答案1
你可以这样做blkarray
:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fourier, erewhon} \usepackage{multirow,bigdelim}
\usepackage{mathtools, blkarray}
\begin{document} \[ \everymath{\displaystyle}
A = \begin{blockarray}{cccccccc}
& K & & L₁ & & L₂ \\[1ex] %
\begin{block}{(c@{}ccccc@{\;}c)c}
\dotsm & ∑_{\smash[b]{\substack{\epsilon_{K, L} ⊂ \partial K\\\epsilon_{K, L}\not ⊂ \partialΩ}}}\frac{\lvert\epsilon_{K, L}\rvert} {d_{K, L}} & \dotsm & -\frac{\lvert\epsilon_{K, L₁}\rvert}{d_{K, L₁}} & \dotsm & -\frac{\lvert\epsilon_{K, L₂}\rvert}{d_{K, L₂}} & \dotsm & K\\ %
& ⋮ & & ⋮ & & ⋮ & & \\%
\dotsm & -\frac{\lvert\epsilon_{L₁,K}\rvert}{d_{L₁,K}} & \dotsm & \smashoperator[l]{∑_{\smash[b]{\substack{\epsilon_{L₁, L} ⊂ \partial L₁ \\ \epsilon_{L₁, L}\not ⊂ \partialΩ}}}}\mkern-8mu \frac{\lvert\epsilon_{L₁, L}\rvert} {d_{L₁, L}} & \dotsm & -\frac{\lvert\epsilon_{L₁, L₂}\rvert}{d_{L₁, L₂}} & \dotsm & L₁ \\ %%
& ⋮ & & ⋮ & & ⋮ & & \\%
\dotsm & -\frac{\lvert\epsilon_{L₂, K}\rvert}{d_{L₂, K}} & \dotsm & -\frac{\lvert\epsilon_{L₂, L₁}\rvert}{d_{L₂, L₁}} & \dotsm & \smashoperator[l]{∑_{\substack{\epsilon_{L₂, L} ⊂ \partial L₂ \\ \epsilon_{L₂, L}\not ⊂ \partialΩ}}}\mkern-8mu \frac{\lvert\epsilon_{L₂, L}\rvert} {d_{L₂, L}} & \dotsm & L₂ \\%
\end{block}
\end{blockarray} \]
\end{document}