horizontal spacing in equations

Question

I am trying to replicate the same exact output as plain math mode would give for a simple polynomial formula but using \matrix because I want to right-align the terms in the polynomial or leave some of them blank. I can't use \alignat either for other reasons.

I have made some progress but the spacing is hard to get right. In the following picture you can see the output in plain math mode, using \alignat and my attempt using \matrix (I use the convenience macro \arr).

As you can see, my version is somewhat off. I think the biggest error is in the spacing around the equals sign.

\documentclass[a4paper,11pt]{article}
\usepackage{amsmath}
\makeatletter
% http://tex.stackexchange.com/questions/2233/
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols r]{%
  \hskip -\arraycolsep
  \let\@ifnextchar\new@ifnextchar
  \array{#1}}
\newcommand{\arr}{\@ifstar\arr@star\arr@nostar}
\newcommand{\arr@nostar}[2][*\c@MaxMatrixCols r]{%
  \begin{bmatrix}[#1]#2\end{bmatrix}%
}
\newcommand{\arr@star}[2][*\c@MaxMatrixCols r]{%
  \begin{matrix}[#1]#2\end{matrix}%
}
\makeatother

\begin{document}

\begin{equation}
  C + 3D = 2
\end{equation}
%
\begin{alignat}{3}
  C  &+{}& 3D &={}& 2 & \notag
\end{alignat}
%
\begin{equation}
  \arr*[@{}r@{\,}c@{\,}r@{\,}c@{\,}r@{}]{
    C  &+{}& 3D &=& 2 \notag
  }
\end{equation}

\end{document}

Answer 1

Would you consider a tabular stack as an option?

Here I set it up with the one-time invocation:

\usepackage{tabstackengine}
\stackMath
\setstacktabulargap{0pt}
\TABbinary

and then I invoke one of various forms of tabular stack using a syntax that mimics the tabular separators.

\documentclass[a4paper,11pt]{article}
\usepackage{amsmath}
\usepackage{tabstackengine}
\stackMath
\setstacktabulargap{0pt}
\TABbinary
\begin{document}

\begin{equation}
  C + 3D = 2
\end{equation}
%
\begin{alignat}{3}
  C  &+{}& 3D &={}& 2 & \notag
\end{alignat}
%
\begin{equation}
  \tabularCenterstack{rcrcr}{
  C&+&3D&=&2
  }
\end{equation}
\begin{equation}
  \tabularCenterstack{rcrcr}{
  C&+&3D&=&2\\
  4C&-&D&=&27
  }
\end{equation}
\end{document}