# Align many matrices the best way possible?

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}}

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

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

• 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 3 '19 at 12:57
• Closely related: tex.stackexchange.com/questions/488427/… – user30471 May 3 '19 at 13:05

\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 3 '19 at 13:46 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} \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)} \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}} \[\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}