☰

在表格线和矩阵重叠中

tag-icon • tag-icon tables matrices line
在表格线和矩阵重叠中

我在表格中写入矩阵，但它离线太近了。打印出来时看起来很糟糕。

在此处输入图片描述

\documentclass[12pt]{beamer}
\usefonttheme{professionalfonts}
\usepackage{siunitx}
\usepackage{newtxtext,newtxmath}
\usepackage{bm}
\mode<presentation>{\usefonttheme{professionalfonts}}
\usepackage[turkish]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{tikz}
\usepackage[loadonly]{enumitem}
\usepackage{enumitem}
\makeatletter%matris boyu
\renewcommand*\env@matrix[1][\arraystretch]{%
    \edef\arraystretch{#1}%
    \hskip -\arraycolsep
    \let\@ifnextchar\new@ifnextchar
    \array{*\c@MaxMatrixCols c}}
\makeatother
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
        \node[shape=circle,draw,inner sep=2pt] (char) {#1};}}
    \begin{document}
    \begin{tabular}{ | c | c | c |}
            \hline
            Örnek1.2a: & Örnek 1.2b: & Örnek1.2c \\
            Tek çözüm & Çözümsüz & Sonsuz çözümlü \\ \hline
            $D_{A,b}=\begin{bmatrix}
            4&3&5\\
            10&6&4\\
            2&4&7\\
            2&5&8
            \end{bmatrix}$ & $D_{A,b}=\begin{bmatrix}
            4&3&5\\
            12&8&6\\
            2&4&7\\
            0&3&6
            \end{bmatrix}$ & $D_{A,b}=\begin{bmatrix}
            6&5&7\\
            13&9&7\\
            2&4&7\\
            1&4&7
            \end{bmatrix}$ \\ \hline
            $R_{A,b}=\begin{bmatrix}
            0&1&0\\
            0&0&1\\
            1&0&0\\
            1&0&0
            \end{bmatrix}$ & $R_{A,b}=\begin{bmatrix}
            0&1&1\\
            0&0&0\\
            0&0&0\\
            1&1&0
            \end{bmatrix}$ & $R_{A,b}=\begin{bmatrix}
            0&0&1\\
            0&0&1\\
            0&1&1\\
            1&1&1
            \end{bmatrix}$ \\ \hline    
        \end{tabular}
    \end{document}

答案1

像这样？

在此处输入图片描述

它是通过makecell宏包\setcellgapes{3pt}和\makegapedcells

\documentclass[12pt]{beamer}
\usefonttheme{professionalfonts}
\usepackage{siunitx}
\usepackage{newtxtext,newtxmath,bm}
\mode<presentation>{\usefonttheme{professionalfonts}}
%\usepackage[turkish]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{makecell}
\setcellgapes{3pt}
\usepackage{tikz}
%\usepackage[loadonly]{enumitem}
%\usepackage{enumitem}
\makeatletter%matris boyu
\renewcommand*\env@matrix[1][\arraystretch]{%
    \edef\arraystretch{#1}%
    \hskip -\arraycolsep
    \let\@ifnextchar\new@ifnextchar
    \array{*\c@MaxMatrixCols c}}
\makeatother
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
        \node[shape=circle,draw,inner sep=2pt] (char) {#1};}}


    \begin{document}
\begin{frame}
\makegapedcells
    \begin{tabular}{ | c | c | c |}
            \hline
            Örnek1.2a: & Örnek 1.2b: & Örnek1.2c \\
            Tek çözüm & Çözümsüz & Sonsuz çözümlü \\ \hline
            $D_{A,b}=\begin{bmatrix}
            4&3&5\\
            10&6&4\\
            2&4&7\\
            2&5&8
            \end{bmatrix}$ & $D_{A,b}=\begin{bmatrix}
            4&3&5\\
            12&8&6\\
            2&4&7\\
            0&3&6
            \end{bmatrix}$ & $D_{A,b}=\begin{bmatrix}
            6&5&7\\
            13&9&7\\
            2&4&7\\
            1&4&7
            \end{bmatrix}$ \\ \hline
            $R_{A,b}=\begin{bmatrix}
            0&1&0\\
            0&0&1\\
            1&0&0\\
            1&0&0
            \end{bmatrix}$ & $R_{A,b}=\begin{bmatrix}
            0&1&1\\
            0&0&0\\
            0&0&0\\
            1&1&0
            \end{bmatrix}$ & $R_{A,b}=\begin{bmatrix}
            0&0&1\\
            0&0&1\\
            0&1&1\\
            1&1&1
            \end{bmatrix}$ \\ \hline
        \end{tabular}
\end{frame}
    \end{document}

答案2

使用{NiceTabular}和nicematrix其键cell-space-limits。

\documentclass[12pt]{beamer}
\usefonttheme{professionalfonts}
\usepackage{siunitx}
\usepackage{newtxtext,newtxmath,bm}
\mode<presentation>{\usefonttheme{professionalfonts}}
\usepackage[T1]{fontenc}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{tikz}
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
        \node[shape=circle,draw,inner sep=2pt] (char) {#1};}}

\usepackage{nicematrix}

\begin{document}
\begin{frame}
    \begin{NiceTabular}{|c|c|c|}[cell-space-limits=4pt]
            \hline
            Örnek1.2a: & Örnek 1.2b: & Örnek1.2c \\
            Tek çözüm & Çözümsüz & Sonsuz çözümlü \\ \hline
            $D_{A,b}=\begin{bmatrix}
            4&3&5\\
            10&6&4\\
            2&4&7\\
            2&5&8
            \end{bmatrix}$ & $D_{A,b}=\begin{bmatrix}
            4&3&5\\
            12&8&6\\
            2&4&7\\
            0&3&6
            \end{bmatrix}$ & $D_{A,b}=\begin{bmatrix}
            6&5&7\\
            13&9&7\\
            2&4&7\\
            1&4&7
            \end{bmatrix}$ \\ \hline
            $R_{A,b}=\begin{bmatrix}
            0&1&0\\
            0&0&1\\
            1&0&0\\
            1&0&0
            \end{bmatrix}$ & $R_{A,b}=\begin{bmatrix}
            0&1&1\\
            0&0&0\\
            0&0&0\\
            1&1&0
            \end{bmatrix}$ & $R_{A,b}=\begin{bmatrix}
            0&0&1\\
            0&0&1\\
            0&1&1\\
            1&1&1
            \end{bmatrix}$ \\ \hline
        \end{NiceTabular}
\end{frame}
\end{document}

您需要多次编译（因为nicematrix在后台使用 PGF/Tikz 节点）。

上述代码的输出

答案3

还有一种可能性（由于原始答案已经被接受，因此答案是分开的），即带有密钥的tblr（新）tabularray包colsep：rowsep

\documentclass[12pt]{beamer}
\usefonttheme{professionalfonts}
\usepackage{newtxtext, newtxmath, bm}
\mode<presentation>{\usefonttheme{professionalfonts}}
\usepackage[turkish]{babel}
\usepackage[T1]{fontenc}
\usepackage{tabularray}

\begin{document}
\begin{frame}
\frametitle{Solution by use of the \texttt{tabularray} package}
    \centering
\begin{tblr}{hlines, vlines, 
             cells={c},
             colsep=4pt, rowsep=3pt
             }
{Örnek1.2a:\\ Tek çözüm}
    & {Örnek 1.2b:\\  Çözümsüz}
        & {Örnek1.2c\\ Sonsuz çözümlü}     \\ 
$D_{A,b}=\begin{bmatrix}
    4&3&5\\
    10&6&4\\
    2&4&7\\
    2&5&8
        \end{bmatrix}$ 
    & $D_{A,b}=\begin{bmatrix}
        4&3&5\\
        12&8&6\\
        2&4&7\\
        0&3&6
            \end{bmatrix}$ 
        & $D_{A,b}=\begin{bmatrix}
                6&5&7\\
                13&9&7\\
                2&4&7\\
                1&4&7
                \end{bmatrix}$ \\
$R_{A,b}=\begin{bmatrix}
    0&1&0\\
    0&0&1\\
    1&0&0\\
    1&0&0
        \end{bmatrix}$ 
    & $R_{A,b}=\begin{bmatrix}
        0&1&1\\
        0&0&0\\
        0&0&0\\
        1&1&0
            \end{bmatrix}$ 
        & $R_{A,b}=\begin{bmatrix}
            0&0&1\\
            0&0&1\\
            0&1&1\\
            1&1&1
            \end{bmatrix}$ \\ 
\end{tblr}
\end{frame}
    \end{document}

在此处输入图片描述

相关内容