Align many matrices the best way possible?

Question

It's my first assignment in linear algebra and working with matrices. How would you align matrices like below the best way?

Code below:

\newenvironment{sysmatrix}[1]
 {\left(\begin{array}{@{}#1@{}}}
 {\end{array}\right)}
\newcommand{\ro}[1]{%
  \xrightarrow{\mathmakebox[\rowidth]{#1}}%
}
\newlength{\rowidth}% row operation width
\AtBeginDocument{\setlength{\rowidth}{4em}}

\begin{equation}
  \begin{array}{rcl}
  {
    \left(\!\!\!\!
    \begin{array}{rrr|r}
     1 &  a & 2 & a \\
     0 &  1 & 0 & \frac{-a^2}{a^2-2} \\
     0 & 3-a & -2 & 2-a 
    \end{array}
    \!\!\right)
  } 
  & 
  \xymatrix@C=15ex{ 
    \ar[r]^-{\small
      \begin{array}{r}
        \mathbf{r}_1 \rightarrow \mathbf{r}_1 - \mathbf{r}_2a \\
         \mathbf{r}_3 \rightarrow \mathbf{r}_3 - \mathbf{r}_2(3-a) \\
      \end{array}        
    } &
  } 
  & 
  {
    \left(\!\!\!\!
    \begin{array}{rrr|r}
     1  &  0 & 2 & -\frac{2a}{a^2-2}  \\
     0 &  1 & 0 & \frac{a^2}{a^2-2} \\
     0 & 0 & -2 & \frac{1}{a^2-2}(-a^2+2a-4)  
    \end{array}
    \!\!\right)
  } 
\end{array}\bigskip
\end{equation}

\begin{equation}
  \begin{array}{rcl}
  {
    \left(\!\!\!\!
    \begin{array}{rrr|r}
     1  &  0 & 2 & -\frac{2a}{a^2-2}  \\
     0 &  1 & 0 & \frac{a^2}{a^2-2} \\
     0 & 0 & -2 & \frac{1}{a^2-2}(-a^2+2a-4)  
    \end{array}
    \!\!\right)
  } 
  & 
  \xymatrix@C=16ex{ 
    \ar[r]^-{\small
      \begin{array}{r}
         \mathbf{r}_1 \rightarrow \mathbf{r}_1 + \mathbf{r}_3 \\
         \mathbf{r}_3 \rightarrow \frac{\mathbf{r}_3}{-2}
      \end{array}        
    } &
  } 
  & 
  {
    \left(\!\!\!\!
    \begin{array}{rrr|r}
     1 &  0 & 0 & -\frac{a^2+4}{a^2-2}  \\
     0 &  1 & 0 & \frac{a^2}{a^2-2} \\
     0 &  0 & 1 & \frac{\frac{a^2}{2}-a+2}{a^2-2}
    \end{array}
    \!\!\right)
  } 
\end{array}
\end{equation}

As always on this site, please make this sniplet into a full (but minimal) document, then it is a lot easier for others to test and give advise. — daleif, May 03 '19 at 12:57
Closely related: https://tex.stackexchange.com/questions/488427/beautiful-matrices-in-latex-when-having-bigger-equations-alignment/488544#488544 — , May 03 '19 at 13:05

score 2 · Accepted Answer · answered May 03 '19 at 13:12

\documentclass[a4paper]{article}
\usepackage[margin=2cm]{geometry}
\usepackage{array}
\usepackage{amsmath}
\begin{document}

\begin{align}
    \left(\begin{array}{ccc|>{\displaystyle}c}
    1 &  a & 2 & a \\
    0 &  1 & 0 & \frac{-a^2}{a^2-2} \\
    0 & 3-a & -2 & 2-a 
    \end{array}\right) 
& \xrightarrow{\small
        \begin{array}{r}
        \mathbf{r}_1 \rightarrow \mathbf{r}_1 - \mathbf{r}_2a \\
        \mathbf{r}_3 \rightarrow \mathbf{r}_3 - \mathbf{r}_2(3-a) \\
        \end{array}} 
    \left(\begin{array}{ccc|>{\displaystyle}c}
    1  &  0 & 2 & -\frac{2a}{a^2-2}  \\
    0 &  1 & 0 & \frac{a^2}{a^2-2} \\
    0 & 0 & -2 & \frac{1}{a^2-2}(-a^2+2a-4)  
    \end{array}\right)  \\[10pt]
\left(\begin{array}{rrr|>{\displaystyle}r}
    1  &  0 & 2 & -\frac{2a}{a^2-2}  \\
    0 &  1 & 0 & \frac{a^2}{a^2-2} \\
    0 & 0 & -2 & \frac{1}{a^2-2}(-a^2+2a-4)  
    \end{array}\right)
&  \xrightarrow[\hphantom{\textstyle~\mathbf{r}_3 \rightarrow \mathbf{r}_3 - \mathbf{r}_2(3-a)}]%
      {\small
        \begin{array}{r}
        \mathbf{r}_1 \rightarrow \mathbf{r}_1 + \mathbf{r}_3 \\
        \mathbf{r}_3 \rightarrow \frac{\mathbf{r}_3}{-2}
        \end{array}}
\left(\begin{array}{rrr|>{\displaystyle}r}
    1 &  0 & 0 & -\frac{a^2+4}{a^2-2}  \\
    0 &  1 & 0 & \frac{a^2}{a^2-2} \\
    0 &  0 & 1 & \frac{\frac{a^2}{2}-a+2}{a^2-2}
    \end{array}\right) 
\end{align}

\end{document}

Exactly what I have been looking for the last week! Thank you! — jubibanna, May 03 '19 at 13:46

score 0 · Answer 2 · answered Jul 21 '20 at 21:59

Another possible solution a bit more fast using spalign package:

\documentclass[a4paper,12pt]{article}
\usepackage[margin=2.2cm]{geometry}
\usepackage{mathtools}
\usepackage{spalign}
\begin{document}
\begin{equation}
\spalignaugmat[c]{1 a 2 a; 0 1 0 \dfrac{-a^2}{a^2-2}; 0 3-a -2 2-a} \xrightarrow{\begin{matrix}
\mathbf{r}_1 \rightarrow \mathbf{r}_1 - \mathbf{r}_2a \
        \mathbf{r}_3 \rightarrow \mathbf{r}_3 - \mathbf{r}_2(3-a) 
\end{matrix}}{} 
  \spalignaugmat{1 0 2 -\dfrac{2a}{a^2-2}; 0 1 0 \dfrac{a^2}{a^2-2}; 0 0 -2 \dfrac{1}{a^2-2}(-a^2+2a-4)}

\end{equation}
\begin{equation}
\spalignaugmat[c]{1 0 2 -\dfrac{2a}{a^2-2}; 0 1 0  \dfrac{a^2}{a^2-2}; 0 0 -2 \dfrac{1}{a^2-2}(-a^2+2a-4)} \xrightarrow{\begin{matrix}
\mathbf{r}_1 \rightarrow \mathbf{r}_1 + \mathbf{r}_3 \
        \mathbf{r}_3 \rightarrow -\frac{1}{2}\mathbf{r}_3
\end{matrix}}{}

  \spalignaugmat{1 0 0 -\dfrac{a^2+4}{a^2-2}; 0 1 0 \dfrac{a^2}{a^2-2}; 0 0 1 \dfrac{\frac{a^2}{2}-a+2}{a^2-2}}

\end{equation}
[\mathbf{S}^{\ast}=\spalignaugmat[c]{1 0 0 -\frac{a^2+4}{a^2-2};0 1 0 \frac{a^2}{a^2-2};  0 0 1 \frac{\frac{a^2}{2}-a+2}{a^2-2}}]
\end{document}

Align many matrices the best way possible?

2 Answers2