How to make fbox around minipage fit the whole minipage inside it?

Question

I have long equation, Ax=b that I want to put a frame around. I can't use mdframed since I need to also compile this with tex4ht, which still does not support mdframed. So I use this trick setup:

\fbox{
\noindent\begin{minipage}{\linewidth}
\scriptsize
\[
\begin{bmatrix}
...
\end{bmatrix}
\]
\end{minipage}
}
\normalsize

This works, but the problem with the above, is that the fbox does not fit the whole minipage. Here is screen

Without minipage and no frame, here is how it looks like:

My question is: How can one automatically have fbox fit the minipage content? The solution has to also compile with not errors with tex4ht. Is there a way to tell Latex to shift the page margin to the left more, just in the minipage? i.e. make the minipage uses custom geometry? I tried this

\fbox{
\noindent\begin{minipage}{\linewidth}
\newgeometry{left=.1in,right=.1in,top=1in,bottom=1in}
\scriptsize
....

But it made things worst. I do get these messages from lualatex, but this is because the equations are too wide. But they are still in the page.

Overfull \hbox (31.09998pt too wide) in paragraph at lines 49--51
[][][] 

Here is the MWE. Sorry for large equation size, but this is the example am working on.

\documentclass[11pt]{report}%
\usepackage{amsmath,mathtools}
\usepackage[paperheight=11in,paperwidth=8.5in,top=.7in,bottom=.7in, 
      left=1.2in, right=.8in]{geometry}
\begin{document}

Therefore, the $Ax=b$ system to solve is%

\fbox{
\noindent\begin{minipage}{\linewidth}
\scriptsize
\[%
\begin{bmatrix}
7 & \left(  -4-\frac{1}{2}h^{3}\right)   & 1 & 0 & \cdots & 0 & 0 & 0\\
\left(  -4+\frac{1}{2}h^{3}\right)   & 6 & \left(  -4-\frac{1}{2}h^{3}\right)
& 1 & 0 & \cdots & 0 & 0\\
1 & \left(  -4+\frac{1}{2}h^{3}\right)   & 6 & \left(  -4-\frac{1}{2}%
h^{3}\right)   & 1 & 0 & \cdots & 0\\
0 & 1 & \left(  -4+\frac{1}{2}h^{3}\right)   & 6 & \left(  -4+\frac{1}{2}%
h^{3}\right)   & 1 & 0 & \cdots\\
0 & 0 & \left(  -4+\frac{1}{2}h^{3}\right)   & 6 & \left(  -4+\frac{1}{2}%
h^{3}\right)   & 1 & 0 & \cdots\\
&  &  &  &  &  &  & \\
&  &  &  &  &  &  & \\
&  &  &  &  &  &  &
\end{bmatrix}%
\begin{bmatrix}
y_{1}\\
y_{2}\\
y_{3}\\
y_{4}\\
\vdots\\
y_{N-2}\\
y_{N-1}\\
y_{N}%
\end{bmatrix}
=%
\begin{bmatrix}
h^{4}e^{h}-2hy_{0}^{\prime}+y_{0}\left(  4-\frac{1}{2}h^{3}\right)  \\
h^{4}e^{2h}-y_{0}\\
h^{4}e^{3h}\\
h^{4}e^{4h}\\
\vdots\\
\\
\\
\end{bmatrix}
\]
\end{minipage}
}
\normalsize

Therefore ...

\end{document}

compiled using lualatex foo.tex TL 2015

Answer 1

You have to avoid minipage, that constrains the size.

\documentclass[11pt]{report}
\usepackage{amsmath,mathtools}
\usepackage[
  letterpaper,
  top=.7in, bottom=.7in, 
  left=1.2in, right=.8in
]{geometry}

\DeclarePairedDelimiter{\paren}{(}{)}

\begin{document}

Therefore, the $Ax=b$ system to solve is
\[
\makebox[\textwidth]{\fbox{%
  \scriptsize$
  \begin{bmatrix}
  7 & \paren*{-4-\frac{1}{2}h^{3}}   & 1 & 0 & \cdots & 0 & 0 & 0\\
  \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4-\frac{1}{2}h^{3}}
  & 1 & 0 & \cdots & 0 & 0\\
  1 & \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4-\frac{1}{2}h^{3}}   & 1 & 0 & \cdots & 0\\
  0 & 1 & \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4+\frac{1}{2}h^{3}}   & 1 & 0 & \cdots\\
  0 & 0 & \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4+\frac{1}{2}h^{3}}   & 1 & 0 & \cdots\\
  &  &  &  &  &  &  & \\
  &  &  &  &  &  &  & \\
  &  &  &  &  &  &  &
  \end{bmatrix}
  \begin{bmatrix}
  y_{1}\\
  y_{2}\\
  y_{3}\\
  y_{4}\\
  \vdots\\
  y_{N-2}\\
  y_{N-1}\\
  y_{N}
  \end{bmatrix}
  =
  \begin{bmatrix}
  h^{4}e^{h}-2hy_{0}^{\prime}+y_{0}\paren*{4-\frac{1}{2}h^{3}}  \\
  h^{4}e^{2h}-y_{0}\\
  h^{4}e^{3h}\\
  h^{4}e^{4h}\\
  \vdots\\
  \\
  \\
  \end{bmatrix}
$}}
\]
Therefore ...

\end{document}

Answer 2

You have

\fbox{
\noindent\begin{minipage}{\linewidth}
\end{minipage}
}

\fbox like \mbox is a horizontal mode construct so \noindent there is not doing anything.

So you have on a line that is \linewidth wide

• A paragraph indent
• A vertical rule of width \fboxrule
• Padding of width \fboxsep
• one inter-word space from the white space after {
• A minipage of width \linewidth
• One inter-word space from the white space before }
• Padding of width fboxsep
• A vertical rule of width \fboxrule
• \parfillskip glue, natural length 0pt most likely.

That does not fit.

You want

\noindent
\fbox{%
\begin{minipage}{\dimexpr\linewidth-2\fboxrule-2\fboxsep}
\end{minipage}%
}

Answer 3

Resize it to \linewidth:

\documentclass[11pt]{report}
\usepackage{mathtools}
\usepackage[
  letterpaper,
  top=.7in, bottom=.7in, 
  left=1.2in, right=.8in
]{geometry}    
\DeclarePairedDelimiter{\paren}{(}{)}

\begin{document}

Therefore, the $Ax=b$ system to solve is
\[
\fbox{\resizebox{\dimexpr\linewidth-2\fboxrule-2\fboxsep}{!}{$
        \begin{bmatrix}
        7 & \paren*{-4-\frac{1}{2}h^{3}}   & 1 & 0 & \cdots & 0 & 0 & 0\\
        \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4-\frac{1}{2}h^{3}}
        & 1 & 0 & \cdots & 0 & 0\\
        1 & \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4-\frac{1}{2}h^{3}}   & 1 & 
        0 & \cdots & 0\\
        0 & 1 & \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4+\frac{1}{2}h^{3}}   & 
        1 & 0 & \cdots\\
        0 & 0 & \paren*{-4+\frac{1}{2}h^{3}}   & 6 & \paren*{-4+\frac{1}{2}h^{3}}   & 
        1 & 0 & \cdots\\
        &  &  &  &  &  &  & \\
        &  &  &  &  &  &  & \\
        &  &  &  &  &  &  &
        \end{bmatrix}
        \begin{bmatrix}
        y_{1}\\
        y_{2}\\
        y_{3}\\
        y_{4}\\
        \vdots\\
        y_{N-2}\\
        y_{N-1}\\
        y_{N}
        \end{bmatrix}
        =
        \begin{bmatrix}
        h^{4}e^{h}-2hy_{0}^{\prime}+y_{0}\paren*{4-\frac{1}{2}h^{3}}  \\
        h^{4}e^{2h}-y_{0}\\
        h^{4}e^{3h}\\
        h^{4}e^{4h}\\
        \vdots\\
        \\
        \\
        \end{bmatrix}
        $}}
\]
Therefore \ldots\hrulefill

\end{document}