两列三个方程

Question

我有一个带有两个小矩阵的大矩阵，我尝试按以下方式对齐它：

我希望矩阵 C 位于 B 下方而不是新行。

现在的代码是：

\begin{center}
\begin{tabular}{ c  c }\hspace*{-1.5cm}
$A^{T}(t)=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\       
    0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\   
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & -\frac{1}{\tau} & 0 \\  
    0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{\tau} \\                      
\end{bmatrix}
\qquad
B^{T}(t)=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} & 0 \\   
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} \\
\end{bmatrix}$
\\
\\
$C(t)=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}$
\end{tabular}
\end{center}

谢谢你！

Answer 1

这仍然太宽了 70pt，但是您还没有说您的页面有多宽，所以也许它可以起作用......

\documentclass[a4paper]{article}

\usepackage{amsmath}
\setcounter{MaxMatrixCols}{20}

\begin{document}

something
\[\setlength\arraycolsep{3pt}
A^{T}(t)=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\       
    0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\   
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & -\frac{1}{\tau} & 0 \\  
    0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{\tau} \\                      
\end{bmatrix}
%
\begin{aligned}
B^{T}(t)&=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} & 0 \\   
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} \\
\end{bmatrix}\\
C(t)&=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{aligned}
\]

\end{document}

Answer 2

除非你的文档的文本块异常宽，或者你愿意使用非常小的字体大小，否则和的矩阵表达式不太B可能C适合表达式的右侧A

我将使用一个简单的align*环境来排版这三个矩阵，一个在另一个下面。

\documentclass{article}
\usepackage{mathtools}
\setcounter{MaxMatrixCols}{13}
\newcommand\trans{^{T\mkern-5mu}} % transpose oper.

\begin{document}
\begin{align*}
A\trans(t) &=
\begin{bmatrix*}[r]
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\       
    0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\   
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & -\frac{1}{\tau} & 0 \\  
    0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{\tau} \\                      
\end{bmatrix*} \\
B\trans(t) &=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} & 0 \\   
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} \\
\end{bmatrix} \\
C(t) &=
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}

\end{document}

Answer 3

mmatrix它几乎不适合文本宽度。我可以使用环境 from nccmath（中型矩阵）来实现，减少\arraycolsep和加载的值geometry以获得更合理的外边距。我还使用了bmatrix*环境 fom mathtools，它接受一个可选参数来对齐列（在我看来，右对齐看起来更好，因为—有些列中有符号）：

    \documentclass{article}
    \usepackage{mathtools, nccmath, bigstrut}
\usepackage[showframe]{geometry}
\setcounter{MaxMatrixCols}{20}

    \begin{document}

\[ \setlength{\arraycolsep}{5pt}
  A^{T}(t) =
\begin{mmatrix}
\begin{bmatrix*}[r]
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \bigstrut[t]\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    1 & 0 & 0 & 0 & 0 & 0 & \mathllap{-\mkern-1.5mu}1 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 & 0 & \mathllap{-\mkern-1.5mu}1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & \mathllap{-\mkern-1.5mu}\frac{1}{\tau} & 0 \\
    0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \mathllap{-\mkern-1.5mu}\frac{1}{\tau}\bigstrut[b] \\
\end{bmatrix*}
\end{mmatrix}
\quad
 \begin{aligned}
B^{T}(t) & =
\begin{mmatrix}
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} & 0 \bigstrut[t]\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} \\
\end{bmatrix}
\end{mmatrix}
\\
\\
C(t) & =
\begin{mmatrix}
\begin{bmatrix}
    0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \bigstrut[t]
\end{bmatrix}
\end{mmatrix}
\end{aligned}
\]

\end{document} 

Answer 4

我按照你的要求做了，用array代替bmatrix+ tabular。顺便说一句，如果你将这三个矩阵与 LaTeX 的默认字体大小一起使用，则必须减小纸张边距。我使用了：

\usepackage{geometry}
\geometry{a4paper,left=5mm,top=5mm,bottom=5mm,right=5mm}

\begin{document}

\begin{center}

$A^{T}(t)=\left[\begin{array}{c c c c c c c c c c c c c}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\       
    0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\        
    0 & 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\   
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\    
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & -\frac{1}{\tau} & 0 \\  
    0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{\tau} \\  
\end{array}\right]
\qquad
\begin{array}{l} {B^{T}(t)= \left[\begin{array}{c c c c c c c c c c c c c}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau} & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\tau}  \end{array}\right]} \\
\\
{C(t)=\left[\begin{array}{c c c c c c c c c c c c c}
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]} \end{array}$

\end{center}

\end{document}

结果是：