问题
嘿，我花了大约 3 个小时尝试用multicolLaTeX 解决这个问题。希望它能帮助一些人：LaTeX 中的数组截断
正如您所看到的，我的环境中的数组在环境中看起来良好的环境中\align*被切断了：multicol\align*align*环境中的阵列
这是我的代码：
\begin{align*}...\end{align*}无切断：
\documentclass[11pt]{article}
\usepackage{header}
\usepackage{pgfplots}
\allowdisplaybreaks
\renewcommand{\arraystretch}{1.5}
\begin{document}
\begin{align*}
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 1 & \frac{9}{5} & 0 & \frac{111}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_3-\frac{9}{5}R_1$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 1 & 0 & \frac{3}{2} & \frac{216}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_3-R_1$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 0 & 0 & \frac{1}{2} & \frac{206}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{Normalizing $R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_2-R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_1+\frac{5}{6}R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & 0 & \frac{171}{3} \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{Rearranging into Row Echelon Form} \\
\left[\begin{array}{cccc|c}
1 & 0 & 0 & 0 & 19 \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 1 & 0 & \frac{171}{3} \\
0 & 0 & 0 & 1 & \frac{412}{5}
\end{array}\right] \\
\end{align*}
\end{document}
\multicol截止时间为：
\documentclass[11pt]{article}
\usepackage{header}
\usepackage{pgfplots}
\usepackage{multicol}
\allowdisplaybreaks
\setlength{\columnsep}{0cm}
\setlength{\columnseprule}{0.4pt}
\renewcommand{\arraystretch}{1.5}
\begin{document}
\begin{multicol*}{2}
\begin{align*}
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 1 & \frac{9}{5} & 0 & \frac{111}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_3-\frac{9}{5}R_1$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 1 & 0 & \frac{3}{2} & \frac{216}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_3-R_1$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 0 & 0 & \frac{1}{2} & \frac{206}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{Normalizing $R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_2-R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_1+\frac{5}{6}R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & 0 & \frac{171}{3} \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{Rearranging into Row Echelon Form} \\
\left[\begin{array}{cccc|c}
1 & 0 & 0 & 0 & 19 \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 1 & 0 & \frac{171}{3} \\
0 & 0 & 0 & 1 & \frac{412}{5}
\end{array}\right] \\
\end{align*}
\end{multicol*}
\end{document}
解决方案
如果你编译的 LaTeX 解决方案是将长文本拆分\begin{align*}...\end{align*}成几个`\begin{align*}...\end{align*}'，那么你会得到类似这样的内容：
\begin{multicol}{2}
...
\begin{align*}...\end{align*}
\begin{align*}...\end{align*}
\begin{align*}...\end{align*}
...
\end{multicol}
因此，正确的代码：
\begin{multicol*}{2}
\begin{align*}
\text{Simplifying} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 1 & \frac{9}{5} & 0 & \frac{111}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_3-\frac{9}{5}R_1$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 1 & 0 & \frac{3}{2} & \frac{216}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\end{align*}
\begin{align*}
\text{$R_3-R_1$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 0 & 0 & \frac{1}{2} & \frac{206}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{Normalizing $R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{$R_2-R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & -\frac{5}{6} & -\frac{35}{3} \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\end{align*}
\begin{align*}
\text{$R_1+\frac{5}{6}R_3$} \\
\left[\begin{array}{cccc|c}
0 & 0 & 1 & 0 & \frac{171}{3} \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 0 & 1 & \frac{412}{5} \\
1 & 0 & 0 & 0 & 19
\end{array}\right] \\ \\
\text{Rearranging into Row Echelon Form} \\
\left[\begin{array}{cccc|c}
1 & 0 & 0 & 0 & 19 \\
0 & 1 & 0 & 0 & -\frac{402}{5} \\
0 & 0 & 1 & 0 & \frac{171}{3} \\
0 & 0 & 0 & 1 & \frac{412}{5}
\end{array}\right] \\
\end{align*}
\end{multicols*}
\end{document}