如何在 LaTex 中组织表格？

Question

你能帮忙整理一下下面的表格吗？表格的最后两行很乱。

\begin{center}
\begin{tabular}{ ||c|c|| } 
    \hline
    \hline
    &\\
    Singular matrix    & Non-singular matrix   \\&\\
    \hline
    \hline
    &\\
     A matrix $A$ is singular if 
    $det(A)=0$ 
      & A matrix $A$ is non-singular 
      $det(A)\ne0$
      \\ &\\
    \hline
    &\\
    If $A$ is singular, then $A^{-1}$ is not defined  & If $A$ is non-singular, then $A^{-1}$ is defined \\&\\
    \hline
    &\\
    Rank of $A <$ order of $ A$. &Rank of $A=$ order of $A$. \\&\\
    \hline
    &\\
    Some rows and columns are linearly dependent. & All rows and columns are linearly independent  \\&\\
    \hline
    &\\
    If $A$ singular, then the system of equations\\ $A\bx=\bb$
    has either no solution or\\infinitely many solutions.
    & If A non-singular, then the system of equations\\& $A\bx=\bb$ has unique solutions.
     \\&\\
    \hline
    &\\
    Example: $
    \left[\begin{array}{ll}
    3 & 6  \\
    1 & 2 \\
    \end{array}\right]
    $ is singular because
    \\
    $\left|\begin{array}{ll}3 & 6 \\ 1 & 2\end{array}\right|=(3 \times 2)-(6 \times 1)$
    $=6-6=0$
     & Example: $
     \left[\begin{array}{ll}
     3 & 2  \\
     1 & -2 \\
     \end{array}\right]
     $ is non-singular because
     \\&
     $\left|\begin{array}{ll}3 & 2 \\ 1 & -2\end{array}\right|=(3 \times (-2))-(2 \times 1)$
     $=-8\ne0$&\\
    \hline
    \hline
\end{tabular}
\end{center}

Answer 1

双线与其他单/双线的交互始终存在问题，正如您在表格角落中看到的那样。要制作具有单双规则的“监禁”表格，请考虑使用包hhlines：

为了制作更加优雅的表格，请考虑囚禁思想的可憎之处，并使用以下方法释放您的文本booktabs：

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{tabularx,tabulary}
\newcolumntype{C}{>{\centering\arraybackslash}X}
\usepackage{hhline,booktabs}
\begin{document}

\renewcommand\arraystretch{2}
\noindent\begin{tabularx}{\linewidth}{||C|C||} 
\hhline{|t:==:t|} Singular matrix    & Non-singular matrix \\
\hhline{|:==:|}    A matrix $A$ is singular if $det(A)=0$  & A matrix $A$ is non-singular  $det(A)\ne0$ \\
\hhline{||-|-||}    If $A$ is singular, then $A^{-1}$ is not defined  & If $A$ is non-singular, then $A^{-1}$ is defined \\
\hhline{||-|-||}    Rank of $A <$ order of $ A$. &Rank of $A=$ order of $A$. \\
\hhline{||-|-||}    Some rows and columns are linearly dependent. & All rows and columns are linearly independent \\
\hhline{||-|-||}    If $A$ singular, then the system of equations $Ax=b$  has either no solution or infinitely many solutions.
                           &  If A non-singular, then the system of equations $Ax=b$ has unique solutions. \\
\hhline{||-|-||}    Example:  & Example: \\
    \renewcommand\arraystretch{1}
    $\left[\begin{array}{ll} 3 & 6  \\  1 & 2 \\\end{array}\right]$ is singular because \bigskip\par
    $\left|\begin{array}{ll}3 & 6 \\ 1 & 2\end{array}\right|=(3 \times 2)-(6 \times 1)$
    $=6-6=0$\par &
    \renewcommand\arraystretch{1}
     $\left[\begin{array}{ll} 3 & 2  \\  1 & -2 \\\end{array}\right]$ is non-singular because \bigskip\par
     $\left|\begin{array}{ll}3 & 2 \\ 1 & -2\end{array}\right|=(3 \times (-2))-(2 \times 1)$
     $=-8\ne0$par\\
\hhline{|b:=b=:b|}
\end{tabularx}

\vspace{1cm}


\renewcommand\arraystretch{1} % return to defeult
\tabcolsep2em
\noindent\begin{tabularx}{\linewidth}{@{}c@{\quad}Xc@{\quad}X@{}} \toprule[1.5pt]\addlinespace 
 \multicolumn{2}{c}{Singular matrix}  &  \multicolumn{2}{c}{Non-singular matrix} \\
 \cmidrule(rl){1-2}\cmidrule(rl){3-4}\addlinespace 
$\bullet$\quad & A matrix $A$ is singular if $det(A)=0$. & 
$\bullet$\quad &  A matrix $A$ is non-singular  $det(A)\ne0$. \\\addlinespace
$\bullet$\quad & If $A$ is singular, then $A^{-1}$ is not defined . &
$\bullet$\quad & If $A$ is non-singular, then $A^{-1}$ is defined. \\\addlinespace
$\bullet$\quad & Rank of $A <$ order of $ A$. &
$\bullet$\quad &  Rank of $A=$ order of $A$. \\\addlinespace
$\bullet$\quad & Some rows and columns are linearly dependent. & 
$\bullet$\quad &  All rows and columns are linearly independent. \\\addlinespace
$\bullet$\quad & If $A$ singular, then the system of equations $Ax=b$  has either no solution or infinitely many solutions. &
$\bullet$\quad &  If A non-singular, then the system of equations $Ax=b$ has unique solutions. \\\addlinespace[3ex]
 \multicolumn{2}{c}{Example:} &  \multicolumn{2}{c}{Example:} \\
 \cmidrule(rl){1-2}\cmidrule(rl){3-4}\addlinespace 
\addlinespace
& \hfil $\left[\begin{array}{ll} 3 & 6  \\  1 & 2 \\\end{array}\right]$ is singular because \bigskip\par
  \hfil   $\left|\begin{array}{ll}3 & 6 \\ 1 & 2\end{array}\right|=(3 \times 2)-(6 \times 1)$
    $=6-6=0$\par & &
 \hfil    $\left[\begin{array}{ll} 3 & 2  \\  1 & -2 \\\end{array}\right]$ is non-singular because \bigskip\par
 \hfil     $\left|\begin{array}{ll}3 & 2 \\ 1 & -2\end{array}\right|=(3 \times (-2))-(2 \times 1)$
     $=-8\ne0$par\\
\bottomrule[1.5pt]
\end{tabularx}
\end{document}

Answer 2

我会重新设计你的表格如下：

为此我使用了以下软件包：

• tabularray对于tblr表格，
• enumitem用于矩阵属性列表
• amsmathtabularray作为用于编写矩阵和行列式的库加载
• 写行列式$det...$用的是 mah 算符 $\det$

通过考虑上述所有内容，MWE 变得更短更清晰：

\documentclass{article}
\usepackage[a4paper,margin=2cm]{geometry}

\usepackage{enumitem}
\usepackage{tabularray}
\UseTblrLibrary{amsmath, varwidth}

\newcommand{\bx}{\mathbf{x}}
\newcommand{\bb}{\mathbf{b}}

\begin{document}
    \begin{center}
    \setlist[itemize]{nosep, 
                      itemsep=0.5ex,
                      leftmargin=*}
\begin{tblr}{hlines, vlines,
             colspec = {X[l] X[l]},
             measure = vbox,
             row{1}  = {font=\bfseries, c},
             row{Z}  = {abovesep=5pt},
             stretch=-1,%<--- remove extra space above and below lists
                        % with nosep option; doc p.51 tabularray
             }
             
Singular matrix    
    & Non-singular matrix    \\
    \begin{itemize}
\item   A matrix $A$ is singular if $\det(A)=0$
\item   If $A$ is singular, then $A^{-1}$ is not defined  
\item   Rank of $A <$ order of $ A$. 
\item   Some rows and columns are linearly dependent. 
\item   If $A$ singular, then the system of equations $A\bx=\bb$
    has either no solution or infinitely many solutions.
    \end{itemize}
    &   \begin{itemize}
    \item   A matrix $A$ is non-singular $\det(A)\ne0$
    \item   If $A$ is non-singular, then $A^{-1}$ is defined
    \item   Rank of $A=$ order of $A$.
    \item   All rows and columns are linearly independent
    \item   If A non-singular, then the system of equations $A\bx=\bb$ has unique solutions.
        \end{itemize}   \\
%   
\textbf{Example:}\par\medskip 
$A = \begin{bmatrix}
 3 & 6  \\
 1 & 2 \\
 \end{bmatrix}$ is singular because
\[ 
\det A = \begin{vmatrix}
     3 & 6 \\
     1 & 2 \\
\end{vmatrix} =(3\times 2) - (6\times 1) = 6-6 = 0
\]
 %%  
    &   \textbf{Example:}\par\medskip
    $A = \begin{bmatrix}
         3 &  2  \\
         1 & -2 \\
     \end{bmatrix}$ is non-singular because
    \[
    \det A = \begin{vmatrix}
         3 &  2  \\
         1 & -2 \\
    \end{vmatrix} = (3\times (-2)) - (2\times 1) = -8\ne0
    \]                  \\
\end{tblr}
    \end{center}
\end{document}

如果矩阵属性和示例之间确实需要界限，我会再次重新考虑。

Answer 3

\documentclass{article}
\usepackage[a4paper,margin=2cm]{geometry}
\usepackage{tabularx}
\newcommand{\bx}{\mathbf{x}}
\newcommand{\bb}{\mathbf{b}}
\newcommand{\pbx}[1]{{\centering #1}}
\newcolumntype{C}{>{\centering\arraybackslash}X}

\begin{document}
\begin{center}
\begin{tabularx}{\textwidth}{ ||C|C|| } 
    \hline
    \hline
    &\\
    Singular matrix    & Non-singular matrix   \\&\\
    \hline
    \hline
    &\\
     A matrix $A$ is singular if 
    $det(A)=0$ 
      & A matrix $A$ is non-singular 
      $det(A)\ne0$
      \\ &\\
    \hline
    &\\
    If $A$ is singular, then $A^{-1}$ is not defined  & If $A$ is non-singular, then $A^{-1}$ is defined \\&\\
    \hline
    &\\
    Rank of $A <$ order of $ A$. &Rank of $A=$ order of $A$. \\&\\
    \hline
    &\\
    Some rows and columns are linearly dependent. & All rows and columns are linearly independent  \\&\\
    \hline
    &\\
    \pbx{If $A$ singular, then the system of equations\\ $A\bx=\bb$
    has either no solution or\\infinitely many solutions.}
    & \pbx{If A non-singular, then the system of equations\\ $A\bx=\bb$ has unique solutions.}
     \\&\\
    \hline
    &\\
    \pbx{Example: $
    \left[\begin{array}{ll}
    3 & 6  \\
    1 & 2 \\
    \end{array}\right]
    $ is singular because
    \\
    $\left|\begin{array}{ll}3 & 6 \\ 1 & 2\end{array}\right|=(3 \times 2)-(6 \times 1)$
    $=6-6=0$}
     & \pbx{Example: $
     \left[\begin{array}{ll}
     3 & 2  \\
     1 & -2 \\
     \end{array}\right]
     $ is non-singular because
     \\
     $\left|\begin{array}{ll}3 & 2 \\ 1 & -2\end{array}\right|=(3 \times (-2))-(2 \times 1)$
     $=-8\ne0$} \\&\\
    \hline
    \hline
\end{tabularx}
\end{center}

\end{document}