El problema
Oye, pasé unas 3 horas intentando resolver este problema multicolen LaTeX. Espero que ayude a algunas personas:corte de matrices en LaTeX
Como puede ver, mis matrices en el \align*entorno están cortadas en el multicolentorno que se veía bien en el \align*entorno:Matrices en un align*entorno.
Aquí está mi código:
\begin{align*}...\end{align*}sin cortar:
\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}
\multicolcon el corte:
\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}
La solución
Si su solución LaTeX compilada es dividir el largo \begin{align*}...\end{align*}en unos pocos `\begin{align*}....\end{align*}' entonces tendrá algo como:
\begin{multicol}{2}
...
\begin{align*}...\end{align*}
\begin{align*}...\end{align*}
\begin{align*}...\end{align*}
...
\end{multicol}
Entonces, código correcto:
\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}