\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{enumerate}
\usepackage{ifthen}
\usepackage{ulem} % for strikethrough \sout{...} command
\usepackage{color}

\usepackage{hyperref}

\usepackage{wasysym}

\setlength{\unitlength}{0.1in}

\setlength{\oddsidemargin}{-0.25in}
\setlength{\evensidemargin}{-0.25in}
\setlength{\topmargin}{-0.25in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\topskip}{0.0in}
\setlength{\textheight}{9.5in}
\setlength{\textwidth}{7in}

\newcommand{\bdot}{\, {\scriptstyle{\bullet}} \,}
\newcommand{\B}[1]{{\bf {#1}}}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}

\newcommand{\myspace}{\parbox[c][0.25in]{0.25in}{\hfill}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}
\newcommand{\force}{{\noindent $\mbox{\;}$}}

\newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}}

\makeatletter
\def\@xfootnote[#1]{%
  \protected@xdef\@thefnmark{#1}%
  \@footnotemark\@footnotetext}
\makeatother

\begin{document}

\pagestyle{empty}

\noindent
\parbox{2.25in}{\bf Math 5230}
\hfill {\Large \bf Homework \#Dos\footnote[I]{Enumeration scheme inspired by class discussion. Thanks for participating!}\footnote[$\smiley$]{If you aren't getting clean answers, double check that you copied the problems correctly.}} \hfill
\parbox{2.25in}{\bf \hfill Due: Fri., Sept. $2^\mathrm{nd}$, 2016}

\vspace{0.2in}

\begin{description}
\item[\#$\alpha$ Row, Column, Null] \quad Let $A=\begin{bmatrix} 1 & 3 & 1 & 6 & 1 & 2 \\ 4 & 12 & -1 & 4 & 1 & 9 \\ 3 & 9 & -2 & -2 & 0 & 7 \\ 2 & 6 & 0 & 4 & -1 & -1 \end{bmatrix}$. Find a bases for $\mathrm{Row}(A)$, $\mathrm{Col}(A)$, and $\mathrm{Null}(A)$. 

\item[\#ii Linear Correspondence] \quad Fill in the missing entries. $A$ is a matrix with real entries and $R$ is its RREF.

\begin{enumerate}[(a)]
\item $ A = \begin{bmatrix} 1 & ? & 1 & ? & ? &  3 & ? & ? \\  0 & ? & 2 & ? & ? &  1 & ? & ? \\  2 & ? & 2 & ? & ? & -1 & ? & ? \\ -1 & ? & 1 & ? & ? &  2 & ? & ? \end{bmatrix} 
     \quad \stackrel{\mathrm{RREF}}{\sim} \quad
      R = \begin{bmatrix} 1 & 3 & 0 & -1 & -1 & 0 & 0 &  2 \\  0 & 0 & 1 &  2 &  1 & 0 & 1 & -1 \\  0 & 0 & 0 &  0 &  0 & 1 & 1 &  1 \\  0 & 0 & 0 &  0 &  0 & 0 & 0 &  0 \end{bmatrix}$

\vspace{0.1in}

\item $ A = \begin{bmatrix} ? & ? & -2 & ? & -2 & ? \\  ? & ? & -4 & ? &  5 & ? \\  ? & ? & -6 & ? &  7 & ? \\  ? & ? & -2 & ? &  0 & ? \end{bmatrix}
     \quad \stackrel{\mathrm{RREF}}{\sim} \quad
      R =  \begin{bmatrix} 0 & 1 & -2 & 0 &  3 & 3 \\  0 & 0 &  0 & 1 & -1 & 2 \\  0 & 0 &  0 & 0 &  0 & 0 \\  0 & 0 &  0 & 0 &  0 & 0 \end{bmatrix}$

\vspace{0.1in}

\item $ A = \begin{bmatrix} 1 & ? & ? & 4 & ? & 0 \\ 2 & ? & ? & 2 & ? & 3 \\ 3 & ? & ? & 8 & ? & 2 \end{bmatrix}
     \quad \stackrel{\mathrm{RREF}}{\sim} \quad
      R =  \begin{bmatrix} ? & 2 & ? & 2 & 1 & ? \\ ? & 0 & ? & 2 & 3 & ? \\ ? & 0 & ? & 0 & 0 & ? \end{bmatrix}$
\end{enumerate}      

\item[\#$\lceil\pi/\sqrt{5}+1\rceil$ Coordinates, Spanning, Bases] \quad Let $\beta = \{ E_{11}, E_{12}, E_{21}, E_{22} \}$ be the standard basis for $\mathbb{R}^{2 \times 2}$.

{}[{\it Note:} As discussed in class, even though $\beta$ is a set, treat it like a list and maintain the order as shown above.]

$
A_1=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad
A_2=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, \quad
A_3=\begin{bmatrix} -1 & -2 \\ -1 & 2 \end{bmatrix}, \quad
A_4=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad
A_5=\begin{bmatrix} 1 & 3 \\ 6 & 10 \end{bmatrix} \quad
$

\begin{enumerate}[(a)]
\item Does $A_4$ belong to $\mathrm{Span} \{A_1,A_2,A_3 \}$?
\item Let $S= \{A_1,A_2,A_3,A_4 \}$ and $W=\mathrm{Span}(S)$. Find a subset of $S$ which forms a basis for $W$.
\item Using your basis found in part (b), find coordinates for $A_1,\dots,A_5$.
\item Extend $\{A_5\}$ to a basis for $W$.
\end{enumerate}
\end{description}

\vspace{1in}

For no reason a tree of $\pi$'s and $\chi$'s\dots

\begin{center}
$\pi$\\
$\pi$\\
$\pi\pi\pi$\\
$\pi\pi\pi$\\
$\pi\pi\pi$\\
$\pi\pi\pi\pi\pi$\\
$\pi\pi\pi\pi\pi$\\
$\pi\pi\pi\pi\pi\pi\pi$\\
$\pi\pi\pi\pi\pi\pi\pi$\\
$\pi\pi\pi\pi\pi\pi\pi\pi\pi$\\
$\chi$
\end{center}

\end{document}