一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
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 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
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 $- \boldsymbol{E}$
高等数学![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
线性代数![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
以独特的视角解析线性代数,让繁复的知识变得直观明了。
特别专题![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。