将 vdot 与方程组中的项对齐

Question

我正在尝试编写一个如下所示的线性方程组：

A_{11}x_1 + A_{12}x_2 + \cdots + A_{1n}x_n = y_1
A_{21}x_1 + A_{22}x_2 + \cdots + A_{2n}x_n = y_2
\vdots      \vdots               \vdots      \vdots
A_{m1}x_1 + A_{m2}x_2 + \cdots + A_{mn}x_n = y_m

其中 vdot 位于项的中心，而不是等号的中心。这是我的尝试：

\documentclass{report}
\usepackage{amsmath}
\begin{document}

\begin{align*}
&A_{11}x_1& + &A_{12}x_2& + \cdots + &A_{1n}x_n& = &y_1& \\
&A_{21}x_1& + &A_{22}x_2& + \cdots + &A_{2n}x_n& = &y_2& \\
&\vdots&  &\vdots& &\vdots& &\vdots& \\
&A_{m1}x_1& + &A_{m2}x_2& + \cdots + &A_{mn}x_n& = &y_m& \\
\end{align*}

\end{document}

有人有一个优雅的解决方案吗？

编辑。这是第二次尝试，使用更适合的array环境。看起来好多了，但仍然不自然。

\documentclass{report}
\begin{document}

\[\begin{array}{ccccccccc}
A_{11}x_1 & + & A_{12}x_1 & + & \cdots & + & A_{1n}x_n & = & y_1 \\
A_{21}x_1 & + & A_{22}x_2 & + & \cdots & + & A_{2n}x_n & = & y_2 \\
\vdots    &   & \vdots    &   &        &   & \vdots    &   & \vdots \\
A_{m1}x_1 & + & A_{m2}x_2 & + & \cdots & + & A_{mn}x_n & = & y_m \\
\end{array}\]

\end{document}

有没有什么办法可以让 vdot 像这样对齐，但又使方程式看起来正常？

Answer 1

我建议做如下修改：

\documentclass{report}
\usepackage{array}
\begin{document}

\[\begin{array}{*{9}{@{}c@{}}}
A_{11}x_1 & {}+{} & A_{12}x_1 &{}+{} & \cdots & {}+{} & A_{1n}x_n & {}\mathrel{=}{} & y_1 \\
A_{21}x_1 & + & A_{22}x_2 & + & \cdots & + & A_{2n}x_n & = & y_2 \\
\vdots    &   & \vdots    &   &        &   & \vdots    &   & \vdots \\
A_{m1}x_1 & + & A_{m2}x_2 & + & \cdots & + & A_{mn}x_n & = & y_m \\
\end{array}\]

\end{document}

Answer 2

我建议您使用 a\makebox来居中\vdots并使用alignat环境来执行多个对齐点：

代码：

\documentclass{report}
\usepackage{amsmath}
\usepackage{calc}

\newcommand*{\MyVdots}[1][$A_{22}x_n$]{\makebox[\widthof{#1}]{$\vdots$}}%
\newcommand*{\Ph}{\hphantom{{}+{}}}%
\begin{document}

\begin{alignat*}{5}
    &A_{11}x_1 &&{}+ A_{12}x_2  &&{}+ \cdots &&{}+ A_{1n}x_n && = y_1 \\
    &A_{21}x_1 &&{}+ A_{22}x_2  &&{}+ \cdots &&{}+ A_{2n}x_n && = y_2 \\
    &\MyVdots  &&\Ph \MyVdots   &&           &&\Ph  \MyVdots &&\Ph \MyVdots[$y_1$] \\
    &A_{m1}x_1 &&{}+ A_{m2}x_2  &&{}+ \cdots &&{}+ A_{mn}x_n && = y_m \\
\end{alignat*}

\end{document}

Answer 3

这里有两个 TABstack，一个周围的空间较小\vdots，另一个周围的空间较大。

\documentclass{report}
\usepackage{tabstackengine}
\stackMath
\begin{document}
\[\TABbinary
\let\svvdots\vdots\def\vdots{\protect\raisebox{-3pt}{$\svvdots$}}
\tabbedCenterstack{
A_{11}x_1 & + & A_{12}x_1 & + & \cdots & + & A_{1n}x_n & = & y_1 \\
A_{21}x_1 & + & A_{22}x_2 & + & \cdots & + & A_{2n}x_n & = & y_2 \\
\vdots    &   & \vdots    &   &        &   & \vdots    &   & \vdots \\
A_{m1}x_1 & + & A_{m2}x_2 & + & \cdots & + & A_{mn}x_n & = & y_m \\
}\]
\[\TABbinary
\let\svvdots\vdots\def\vdots{\protect\raisebox{3pt}{$\svvdots$}}
\tabbedShortstack{
A_{11}x_1 & + & A_{12}x_1 & + & \cdots & + & A_{1n}x_n & = & y_1 \\
A_{21}x_1 & + & A_{22}x_2 & + & \cdots & + & A_{2n}x_n & = & y_2 \\
\vdots    &   & \vdots    &   &        &   & \vdots    &   & \vdots \\
A_{m1}x_1 & + & A_{m2}x_2 & + & \cdots & + & A_{mn}x_n & = & y_m \\
}\]
\end{document}