2015年考研数二第22题解析:矩阵、逆矩阵

题目

设矩阵 $A = \begin{bmatrix}
a & 1 & 0\\
1 & a & -1\\
0 & 1 & a
\end{bmatrix}$, 且 $A^{3} = O$.

$(Ⅰ)$ 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

由题可知:

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

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

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

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

$$
|A| = 0 \Rightarrow
$$

$$
a^{3} – a + a = 0\Rightarrow
$$

$$
a = 0.
$$

第 $(Ⅱ)$ 问

由第 $(Ⅰ)$ 问可知:

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

又:

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

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

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

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

$$
(E-A)X(E – A^{2})=E \Rightarrow
$$

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

$$
X = (E-A)^{-1}(E – A^{2})^{-1}.
$$

接着:

$$
E-A \Rightarrow
$$

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

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

$$
A^{2} \Rightarrow
$$

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

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

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

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

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

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

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

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

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

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

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

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

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

于是:

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

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

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