实际上我想让这两个矩阵并排放置,代码如下:
\immediate\write18{makeindex \jobname.nlo -s nomencl.ist -o \jobname.nls}
\documentclass[10pt]{elsarticle}
\DeclareGraphicsExtensions{.pdf,.gif,.jpg}
\usepackage{lineno}
\modulolinenumbers[5]
\usepackage{graphicx}
\usepackage{amssymb,amsmath,nccmath}
\usepackage{cclicenses}
\usepackage{makecell}
\usepackage{lscape,array}
\newcolumntype{C}[1]{>{\centering\arraybackslash}p{#1}}
\usepackage[thin, , thinc]{esdiff}
\usepackage{subcaption}
\usepackage{caption}
\usepackage{framed}
\usepackage{nomencl}
\usepackage[hmargin=2.6 cm,vmargin=3.25cm]{geometry}
\usepackage{setspace}\doublespacing
\usepackage[font=small,skip=0pt]{caption}
\makenomenclature
\setlength{\nomitemsep}{-\parskip}\usepackage{booktabs,multirow}
\renewcommand{\cellset}{\renewcommand{\arraystretch}{0.5}}
\journal{Journal of \LaTeX\ Templates}
\bibliographystyle{elsarticle-num}
\begin{document}
\begin{align*}
\begin{bmatrix}
1
\end{bmatrix}
=
\begin{bmatrix}
\mfrac{A}{B} & E & \mfrac{C}{D} & 0 \\
-K & \mfrac{-Z}{Y} & T & \mfrac{W}{V} \\
\mfrac{XY}{MOP}& -\mfrac{US}{TV} & \mfrac{1}{B}(\mfrac{MN}{AB}-\mfrac{LA}
{BC}) & RR \\
\mfrac{A}{B} & \mfrac{C}{D} & -E & \mfrac{1}{C}(\mfrac{DE}{GH}-\mfrac{IJ}
{KL}) \\
\end{bmatrix}
\end{align*}
\begin{align*}
\begin{bmatrix}
2
\end{bmatrix}
=
\begin{bmatrix}
A & 0 & 0 & 0\\
0 & A & 0 & 0\\
\mfrac{AA}{BB} & 0 & \mfrac{1}{CCr} & 0 \\
0 & \mfrac{-DE}{BCD} & 0 & \mfrac{1}{FG}
\end{bmatrix}
\end{align*}\par
\end{document}
答案1
与建议的非常相似土拨鼠在上述评论中:
\documentclass[10pt]{elsarticle}
\usepackage[hmargin=2.6 cm,
vmargin=3.25cm]{geometry}
\usepackage{amssymb, nccmath}
\begin{document}
\[\renewcommand\arraystretch{2}
[1] = \begin{bmatrix}
\mfrac{A}{B} & E & \mfrac{C}{D} & 0 \\
-K & \mfrac{-Z}{Y} & T & \mfrac{W}{V} \\
\mfrac{XY}{MOP} & -\mfrac{US}{TV} & \mfrac{1}{B}\left(\mfrac{MN}{AB}-\mfrac{LA}{BC}\right)
& RR \\
\mfrac{A}{B} & \mfrac{C}{D} & -E & \mfrac{1}{C}\left(\mfrac{DE}{GH}-\mfrac{IJ}{KL}\right) \\
\end{bmatrix},
\qquad
[2] = \begin{bmatrix}
A & 0 & 0 & 0 \\
0 & A & 0 & 0 \\
\mfrac{AA}{BB} & 0 & \mfrac{1}{CCr} & 0 \\
0 & \mfrac{-DE}{BCD} & 0 & \mfrac{1}{FG}
\end{bmatrix}
\]