如何在方程组中排版斜点？

Question

我尝试在乳胶中编写向后替换上三角系统

格式如下

我尝试编写代码，但它不能正常工作

    \begin{equation*}
    \begin{split}
        &a_{1,1}x_{1}  + a_{1,2}x_{2} +\;.....+ a_{1},{n}X_{n} = b_{1}\\
               &  a_{2,2}x_{2} +\;.....+a_{2},{n}X_{n}  = b_{2}\\
      \end{split}
     \end{equation*}

请帮忙看看哪里出了问题，我尝试&在项目中使用另一个代码，它可以正常工作。为什么这里不能正常工作？:(

Answer 1

基于的解决方案array：

\documentclass{article}
\usepackage{amsmath,array}

\begin{document}

\begin{equation*}
\setlength{\arraycolsep}{0pt}% so we don't need too many @{}
\newcommand{\LD}{\ldots}%
\newcommand{\DF}[1]{\multicolumn{#1}{c}{\dotfill}}%
\newcommand{\DD}{\multicolumn{2}{r}{\ddots}}%
\newcommand{\VD}{\multicolumn{1}{c}{\vdots}}%
\begin{array}{*{5}{r>{{}}c<{{}}}l}
a_{1,1}x_1 &+& a_{1,2}x_2 &+& \DF{3}     &+& a_{1,n}x_n &=& b_1 \\[1ex]
           & & a_{2,2}x_2 &+& \DF{3}     &+& a_{2,n}x_n &=& b_2 \\
           & & \DD          &            & & && \VD     & & \VD \\
           & &            & & a_{i,i}x_i &+& \LD        &+& a_{i,n}x_n &=& b_i \\
           & &            & & \DD          & && \VD     & & \VD \\
           & &            & &            && & & a_{n,n}x_n &=& b_n
\end{array}
\end{equation*}

\end{document}

Answer 2

这是一个array基于 的解决方案：

\documentclass{article}
\usepackage{newtxtext,newtxmath} % Times Roman clone - optional
\usepackage{array}               % for '\newcolumntype' macro
\newcolumntype{C}{>{{}}c<{{}}}   % col. type for bin. and rel. operators
\newcommand\myddots{\multicolumn{1}{r}{\ddots}} % handy shortcut macro
\begin{document}

\[
\setlength\arraycolsep{0pt}
\begin{array}{*{5}{cC}c}
a_{1,1}x_1 &+& a_{1,2}x_2 &+& \multicolumn{3}{c}{\dotfill} &+& a_{1,n}x_n &=& b_1 \\
&& a_{2,2}x_2 &+& \multicolumn{3}{c}{\dotfill} &+& a_{2,n}x_n &=& b_2 \\
&& \myddots  && && \phantom{a_{i,n}x_n} && \vdots && \vdots\\
&&&& a_{i,i}x_i &+& \dotfill &+& a_{i,n}x_n &=& b_i \\
&&&& \myddots &&&& \vdots && \vdots\\
&&&&&&&& a_{n,n}x_n &=& b_n
\end{array}
\]
\end{document}

Answer 3

好的，我找到了解决方案

     \[    
       \begin{array}{@{}*{7}{c@{}}}
        & a_{1,1}x_{1}  + a_{1,2}x_{2} +\;.....+ a_{1},{n}X_{n} = b_{1}\\
          & a_{2,2}x_{2} +\;.....+a_{2},{n}X_{n}  = b_{2}\\         
          \end{array}               
   \]    

Answer 4

基于 的解决方案alignedat，借助于eqparbox：

\documentclass{article}
\usepackage{mathtools}
\usepackage{eqparbox} 

\begin{document}

  \begin{equation*}
  \begin{alignedat}{2}
   a_{1,1}x_{1} + a_{1,2}x_{2} & +{}&\cdots\cdots \cdots \cdots + a_{1,n}X_{n} & = b_{1}\\
         a_{2,2}x_{2} & +{} &\cdots\cdots \cdots\cdots +a_{2,n}X_{n} & = b_{2}\\[-1.5ex]
        &\ddots && \vdotswithin{=}\\[-1ex]
        & & a_{i, i}x_{i}\eqparbox{M}{${} + \dots + a_{i, n}X_{n} $} & = b_{i} \\[-1.5ex]
        & &\eqparbox{M}{\raisebox{1ex}{$\ddots $}} & \vdotswithin{=}\\[-2ex]
        & & a_{n, n}X_{n} & =b_{n}
  \end{alignedat}
 \end{equation*}
\end{document}