我有一份关于线性代数的大型文档（共 7 章）。在文档中我进行行/列运算的部分，我将矩阵放在小页面内的图形中。我在一组“矩阵行运算”和下一组“矩阵行运算”之间获得了大量垂直空间。我想知道是否可以控制垂直空间。我将我的 MWE 作为两个单独的文档（序言和章节）发布在下方。我很抱歉在序言中包含了可能不相关的包。
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%
我们如何找到 P？\ 答案:\ 通过基本行运算将 $(A_{3x4}|I_3)$ 简化为行标准形式
\begin{figure}[H]
\centering
\begin{minipage}{0.3\linewidth}
\begin{equation*}
\begin{pmatrix}[cccc|ccc]
2 & -1 & 0 & 1 & 1 & 0 & 0\\
1 & 2 & 1 & -1 & 0 & 1 & 0\\
2 & 9 & 4 & -5 & 0 & 0 & 1
\end{pmatrix}
\end{equation*}
\end{minipage}
\begin{minipage}{0.3\linewidth}
$\rowops {,R_1\rightarrow R_1-2r_2,R_3\rightarrow R_3-2R_2}$
\end{minipage}%
\begin{minipage}{0.3\linewidth}
\begin{equation*}
\begin{pmatrix}[cccc|ccc]
0 & -5 & -2 & 3 & 1 & -2 & 0\\
1 & 2 & 1 & -1 & 0 & 1 & 0\\
0 & 5 & 2 & -3 & 0 & -2 & 1\\
\end{pmatrix}
\end{equation*}
\end{minipage}
\end{figure}
\begin{figure}[H]
\centering
\begin{minipage}{0.33\linewidth}
\begin{equation*}
\begin{pmatrix}[cccc|ccc]
0 & -5 & -2 & 3 & 1 & -2 & 0\\
1 & 2 & 1 & -1 & 0 & 1 & 0\\
0 & 5 & 2 & -3 & 0 & -2 & 1\\
\end{pmatrix}
\end{equation*}
\end{minipage}
\begin{minipage}{0.33\linewidth}
$\rowops {,R_3\rightarrow R_3-R_1,R_1\rightarrow\frac{1}{5}R_1}$
\end{minipage}%
\begin{minipage}{0.33\linewidth}
\begin{equation*}
\begin{pmatrix}[cccc|ccc]
0 & 1 &\frac{2}{5} & -\frac{3}{5} & -\frac{1}{5} &\frac{2}{5} & 0\\
1 & 2 & 1 & -1 & 0 & 1 & 0\\1 & 2 & 1 & -1 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1 & -4 & 1
\end{pmatrix}
\end{equation*}
\end{minipage}
\end{figure}
\begin{figure}[H]
\centering
\begin{minipage}{0.36\textwidth}
\begin{equation*}
\begin{pmatrix}[cccc|ccc]
0 & 1 &\frac{2}{5} & -\frac{3}{5} & -\frac{1}{5} &\frac{2}{5} & 0\\
1 & 2 & 1 & -1 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1 & -4 & 1
\end{pmatrix}
\end{equation*}
\end{equation*}
\end{minipage}%
\begin{minipage}{0.36\textwidth}
\hspace{2ex}$\rowops {,R_2\leftrightarrow R_1,R_1\rightarrow R_1-2R_2}$
\end{minipage}%
\begin{minipage}{0.36\textwidth}
\begin{equation*}
\begin{pmatrix}[cccc|ccc]
1 & 0 &\frac{1}{5} & -\frac{1}{5} & -\frac{2}{5} &\frac{1}{5} & 0\\
0 & 1 &\frac{2}{5} & -\frac{3}{5} & -\frac{1}{5} &\frac{2}{5} & 0\\
0 & 0 & 0 & 0 & 1 & -4 & 1
\end{pmatrix}
\end{equation*}
\end{minipage}
\end{figure}
\medskip
因此
$P=\begin{pmatrix*}[r]
-\frac{2}{5} &\frac{1}{5} & 0\\
-\frac{1}{5} &\frac{2}{5} & 0\\
1 & -4 & 1
\end{pmatrix*}$\\
和
$PA=\begin{pmatrix*}[r]
1 & 0 &\frac{2}{5} &\frac{1}{5}\\
0 & 1 &\frac{2}{5} & -\frac{3}{5}\\
0 & 0 & 0 & 0
\end{pmatrix*}$\\