4

I have the following matrices, I'd like to align them under each other more beautifully. Can anyone recommend a good way to do this? enter image description here

$
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
-1 & 2 & 0 & 4z-5 \\
1 & 0 & -1 & z-3 \\
1 & 2 & 0 & 8z-7
\end{matrix}\right]
R_2-(-1)R_1->R_2 
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 3 & 1 & 10z-7 \\
1 & 0 & -1 & z-3 \\
1 & 2 & 0 & 8z-7
\end{matrix}\right]
R_3-1R_1->R_3
$ \bigskip

$
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 3 & 1 & 10z-7 \\
0 & -1 & -2 & -5z-1 \\
1 & 2 & 0 & 8z-7
\end{matrix}\right]
R_4-1R_1->R_4
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 3 & 1 & 10z-7 \\
0 & -1 & -2 & -5z-1 \\
0 & 1 & -1 & 2z-5
\end{matrix}\right]
R_2/(3)->R_2
$ \bigskip

$
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 1 & \frac{1}{3} & \frac{10z-7}{3} \\
0 & -1 & -2 & -5z-1 \\
0 & 1 & -1 & 2z-5
\end{matrix}\right]
R_3-(-1)R_2->R_3
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 1 & \frac{1}{3} & \frac{10z-7}{3} \\
0 & 0 & \frac{-5}{3} & \frac{-5z-10}{3} \\
0 & 1 & -1 & 2z-5
\end{matrix}\right]
R_4-1R_2->R_4
$ \bigskip

$
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 1 & \frac{1}{3} & \frac{10z-7}{3} \\
0 & 0 & \frac{-5}{3} & \frac{-5z-10}{3} \\
0 & 0 & \frac{-4}{3} & \frac{-4z-8}{3}
\end{matrix}\right]
R_3/((-5)/3)->R_3
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 1 & \frac{1}{3} & \frac{10z-7}{3} \\
0 & 0 & 1 & z+2 \\
0 & 0 & \frac{-4}{3} & \frac{-4z-8}{3}
\end{matrix}\right]
R_4-((-4)/3)R_3->R_4
$ \bigskip

$
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 1 & \frac{1}{3} & \frac{10z-7}{3} \\
0 & 0 & 1 & z+2 \\
0 & 0 & 0 & 0
\end{matrix}\right]
R_2-(1/3)R_3->R_2
\left[\begin{matrix}
1 & 1 & 1 & 6z-2 \\
0 & 1 & 0 & 3z-3 \\
0 & 0 & 1 & z+2 \\
0 & 0 & 0 & 0
\end{matrix}\right]
R_1-1R_3->R_1
$ \bigskip

$
\left[\begin{matrix}
1 & 1 & 0 & 5z-4 \\
0 & 1 & 0 & 3z-3 \\
0 & 0 & 1 & z+2 \\
0 & 0 & 0 & 0
\end{matrix}\right]
R_1-1R_2->R_1
\left[\begin{matrix}
1 & 0 & 0 & 2z-1 \\
0 & 1 & 0 & 3z-3 \\
0 & 0 & 1 & z+2 \\
0 & 0 & 0 & 0
\end{matrix}\right]
$ 

$
\left[\begin{matrix}
x_1 & = & 2z-1 \\
x_2 & = & 3z-3 \\
x_3 & = & z+2
\end{matrix}\right]
$
  • Please provide a minimal working example (with \documentclass, \begin{document}, etc.). You can probably put everything in a large array of three columns (for the desired alignment). And use the bmatrix environment from amsmath to make the code a bit shorter and nicer. – frougon May 15 at 18:56
3

I thought I'd use nicematrix package: https://ctan.org/pkg/nicematrix . You can see that the matrices are all the same. On each row you can make the transformations by row and by column as in the minimal example I showed you. I don't know if this can correspond to what you are looking for.

enter image description here

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb}
\usepackage{nicematrix}

\usepackage[left=.1in, right=.5in]{geometry}

\begin{document}
\[
\begin{matrix}
\begin{bNiceArrayC}{CCCC}
1 & 1 & 1 & 6z-2 & \\
-1 & 2 & 0 & 4z-5 & R_2-(-1)R_1\rightarrow R_2 \\
1 & 0 & -1 & z-3 & \\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} & \hspace{4cm} \begin{bNiceArrayC}{CCCC}
1 & 1 & 1 & 6z-2 & \\
0 & 3 & 1 & 10z-7 & \\
1 & 0 & -1 & z-3 & R_3-1R_1\rightarrow R_3\\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} \\[2cm]
\begin{bNiceArrayC}{CCCC}
1 & 1 & 1 & 6z-2 & \\
-1 & 2 & 0 & 4z-5 & R_2-(-1)R_1\rightarrow R_2 \\
1 & 0 & -1 & z-3 & \\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} & \hspace{4cm} \begin{bNiceArrayC}{CCCC}
1 & 1 & 1 & 6z-2 & \\
0 & 3 & 1 & 10z-7 & \\
1 & 0 & -1 & z-3 & R_3-1R_1\rightarrow R_3\\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} \\[2cm]
\begin{bNiceArrayC}{CCCC}
1 & 1 & 1 & 6z-2 & \\
-1 & 2 & 0 & 4z-5 & R_2-(-1)R_1\rightarrow R_2 \\
1 & 0 & -1 & z-3 & \\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} & \hspace{4cm} \begin{bNiceArrayC}{CCCC}
1 & 1 & 1 & 6z-2 & \\
0 & 3 & 1 & 10z-7 & \\
1 & 0 & -1 & z-3 & R_3-1R_1\rightarrow R_3\\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} \\
\end{matrix}
\]
\end{document}

Or you can adapt this code using your matrix(s):

enter image description here

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb}
\usepackage{nicematrix}

\usepackage[left=.1in, right=.5in]{geometry}

\begin{document}
\[
\begin{matrix}
\begin{bNiceArrayC}{CCC|C}
1 & 1 & 1 & 6z-2 & \\
-1 & 2 & 0 & 4z-5 & R_2-(-1)R_1\rightarrow R_2 \\
1 & 0 & -1 & z-3 & \\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} & \hspace{4cm} \begin{bNiceArrayC}{CCC|C}
1 & 1 & 1 & 6z-2 & \\
0 & 3 & 1 & 10z-7 & \\
1 & 0 & -1 & z-3 & R_3-1R_1\rightarrow R_3\\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} \\[2cm]
\begin{bNiceArrayC}{CCC|C}
1 & 1 & 1 & 6z-2 & \\
-1 & 2 & 0 & 4z-5 & R_2-(-1)R_1\rightarrow R_2 \\
1 & 0 & -1 & z-3 & \\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} & \hspace{4cm} \begin{bNiceArrayC}{CCC|C}
1 & 1 & 1 & 6z-2 & \\
0 & 3 & 1 & 10z-7 & \\
1 & 0 & -1 & z-3 & R_3-1R_1\rightarrow R_3\\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} \\[2cm]
\begin{bNiceArrayC}{CCC|C}
1 & 1 & 1 & 6z-2 & \\
-1 & 2 & 0 & 4z-5 & R_2-(-1)R_1\rightarrow R_2 \\
1 & 0 & -1 & z-3 & \\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} & \hspace{4cm} \begin{bNiceArrayC}{CCC|C}
1 & 1 & 1 & 6z-2 & \\
0 & 3 & 1 & 10z-7 & \\
1 & 0 & -1 & z-3 & R_3-1R_1\rightarrow R_3\\
1 & 2 & 0 & 8z-7 & \\
\end{bNiceArrayC} \\
\end{matrix}
\]
\end{document}
  • 1
    This is really beautiful! Thank you sebastiano! – jubibanna May 15 at 21:56
0

Here is a TikZ solution:

Output

\documentclass{article}
\usepackage{amsmath,showframe}
\usepackage{mathtools}  % for mathmakebox
\usepackage{tikz}
\usetikzlibrary{matrix,positioning}
% Having all the cells to be equal in width is taken from Gonzalo Medina:
% https://tex.stackexchange.com/a/191240/167081

% The next is used to make the arrow fit the longest formula width.
% Copied from: Stefan Kottwitz, https://tex.stackexchange.com/a/6840/167081
\newlength{\arrow}
\settowidth{\arrow}{\tiny$R_4-(\frac{-4}{3})R_3$}% Longest formula
\newcommand*{\matrixArrow}[1]{\xrightarrow{\mathmakebox[\arrow]{#1}}}

\newcommand\mymatrix[3]{%
\noindent
    \begin{tikzpicture}
        \matrix[ampersand replacement=\&,matrix of math nodes,left delimiter = {[},
                right delimiter ={]}, inner xsep=0pt,inner ysep=3pt, outer xsep=0pt,
                align = center,font=\footnotesize,
                every node/.style={anchor=base,text depth=.5ex,text height=2ex,text width=2.6em}, 
                nodes={execute at begin node=$, execute at end node=$}] (m){#1};
                \node[right=0.5ex of m] (b) {$\matrixArrow{\textcolor{red}{#2}#3}$};
    \end{tikzpicture}
}

\begin{document}
    % How to use:
        % Place the matrix (delimited by \&) in the first argument.
        % Place the row that will be changed in the second (to be shown in red).
        % Place rest of the formula of the row reduction in the 3rd and last arugment.
    % Row 1
    \mymatrix{
        1   \& 1 \& 1   \& 6z - 2   \\
        -1  \& 2 \& 0   \& 4z - 5   \\
        1   \& 0 \& -1  \& z - 3    \\
        1   \& 2 \& 0   \& 8z - 7   \\
    }{R_2}{- (-1)R_1}
    \mymatrix{
        1 \& 1 \& 1     \& 6z - 2   \\
        0 \& 3 \& 1     \& 10z - 7  \\
        1 \& 0 \& -1    \& z - 3    \\
        1 \& 2 \& 0     \& 8z - 7   \\
    }{R_3}{- 1R_1}\\[2ex]
    % Row 2
    \mymatrix{
        1 \& 1  \& 1    \& 6z - 2   \\
        0 \& 3  \& 1    \& 10z - 7  \\
        0 \& -1 \& -2   \& -5z - 1  \\
        1 \& 2  \& 0    \& 8z - 7   \\
    }{R_4}{- R_1}
    \mymatrix{
        1 \& 1  \& 1    \& 6z - 2   \\
        0 \& 3  \& 1    \& 10z - 7  \\
        0 \& -1 \& -2   \& -5z - 1  \\
        0 \& 1  \& -1   \& 2z - 5   \\
    }{R_2}{/3}\\[2ex]
    % Row 3
    \mymatrix{
        1 \& 1  \& 1            \& 6z - 2           \\
        0 \& 1  \& \frac{1}{3}  \& \frac{10z-7}{3}  \\
        0 \& -1 \& -2           \& -5z - 1          \\
        0 \& 1  \& -1           \& 2z - 5           \\
    }{R_3}{-(-1)R_2}
    \mymatrix{
        1 \& 1 \& 1             \& 6z - 2           \\
        0 \& 1 \& \frac{1}{3}   \& \frac{10z-7}{3}  \\
        0 \& 0 \& \frac{-5}{3}  \& \frac{-5z-10}{3} \\
        0 \& 1 \& -1            \& 2z - 5           \\
    }{R_4}{-1R_2}\\[2ex]
    % Row 4
    \mymatrix{
        1 \& 1 \& 1             \& 6z-2             \\
        0 \& 1 \& \frac{1}{3}   \& \frac{10z-7}{3}  \\
        0 \& 0 \& \frac{-5}{3}  \& \frac{-5z-10}{3} \\
        0 \& 0 \& \frac{-4}{3}  \& \frac{-4z-8}{3}  \\
    }{R_3}{/(\frac{-5}{3})}
    \mymatrix{
        1 \& 1 \& 1             \& 6z - 2           \\
        0 \& 1 \& \frac{1}{3}   \& \frac{10z-7}{3}  \\
        0 \& 0 \& \frac{-5}{3}  \& \frac{-5z-10}{3} \\
        0 \& 1 \& -1            \& 2z - 5           \\
    }{R_4}{-(\frac{-4}{3})R_3}\\[2ex]
    % Row 5
    \mymatrix{
        1 \& 1 \& 1             \& 6z - 2           \\
        0 \& 1 \& \frac{1}{3}   \& \frac{10z-7}{3}  \\
        0 \& 0 \& 1             \& z + 2            \\
        0 \& 0 \& 0             \& 0                \\
    }{R_1}{-1R_3}
    \mymatrix{
        1 \& 1 \& 0 \& 5z - 4   \\
        0 \& 1 \& 0 \& 3z - 3   \\
        0 \& 0 \& 1 \& z + 2    \\
        0 \& 0 \& 0 \& 0        \\
    }{R_1}{-1R_2}\\[2ex]
    % Row 6 One matrix only
    \[
        \begin{matrix}
            x_1 & = & 2z-1 \\
            x_2 & = & 3z-3 \\
            x_3 & = & z+2
        \end{matrix}
    \]
\end{document}
  • This is insane! Thank you! Really what I was looking for! – jubibanna May 16 at 3:32

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