使用数组使群组演示看起来更好

Question

我写过一个类似的演示文稿，但我想找到一种方法让它看起来更好。

$$\left\langle \begin{array}[t]{l|cl}
A_{13}, A_{32}, A_{21} &(A_{13}-t_{13,1})(A_{13}-t_{13,2})=0&\\
&(A_{32}-t_{32,1})(A_{32}-t_{32,2})(A_{32}-t_{32,3})=0& A_{13}A_{32}A_{23}=1\\
&(A_{23}-t_{23,1})(A_{23}-t_{23,2})(A_{23}-t_{23,3})=0&
\end{array}
 \right\rangle$$

Answer 1

也许像这样的事情会看起来更好：

\documentclass{beamer}
\usepackage{amsmath}
\begin{document}
  \begin{frame}
    \[\Biggl\langle 
       \begin{array}{l|cl}
                        & \mathcal{B} = 0 \\
            \mathcal{A} & \mathcal{C} =0  & \mathcal{D}=1 \\
                        & \mathcal{E} =0  &                                             
        \end{array}
     \Biggr\rangle\]
where
\[
\begin{aligned}
  \mathcal{A} &= A_{13}, A_{32}, A_{21}\\
  \mathcal{B} &= (A_{13}-t_{13,1})(A_{13}-t_{13,2})\\
  \mathcal{C} &= (A_{32}-t_{32,1})(A_{32}-t_{32,2})(A_{32}-t_{32,3})\\
  \mathcal{D} &= A_{13}A_{32}A_{23}\\
  \mathcal{E} &= (A_{23}-t_{23,1})(A_{23}-t_{23,2})(A_{23}-t_{23,3})
\end{aligned}
\]
  \end{frame}
\end{document}