借助二次方程求解未知矩阵

一、题目

已知 $n$ 阶矩阵 $\boldsymbol{A}$ 和 $\boldsymbol{B}$ 满足：

$$
\begin{cases}
\boldsymbol{A} = \frac{1}{3} (\boldsymbol{B} + \boldsymbol{E}) \\ \\
\boldsymbol{A} ^{2} = \boldsymbol{A}
\end{cases}
$$

则：

$$
\boldsymbol{B} = ?
$$

难度评级：

二、解析

首先，由题可知：

$$
\begin{aligned}
& \boldsymbol{A} ^{2} = \boldsymbol{A} \\ \\
\Rightarrow & \left[ \frac{1}{3} (\boldsymbol{B} + \boldsymbol{E}) \right] ^{2} = \frac{1}{3} (\boldsymbol{B} + \boldsymbol{E}) \\ \\
\Rightarrow & \frac{1}{9} (\boldsymbol{B} ^{2} + 2 \boldsymbol{B} + \boldsymbol{E}) = \frac{1}{3} (\boldsymbol{B} + \boldsymbol{E}) \\ \\
\Rightarrow & \frac{1}{9} (\boldsymbol{B} ^{2} + 2 \boldsymbol{B} + \boldsymbol{E}) = \frac{1}{9} (3 \boldsymbol{B} + 3 \boldsymbol{E}) \\ \\
\Rightarrow & \boldsymbol{B} ^{2} + 2 \boldsymbol{B} + \boldsymbol{E} = 3 \boldsymbol{B} + 3 \boldsymbol{E} \\ \\
\Rightarrow & \textcolor{orange}{ \boldsymbol{B} ^{2} = \boldsymbol{B} + 2 \boldsymbol{E} }
\end{aligned}
$$

接着，我们借助方程的思想，令：

$$
\textcolor{yellow}{
\textcolor{red}{\Large{\boldsymbol{\star}}}
\begin{cases}
\boldsymbol{B} = x \\
\boldsymbol{E} = 1
\end{cases}
}
$$

则：

$$
\begin{aligned}
& \textcolor{orange}{ \boldsymbol{B} ^{2} = \boldsymbol{B} + 2 \boldsymbol{E} } \\ \\
\Rightarrow & x ^{2} = x + 2 \\ \\
\Rightarrow & x ^{2} – x – 2 = 0 \\ \\
\Rightarrow & \begin{cases}
a = 1 \\
b = -1 \\
c = -2
\end{cases} \\ \\
\Rightarrow & x = \frac{-b \pm \sqrt{b ^{2} – 4ac}}{2a} \\ \\
\Rightarrow & x = \frac{1 \pm \sqrt{1 + 8}}{2} \\ \\
\Rightarrow & x = \frac{1 \pm 3}{2} \\ \\
\Rightarrow & \begin{cases}
x = 2 \\
x = -1
\end{cases} \\ \\
\Rightarrow & \textcolor{springgreen}{\begin{cases}
\boldsymbol{B} = 2 \boldsymbol{E} \\
\boldsymbol{B} = – \boldsymbol{E}
\end{cases} }
\end{aligned}
$$

综上可知，$n$ 阶矩阵 $\boldsymbol{B}$ 等于 $2 \boldsymbol{E}$ 或者等于 $- \boldsymbol{E}$

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