\documentclass[11pt]{article}
\usepackage{linalg}
\usepackage[pdftex]{graphicx}
\parindent=0in
\parskip=1ex
\begin{document}

\begin{center}\large\bf
Linear Algebra 2270-2 \\
\normalsize\rm
Due in Week 5
\end{center}

The fifth week finishes the work from chapter 1 and starts chapter 2. Here's the
list of problems, followed by a few answers.

\begin{description}
\item[{\bf Section 1.9}.] Exercises 15, 25, 31, 39

\item[{\bf Section 2.1}.] Exercises 13, 18, 23, 28, 29

\item[{\bf Section 2.2}.] Exercises 11, 23, 24, 35

\item[{\bf Section 2.3}.] Exercises 7, 13, 14, 17, 19, 20, 21, 33, 35
\end{description}


\subsection*{Some Answers}

{\bf 2.1-18}: The first two columns of $AB$ are $Ab_1$ and $Ab_2$. They are equal since $b_1$ and $b_2$ are equal.

{\bf 2.1-28}: Since the inner product $u^T v$ is a real number, it equals its transpose. That is,
$u^T v = (u^T v)^T = v^T (u^T )^T = v^T u$. Applied here is Theorem 3(d), regarding the transpose of a product of matrices, and
 Theorem 3(a). The outer product $uv^T$ is an n�n matrix. By Theorem 3, $(uv^T )^T = (v^T )^T u^T = vu^T$.

{\bf 2.2-24}: If the equation $Ax = b$ has a solution for each $b$ in $\Rr^n$ , then $A$ has a pivot position in each row, by
Theorem 4 in Section 1.4. Since $A$ is square, then the pivots must be on the diagonal of $A$. It follows that $A$
is row equivalent to the identity matrix $I_n$ . By Theorem 7, matrix $A$ is invertible.

{\bf 2.3-14}: If A is lower triangular with nonzero entries on the diagonal, then these $n$ diagonal entries can be
used as pivots to produce zeros below the diagonal. Thus A has $n$ pivots and so it is invertible, by the
Invertible Matrix Theorem. If one of the diagonal entries in $A$ is zero, then $A$ will have fewer than $n$ pivots and hence $A$ will be singular.


{\bf 2.3-20}: By the box following the Invertible Matrix Theorem, $E$ and $F$ are invertible and are inverses. So $FE = I = EF$, and therefore matrices $E$ and
$F$ commute.

\end{document}
