{bmatrix} 3 & 1 & -2\\ 1 & 1 & -1\\ 2 & 1 & -1 \end{bmatrix}.$$