\documentclass[12pt]{article}
\usepackage{fancyhdr,fancybox,amssymb,epsfig,amsmath,latexsym,bm}
\topmargin=-.5in
\headsep=0.2in
\oddsidemargin=0in
\textwidth=6.7in
\textheight=9.2in
\footskip=.5in

% Change this line to \def\bold{\mathbf} for ``upright boldface.''
\def\bold{\bm}

% Code to make an augmented matrix
\makeatletter
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
  \hskip -\arraycolsep
  \let\@ifnextchar\new@ifnextchar
  \array{#1}}
\makeatother

\newcommand{\C}{\mathbb{C}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\def\0{\bold{0}}
\def\A{\bold{A}}
\def\B{\bold{B}}
\def\D{\bold{D}}
\def\E{\bold{E}}
\def\I{\bold{I}}
\def\M{\bold{M}}
\def\U{\bold{U}}
\def\b{\bold{b}}
\def\c{\bold{c}}
\def\d{\bold{d}}
\def\e{\bold{e}}
\def\n{\bold{n}}
\def\u{\bold{u}}
\def\v{\bold{v}}
\def\x{\bold{x}}
\def\w{\bold{w}}
\def\<{\langle}
\def\>{\rangle}


\DeclareMathOperator{\Poly}{Poly}
\DeclareMathOperator{\Span}{Span}
\DeclareMathOperator{\rank}{rank}
\newcommand{\vv}[2]{\begin{bmatrix} #1 \\ #2 \end{bmatrix}}
\newcommand{\vvv}[3]{\begin{bmatrix} #1 \\ #2 \\ #3 \end{bmatrix}}

\renewcommand{\footrulewidth}{1pt}
\newenvironment{proof}[1][Proof.]{\smallskip\begin{trivlist}
    \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip}
\newenvironment{sol}[1][Solution.]{\smallskip\begin{trivlist}
    \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip}
\newenvironment{fminipage}%
{\setlength{\fboxsep}{15pt}\begin{Sbox}\begin{minipage}}%
{\end{minipage}\end{Sbox}\fbox{\TheSbox}}
\newcommand{\qed}{\hfill \ensuremath{\Box}}
\newcommand{\disp}{\displaystyle}


\pagestyle{fancy} \lhead{{\sf Homework 2 $|$ Due January 24 (Friday) }}
\rhead{\thepage} \cfoot{{\sf Math 3110 $|$ Linear Algebra $|$ Spring 2014 
    $|$ M.~Macauley}}


\begin{document}

$\;$

\noindent\emph{Read}: Strang, Section 1.3, 2.1, 2.2, 2.3. \\

% Source: Wrote all problems from scratch.

\begin{enumerate}

\item In this problem, we will solve a system $\A\x=\b$ using the
  method of \emph{elimination}, also called \emph{row-reduction}. The
  system is shown in matrix form on the left, and the corresponding
  \emph{augmented matrix} is shown at right, which is a useful
  notation.
  \[
  \begin{bmatrix} 1 & -2 \\ -4 & 2 \end{bmatrix}
  \vv{x}{y}=\vv{3}{0}\,\qquad\qquad[\A\;|\;\b\,]
    =\begin{bmatrix}[cc|c] 1 & -2 & 3 \\ -4 & 2 & 0 \end{bmatrix}\,.
  \]
  Carry out the following steps for this system.
  \begin{enumerate}
    \item Perform a single elimination step to convert the equation
      $\A\x=\b$ into an equation $\U\x=\c$, where $\U$ is an
      \emph{upper triangular matrix}. What is the corresponding
      elementary matrix, $\E_{21}$? Notice that $\U=\E_{21}\A$.
    \item Now, perform a single elimination step to remove the entry
      of $\U$ \emph{above} the main diagonal. You will end up with a
      equation of the form $\D\x=\d$, where $\D$ is a \emph{diagonal
        matrix}.
    \item What is $\D^{-1}$, the \emph{inverse matrix} of $\D$? (You
      should be able to do this by inspection.)  Multiply both sides
      of the equation $\D\x=\d$ by $\D^{-1}$ to solve for $\x$.
    \item The solution you obtained in Part~(c) is actually the vector
      $\x=\A^{-1}\b$. Express $\A^{-1}$ as a product of matrices that
      you found above ($\E_{12}$, $\E_{21}$, and $\D^{-1}$).
  \end{enumerate}



\item Consider the following $3\times 3$ system, $\A\x=\b$. The
  augmented matrix $[\A\;|\;\b\,]$ is not shown.
  \[
  \begin{bmatrix} 1 & 3 & 4 \\ 2 & 5 & 6 \\ -2 & -7 & -9\end{bmatrix}
    \vvv{x}{y}{z}=\vvv{1}{-2}{4}
  \]
  \begin{enumerate}
    \item Perform a sequence of elimination steps to convert the
      equation $\A\x=\b$ into an equation $\U\x=\c$, where $\U$ is an
      \emph{upper triangular matrix}. For each step, write down the
      corresponding elementary matrix, $\E_{ij}$.
    \item Express $\U$ as a product of elementary matrices and
      $\A$. (Note that $\A$ should be the rightmost matrix in the
      expression.)
    \item Continue the elimination process to eliminate the entries of
      $\U$ above the main diagonal. You will end up with a equation of
      the form $\D\x=\d$, where $\D$ is a diagonal matrix.
    \item Find $\D^{-1}$ and multiply it to both sides of the equation
      $\D\x=\d$ to solve for $\x$.
    \item Note that the solution you obtained in Part~(d) is
      $\x=\A^{-1}\b$. Express $\A^{-1}$ as a product of elementary
      matrices and $\D^{-1}$.
  \end{enumerate}



\item In both of the following equations, the matrix $\A$ is not
  invertible, so there will either be no solution, or infinitely many
  solutions to the equation $\A\x=\b$.
  \[
  \begin{bmatrix} 1 & -2 \\ -3 & 6 \end{bmatrix}\vv{x}{y}=\vv{-3}{1}\,,
  \qquad\qquad
  \begin{bmatrix} 1 & -2 \\ -3 & 6 \end{bmatrix}\vv{x}{y}=\vv{1}{-3}\,.
  \]
  Carry out the following steps for each equation above.
  \begin{enumerate}
  \item Graph the two linear equations of the system (this is the
    ``row picture'').
  \item Use the eliminaton method until you get an unavoidable zero in
    a pivot location, in which case you have to stop. At this point
    you will have an equation $\U\x=\c$, where $\U$ is
    upper-triangular.
  \item Write out the two equations that the system $\U\x=\c$
    represents. Are there no solutions, or infinitely many solutions?
    Why?
  \end{enumerate}



\item We will solve the equation $\A\x=\0$ for each of the three
  matrices $\A$ shown below.
  \[
  \begin{bmatrix} 1 & -2 & 1 \\ -1 & 5 & 6 \\ 5 & -4 & 10 \\ \end{bmatrix}
  \,\qquad\qquad\qquad
  \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & 0 \\ 1 & -1 & 1 \\ \end{bmatrix}
  \,\qquad\qquad\qquad
  \begin{bmatrix} 1 & -2 & 1 \\ -1 & 2 & -1 \\ 3 & -6 & 3 \\ \end{bmatrix}
  \]
  Recall that if $\A$ is not invertible, then there will be a non-zero
  solution to the equation $\A\x=\0$. Carry out the following steps
  for each equation above.
  \begin{enumerate}
  \item Use the elimination method to row-reduce $\A$ to an
    upper-triangular matrix $\U$, and if possible, to a diagonal
    matrix $\D$. (Show your steps, but you do not need to write out
    each individual elementary matrix like you did in Problems 1 and
    2.)
  \item Describe the set of all solutions to $\U\x=\0$ (and hence to
    $\A\x=\0$) as best you can. Give a geometric interpretation (e.g.,
    a point, line, plane, etc.)
  \end{enumerate}



\item Find the inverse of the following matrices by row-reducing the
  corresponding \emph{augmented matrix} until the left half is the
  identity matrix. 
  \[
    [\A\;|\;\I\,]
    =\begin{bmatrix}[cc|cc] 1 & -2 & 1 & 0 \\ 3 & -7 & 0 & 1\end{bmatrix}
    \,\qquad\qquad
    [\B\;|\;\I\,]=\begin{bmatrix}[ccc|ccc] 3 & 1 & 4 & 1 & 0 & 0 \\ 
    5 & 2 & 6 & 0 & 1 & 0 \\ -7 & -2 & -9 & 0 & 0 & 1\end{bmatrix}
  \]
  When you do this, the augmented matrices that you end up with will
  be $[\I\;|\;\A^{-1}]$ and $[\I\;|\;\B^{-1}]$,
  resepectively. [\emph{Hint}: The matrix $\B^{-1}$ has only integer
    entries -- by scaling appropriately, you can avoid fractions
    altogether in the elimination process.]


\end{enumerate}

\end{document}
