# 借助二次方程求解未知矩阵

## 一、题目

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

$$\boldsymbol{B} = ?$$

## 二、解析

\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}

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

\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}