一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
已知矩阵 $\boldsymbol{K}$ $=$ $\boldsymbol{A K}$ $+$ $\boldsymbol{B}$, 且:
$$
\begin{aligned}
\boldsymbol{A} & = \begin{bmatrix}
0 & 1 & 0 \\
– 1 & -1 & 0 \\
0 & 0 & 2
\end{bmatrix} \\ \\
\boldsymbol{B} & = \begin{bmatrix}
1 & – 1 \\
2 & 0 \\
3 & 1
\end{bmatrix}
\end{aligned}
$$
则 $\boldsymbol{K}$ $=$ $?$
难度评级:
二、解析 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
方法一:使用逆矩阵解题
如果先不考虑矩阵的可逆性,那么,由 $\boldsymbol{K}$ $=$ $\boldsymbol{A K}$ $+$ $\boldsymbol{B}$, 可得:
$$
\begin{aligned}
& \boldsymbol{K} – \boldsymbol{AK} = B \\
\Rightarrow & \boldsymbol{EK} – \boldsymbol{AK} = B \\
\Rightarrow & \left( \boldsymbol{E} – \boldsymbol{A} \right) \boldsymbol{K} = \boldsymbol{B} \\
\Rightarrow & \boldsymbol{K} = \left( \boldsymbol{E} – \boldsymbol{A} \right) ^{-1} \boldsymbol{B}
\end{aligned}
$$
所以,接下来我们要做的就是看看矩阵 $\boldsymbol{E} – \boldsymbol{A}$ 是否可逆。
由于:
$$
\begin{aligned}
\boldsymbol{E} – \boldsymbol{A} \\ \\
& = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} – \begin{bmatrix}
0 & 1 & 0 \\
– 1 & -1 & 0 \\
0 & 0 & 2
\end{bmatrix} \\ \\
& = \begin{bmatrix}
1 & -1 & 0 \\
1 & 2 & 0 \\
0 & 0 & -1
\end{bmatrix}
\end{aligned}
$$
且:
$$
\begin{aligned}
\begin{vmatrix} \boldsymbol{E} – \boldsymbol{A} \end{vmatrix} \\
& = -2-1 \\
& =-3 \\
& \neq 0
\end{aligned}
$$
于是可知,矩阵 $\boldsymbol{E} – \boldsymbol{A}$ 可逆,即下式成立:
$$
\boldsymbol{K} = \left( \boldsymbol{E} – \boldsymbol{A} \right) ^{-1} \boldsymbol{B}
$$
又根据荒原之梦考研数学的《初等变换求逆法 – zhaokaifeng.com》可知:
$$
\begin{aligned}
& (\boldsymbol{E} – \boldsymbol{A} \vdots E) \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 & \vdots & 1 & 0 & 0 \\
1 & 2 & 0 & \vdots & 0 & 1 & 0 \\
0 & 0 & -1 & \vdots & 0 & 0 & 1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 & \vdots & 1 & 0 & 0 \\
0 & 3 & 0 & \vdots & -1 & 1 & 0 \\
0 & 0 & -1 & \vdots & 0 & 0 & 1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 & \vdots & 1 & 0 & 0 \\
0 & 3 & 0 & \vdots & -1 & 1 & 0 \\
0 & 0 & 1 & \vdots & 0 & 0 & -1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 & \vdots & 1 & 0 & 0 \\
0 & 1 & 0 & \vdots & \frac{-1}{3} & \frac{1}{3} & 0 \\
0 & 0 & 1 & \vdots & 0 & 0 & -1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
1 & 0 & 0 & \vdots & \frac{2}{3} & \frac{1}{3} & 0 \\
0 & 1 & 0 & \vdots & \frac{-1}{3} & \frac{1}{3} & 0 \\
0 & 0 & 1 & \vdots & 0 & 0 & -1
\end{bmatrix} \\ \\
\end{aligned}
$$
所以:
$$
(\boldsymbol{E} – \boldsymbol{A})^{-1} = \begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\ \\
\frac{-1}{3} & \frac{1}{3} & 0 \\ \\
0 & 0 & -1
\end{bmatrix}
$$
于是:
$$
\begin{aligned}
& \boldsymbol{K} \\ \\
= & \left( \boldsymbol{E} – \boldsymbol{A} \right) ^{-1} \boldsymbol{B} \\ \\
\Rightarrow & \begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\ \\
\frac{-1}{3} & \frac{1}{3} & 0 \\ \\
0 & 0 & -1
\end{bmatrix} \begin{bmatrix}
1 & – 1 \\
2 & 0 \\
3 & 1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
\frac{4}{3} & \frac{-2}{3} \\ \\
\frac{1}{3} & \frac{1}{3} \\ \\
-3 & -1
\end{bmatrix}
\end{aligned}
$$
综上可知:
$$
\textcolor{springgreen}{
\boldsymbol{K} = ( \boldsymbol{\alpha_{1}}, \boldsymbol{\alpha_{2}} ) = \begin{bmatrix}
\frac{4}{3} & \frac{-2}{3} \\ \\
\frac{1}{3} & \frac{1}{3} \\ \\
-3 & -1
\end{bmatrix}
}
$$
方法二:使用方程组解题
由题目已知条件,可得:
$$
\begin{aligned}
& \boldsymbol{K} = \boldsymbol{A K} + \boldsymbol{B} \\ \\
\Rightarrow & \boldsymbol{K} – \boldsymbol{AK} = \boldsymbol{B} \\ \\
\Rightarrow & (\boldsymbol{E} – \boldsymbol{A}) \boldsymbol{K} = \boldsymbol{B}
\end{aligned}
$$
如果我们将 $(\boldsymbol{E} – \boldsymbol{A}) \boldsymbol{K}$ $=$ $\boldsymbol{B}$ 看作是一个方程组,那么,我们要求解的矩阵 $\boldsymbol{K}$ 其实就是这个方程组的未知数。
分析可知,$\boldsymbol{K}$ 和 $\boldsymbol{B}$ 都是三行两列的矩阵,因此,如果我们设列向量 $\boldsymbol{\alpha _{1}}$ $=$ $\begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{bmatrix}$, $\boldsymbol{\alpha_{2}}$ $=$ $\begin{bmatrix}
y_{1} \\
y_{2} \\
y_{3}
\end{bmatrix}$, $\boldsymbol{\beta_{1}}$ $=$ $\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}$, $\boldsymbol{\beta_{2}}$ $=$ $\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}$, 且令:
$$
\begin{aligned}
\boldsymbol{K} & = (\boldsymbol{\alpha_{1}}, \boldsymbol{\alpha_{2}}) \\
\boldsymbol{B} & = (\boldsymbol{\beta_{1}}, \boldsymbol{\beta_{2}})
\end{aligned}
$$
由于矩阵 $\boldsymbol{B}$ 是已知的,因此,要求解 $(\boldsymbol{E} – \boldsymbol{A}) \boldsymbol{K}$ $=$ $\boldsymbol{B}$ 中的矩阵 $\boldsymbol{K}$, 其实就是分别求解下面两个方程组中的矩阵 $\boldsymbol{\alpha_{1}}$ 和 $\boldsymbol{\alpha_{2}}$:
$$
\begin{aligned}
(\boldsymbol{E} – \boldsymbol{A}) \boldsymbol{\alpha_{1}} & = \boldsymbol{\beta_{1}} \\
(\boldsymbol{E} – \boldsymbol{A}) \boldsymbol{\alpha_{2}} & = \boldsymbol{\beta_{2}}
\end{aligned}
$$
又因为:
$$
\boldsymbol{E} – \boldsymbol{A} = \begin{bmatrix}
1 & -1 & 0 \\
1 & 2 & 0 \\
0 & 0 & -1
\end{bmatrix}
$$
于是:
$$
\begin{aligned}
& (\boldsymbol{E} – \boldsymbol{A}) \boldsymbol{\alpha_{1}} = \boldsymbol{\beta_{1}} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 \\
1 & 2 & 0 \\
0 & 0 & -1
\end{bmatrix} \begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{bmatrix} = \begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}
\end{aligned}
$$
接着,有:
$$
\begin{aligned}
& \begin{bmatrix}
1 & -1 & 0 & \vdots & 1 \\
1 & 2 & 0 & \vdots & 2 \\
0 & 0 & -1 & \vdots & 3
\end{bmatrix} \\ \\
= & \begin{bmatrix}
1 & -1 & 0 & \vdots & 1 \\
0 & 3 & 0 & \vdots & 1 \\
0 & 0 & -1 & \vdots & 3
\end{bmatrix} \\ \\
= & \begin{bmatrix}
1 & -1 & 0 & \vdots & 1 \\
0 & 1 & 0 & \vdots & \frac{1}{3} \\
0 & 0 & 1 & \vdots & -3
\end{bmatrix} \\ \\
= & \begin{bmatrix}
\textcolor{black}{\colorbox{pink}{1}} & 0 & 0 & \vdots & \textcolor{springgreen}{\frac{4}{3}} \\
0 & \textcolor{black}{\colorbox{pink}{1}} & 0 & \vdots & \textcolor{springgreen}{\frac{1}{3}} \\
0 & 0 & \textcolor{black}{\colorbox{pink}{1}} & \vdots & \textcolor{springgreen}{-3}
\end{bmatrix}
\end{aligned}
$$
于是可知:
$$
\textcolor{springgreen}{
\boldsymbol{\alpha_{1}} = \begin{bmatrix}
\frac{4}{3} \\ \\
\frac{1}{3} \\ \\
-3
\end{bmatrix}
}
$$
Note
在利用初等行变换化简线性方程组 $\boldsymbol{A} \boldsymbol{x}$ $=$ $\boldsymbol{\beta}$ 中的矩阵时,一般不是只对系数矩阵。
zhaokaifeng.com
类似的:
$$
\begin{aligned}
& (\boldsymbol{E} – \boldsymbol{A}) \boldsymbol{\alpha_{2}} = \boldsymbol{\beta_{2}} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 \\
1 & 2 & 0 \\
0 & 0 & -1
\end{bmatrix} \begin{bmatrix}
y_{1} \\
y_{2} \\
y_{3}
\end{bmatrix} = \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}
\end{aligned}
$$
接着,有:
$$
\begin{aligned}
& \begin{bmatrix}
1 & -1 & 0 & \vdots & -1 \\
1 & 2 & 0 & \vdots & 0 \\
0 & 0 & -1 & \vdots & 1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 & \vdots & -1 \\
0 & 3 & 0 & \vdots & 1 \\
0 & 0 & -1 & \vdots & 1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 & \vdots & -1 \\
0 & 3 & 0 & \vdots & 1 \\
0 & 0 & 1 & \vdots & -1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
1 & -1 & 0 & \vdots & -1 \\
0 & 1 & 0 & \vdots & \frac{1}{3} \\
0 & 0 & 1 & \vdots & -1
\end{bmatrix} \\ \\
\Rightarrow & \begin{bmatrix}
\textcolor{black}{\colorbox{pink}{1}} & 0 & 0 & \vdots & \textcolor{springgreen}{\frac{-2}{3}} \\
0 & \textcolor{black}{\colorbox{pink}{1}} & 0 & \vdots & \textcolor{springgreen}{\frac{1}{3}} \\
0 & 0 & \textcolor{black}{\colorbox{pink}{1}} & \vdots & \textcolor{springgreen}{-1}
\end{bmatrix}
\end{aligned}
$$
于是可知:
$$
\textcolor{springgreen}{
\boldsymbol{\alpha_{2}} = \begin{bmatrix}
\frac{-2}{3} \\ \\
\frac{1}{3} \\ \\
-1
\end{bmatrix}
}
$$
综上可知:
$$
\textcolor{springgreen}{
\boldsymbol{K} = ( \boldsymbol{\alpha_{1}}, \boldsymbol{\alpha_{2}} ) = \begin{bmatrix}
\frac{4}{3} & \frac{-2}{3} \\ \\
\frac{1}{3} & \frac{1}{3} \\ \\
-3 & -1
\end{bmatrix}
}
$$
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。