你擅长做矩阵乘法，还是矩阵加减法？

一、题目

已知矩阵 $\boldsymbol{A}$ $=$ $\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & -1 \\
0 & 1 & 0
\end{bmatrix}$, 单位矩阵 $\boldsymbol{E}$ $=$ $\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}$, 则：

$$
\begin{vmatrix} \boldsymbol{E} – \boldsymbol{A}^{2} \end{vmatrix} = ?
$$

二、解析

解法 1：用矩阵乘法求解

首先，由题可得：

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

于是：

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

因此：

$$
\begin{vmatrix} \boldsymbol{E} – \boldsymbol{A}^{2} \end{vmatrix} = \textcolor{springgreen}{1}
$$

解法 2：用矩阵加减法求解

首先，根据行列式的性质可知：

$$
\begin{aligned}
\begin{vmatrix} \boldsymbol{E} – \boldsymbol{A}^{2} \end{vmatrix} & = \begin{vmatrix} \begin{pmatrix} \boldsymbol{E} – \boldsymbol{A} \end{pmatrix} \times \begin{pmatrix} \boldsymbol{E} + \boldsymbol{A} \end{pmatrix} \end{vmatrix} \\ \\
& = \begin{vmatrix} \boldsymbol{E} – \boldsymbol{A} \end{vmatrix} \times \begin{vmatrix} \boldsymbol{E} + \boldsymbol{A} \end{vmatrix} \end{aligned}
$$

又因为：

$$
\begin{aligned}
& \boldsymbol{E} – \boldsymbol{A} = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} – \begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & -1 \\
0 & 1 & 0
\end{bmatrix} = \begin{bmatrix}
1 & -1 & 0 \\
-1 & 1 & 1 \\
0 & -1 & 1
\end{bmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \begin{vmatrix} \boldsymbol{E} – \boldsymbol{A} \end{vmatrix} = \textcolor{magenta}{1} \\ \\
& \boldsymbol{E} + \boldsymbol{A} = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} + \begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & -1 \\
0 & 1 & 0
\end{bmatrix} = \begin{bmatrix}
1 & 1 & 0 \\
1 & 1 & -1 \\
0 & 1 & 1
\end{bmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \begin{vmatrix} \boldsymbol{E} + \boldsymbol{A} \end{vmatrix} = \textcolor{#00bffe}{1}
\end{aligned}
$$

因此：

$$
\begin{vmatrix} \boldsymbol{E} – \boldsymbol{A}^{2} \end{vmatrix} = \textcolor{magenta}{1} \times \textcolor{#00bffe}{1} = \textcolor{springgreen}{1}
$$

解法 3：用矩阵特征值求解

由于 $\boldsymbol{A}$ $=$ $\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & -1 \\
0 & 1 & 0
\end{bmatrix}$, 所以矩阵 $\boldsymbol{A}$ 的特征值为：

$$
\begin{aligned}
& \begin{vmatrix}
\lambda \boldsymbol{E} – \boldsymbol{A}
\end{vmatrix} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \begin{vmatrix}
\lambda & -1 & 0 \\
-1 & \lambda & 1 \\
0 & -1 & \lambda
\end{vmatrix} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \lambda^{3} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \lambda_{1} = 0, \ \lambda_{2} = 0, \ \lambda_{3} = 0
\end{aligned}
$$

接着，根据「荒原之梦考研数学」的《矩阵运算与特征值运算的映射关系》这篇文章可知，矩阵 $\boldsymbol{E} – \boldsymbol{A}^{2}$ 的特征值为：

$$
\lambda_{a} = 1, \ \lambda_{b} = 1, \ \lambda_{c} = 1
$$

因此：

$$
\begin{aligned}
\begin{vmatrix} \boldsymbol{E} – \boldsymbol{A}^{2} \end{vmatrix} & = \lambda_{a} \times \lambda_{b} \times \lambda_{c} \\ \\
& = 1 \times 1 \times 1 \\ \\
& = \textcolor{springgreen}{1}
\end{aligned}
$$

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