Issues with system of equations: systeme fix and align environment

Question

I'm writing some notes on linear algebra and to ease my work I've decided to use the systeme package. When I discovered that systeme has no built-in feature of writing systems of equations with dots and non-numeric coefficients, I've decided to use the fix suggested in this answer.

I need to write the full process of row-reduction of a specific system of equations, and I've done it before using the cases environment inside an align* environment. But the cases environment won't align the coefficients as desired, although each system is just where I wanted it to be. The fix suggested in this post yields an error when used inside the align* environment. What can I do?

A test sample for the error:

\documentclass{article}

\usepackage{amsmath}
\usepackage{xstring}

\def\typesystem#1{%
    \begingroup\expandarg
    \baselineskip=1.5\baselineskip% 1.5 to enlarge vertical space between lines
    \StrSubstitute{\noexpand#1}+{&+&}[\tempsystem]%
    \StrSubstitute\tempsystem={&=&}[\tempsystem]%
    \StrSubstitute\tempsystem,{\noexpand\cr}[\tempsystem]%
    \vcenter{\halign{&$\hfil\strut##$&${}##{}$\cr\tempsystem\crcr}}%
    \endgroup
}

\begin{document}

\begin{align*}
    \left\{\typesystem{3x + 2y = 1, x - 3y = 2}\right.
\end{align*}

\end{document}

It yields:

! Argument of \@xs@next has an extra }.
<inserted text> 
                \par 
l.20 \end{align*}

My situation in which each system is where it's desired but the coefficients inside them are not aligned (using align* and cases):

Answer 1

I'd stay with systeme as the code for \typesystem is very faulty.

Anyway, an additional pair of braces will do:

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

\newcommand\typesystem[1]{%
    \begingroup\expandarg
    \linespread{1.5}\selectfont
    \StrSubstitute{\noexpand#1}+{&+&}[\tempsystem]%
    \StrSubstitute\tempsystem-{&-&}[\tempsystem]%
    \StrSubstitute\tempsystem={&=&}[\tempsystem]%
    \StrSubstitute\tempsystem{\noexpand\missing}{{}&&}[\tempsystem]%
    \StrSubstitute\tempsystem,{\noexpand\cr}[\tempsystem]%
    \left\{\vcenter{\halign{&$\hfil\strut##$&${}##{}$\cr\tempsystem\crcr}}\right.%
    \endgroup
}
\newcommand\minus{{-}}
\newcommand{\missing}{}

\begin{document}
\begin{alignat*}{2}
{\typesystem{
  2x_1 - x_2 - x_3 = 1,
  4x_1 - 3x_2 + x_3 = 0,
  \minus x_1 + x_2 + 2x_3 = \minus2
}}
&& {}\xrightarrow{L_2\rightarrow L_2 - 2L_1}{}
&{\typesystem{
  2x_1 - x_2 - x_3 = 1,
{} - x_2 + 3x_3 = \minus2,
  \minus x_1 + x_2 + 2x_3 = \minus2
}}
\\
&& {}\xrightarrow{L_1\leftrightarrow L_3}{}
&{\typesystem{
  \minus x_1 +x_2 +2x_3 = \minus2,
{} - x_2 +3x_3 = \minus2,
  2x_1 -x_2 -x_3 = 1
}}
\\
&& {}\xrightarrow{L_3\rightarrow L_3 + 2L_1}{}
&{\typesystem{
  \minus x_1 +x_2 +2x_3 = \minus2,
{}  - x_2 +3x_3 = \minus2,
\missing  x_2 +3x_3 = \minus3
}}
\end{alignat*}

\end{document}

With systeme:

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

\begin{document}
\begin{alignat*}{2}
\systeme{
  2x_1 - x_2 - x_3 = 1,
  4x_1 - 3x_2 + x_3 = 0,
  -x_1 + x_2 + 2x_3 = -2
}
&& {}\xrightarrow{L_2\rightarrow L_2 - 2L_1}{}
&\systeme{
  2x_1 - x_2 - x_3 = 1,
  -x_2 + 3x_3 = -2,
  -x_1 + x_2 + 2x_3 = -2
}
\\
&& {}\xrightarrow{L_1\leftrightarrow L_3}{}
&\systeme{
  -x_1 +x_2 +2x_3 = -2,
  -x_2 +3x_3 = -2,
  2x_1 -x_2 -x_3 = 1
}
\\
&& {}\xrightarrow{L_3\rightarrow L_3 + 2L_1}{}
&\systeme{
  -x_1 +x_2 +2x_3 = -2,
  -x_2 +3x_3 = -2,
  x_2 +3x_3 = -3
}
\end{alignat*}

\end{document}

Answer 2

Here is a tabstackengine alternative for presenting systems of equations. However, because we are using a space as the term separator, the syntax is rigid and one cannot freely throw in stray blank spaces. In general, things to remember are:

1. if one wants the operators (+, -, =) to line up, then they can not attached to either term (i.e., must form their own column); If right alignment of the terms is all that is required, without operator alignment, then operators can be placed next to leading or trailing terms as long as you are consistent in the syntax across all equations of the system;

2. {} is used as a placeholder for an empty term;

3. In order to achieve this syntax, leading minus signs will be treated as a binary operation, unless presented as {-};

4. stray spaces are a no-no, including leading and trailing spaces, as well as those surrounding the comma , equation separator.

Here is the MWE.

In response to barbara's comment, here is what the result looks like if I group the leading minus signs as {-} instead of -, to prevent them from being treated as binary operators. Additionally, to show flexibility of the solution, the inter-line baselineskip can be set in the preamble, e.g., \setstackgap{L}{1.2\normalbaselineskip}

\documentclass{article}
\usepackage{amsmath}
\usepackage{tabstackengine}
\def\typesystem#1#2{\savestack{#1}{\setstackEOL{,}\setstackTAB{ }
  $\left\{\ensurestackMath{\tabbedCenterstack[r]{#2}}\right.$}}
\TABbinary
\stackMath
\setstackgap{S}{12pt}
\setstackgap{L}{1.2\normalbaselineskip}
\begin{document}
\typesystem{\systemA}{2x_1  - x_2  - x_3  =    1,4x_1 - 3x_2 +  x_3 =    0,{-}x_1 + x_2 + 2x_3 = {-2}}
\typesystem{\systemB}{2x_1  - x_2  - x_3  =    1,{}   - x_2  + 3x_3 = {-}2,{-}x_1 + x_2 + 2x_3 = {-2}}
\typesystem{\systemC}{{-}x_1 + x_2 + 2x_3 = {-2},{}   - x_2  + 3x_3 = {-}2,2x_1   - x_2 - x_3  =  1}
\typesystem{\systemD}{{-}x_1 +x_2  +2x_3  = {-2},{}   -x_2   +3x_3  = {-}2,{} x_2 +3x_3        = {-3}}
\typesystem{\systemE}{{-}x_1 +x_2  +2x_3  = {-2},{}   -x_2   +3x_3  = {-}2,{} {}  +6x_3        = {-5}}

\systemA\quad
\alignShortunderstack{%
  \xrightarrow{L_2\rightarrow L_2 - 2L_1}&\systemB\\
  \xrightarrow{L_1\leftrightarrow L_3}&\systemC\\
  \xrightarrow{L_3\rightarrow L_3 + 2L_1}&\systemD\\
  \xrightarrow{L_3\rightarrow L_3 + L_2}&\systemE
}
\end{document}