\documentstyle[12pt]{article}

\pagestyle{headings}

\newcounter{xprobno}
\newcommand{\xprob}[1]{\addtocounter{xprobno}{1}
   \par\vspace{.25in}\noindent\thexprobno . (#1 points) }
\newcommand{\reals}{{\bf R}}


\begin{document}

\begin{flushright}
  A. Miller\\
  M340 Final Exam \\
  May 13, 1993
\end{flushright}


{\bf There is a total of 150 points.   Show all work. }


\xprob{15} What is the determinant of $(AB)^T$?
$$A=
\left[
\begin{array}{rrrr}
 1 & 0 & 0 & 0   \\
 6 &-1 & 0 & 0   \\
 7 & 0 & 2 & 0   \\
 2 & 1 & 0 & 3   \\
\end{array}
\right]
\;\;\;\;
B=
\left[
\begin{array}{rrrr}
 2 & 1 & 1 & 4   \\
 0 &-1 & 2 & 0   \\
 0 & 0 & 1 &-3   \\
 0 & 0 & 0 & 4   \\
\end{array}
\right]
$$



\xprob{15} Let $A$ be an $m\times n$ matrix and
$X=
\left[
\begin{array}{c}
 x_1    \\
 \vdots \\
 x_n    \\
\end{array}
\right]
$
an $n\times 1$ column vector.  Prove that
$$AX=x_1{\rm col}_1(A)+\ldots +x_n{\rm col}_n(A).$$

\xprob{15}
$$A=
\left[
\begin{array}{rrr}
 1 & -2 &  0   \\
 0 &  0 &  2   \\
 0 &  0 & -1   \\
\end{array}
\right]
$$

Find an invertible matrix $P$ and diagonal
matrix $D$ with $A=PDP^{-1}$.


\xprob{15} Use Cramer's rule to solve:
$$       x - y = 2 $$
$$      2x + y = 3 $$





\xprob{15} Suppose $A$ is an invertible matrix and $\lambda$ is an
eigenvalue of $A$.

  (a) Prove that $\lambda\not=0$.


 (b) Prove that  $1\over\lambda$ is an eigenvalue of $A^{-1}$.

\newpage

\xprob{15}  Suppose $L:V\to W$ is a linear transformation,
the nullspace of $L$ has only zero vector in it
(null($L)=\{{\bf 0}\}$), and
$v_1,\ldots,v_n$ are linearly independent
vectors in $V$.  Prove that $L(v_1),\ldots,L(v_n)$ are linearly independent.

\xprob{20}
$$A=
\left[
\begin{array}{rrrrr}
          0 &  -1 &   1 &   0 &  -1     \\
          0 &   1 &  -1 &   0 &   1     \\
          0 &   1 &  -1 &   1 &  -2     \\
\end{array}
\right]
$$

     (a) Find a basis for the range space of $A$.

     (b) Find a basis for the null space of $A$.


\xprob{15} Suppose $A$ is a $13\times 12$ matrix, and
 $B$ is a $12\times 13$ matrix, and $C=AB$.  Prove that
 $C$ is not invertible.



\xprob{25} Let $\reals^{4\times 1}$ have the inner product defined by
$$\langle \left[
\begin{array}{r}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4
\end{array}
\right],\left[
\begin{array}{r}
 y_1 \\
 y_2 \\
 y_3 \\
 y_4
\end{array}
\right]\rangle=x_1y_1+2x_2y_2+2x_3y_3+x_4y_4$$
and use this inner product in the following problem.  Let
$$v_1=
\left[
\begin{array}{r}
 -1 \\
 1 \\
 1 \\
 0
\end{array}
\right]
\;\;\;\;
v_2=
\left[
\begin{array}{r}
 0    \\
 -1  \\
 1  \\
 1
\end{array}
\right]
\;\;\;\;
v_3=
\left[
\begin{array}{r}
 1    \\
 1  \\
 1  \\
 1
\end{array}
\right]
$$
 Find a  vector $w\in span\{v_1,v_2,v_3\}$ other than
the zero vector such that $w$ is orthogonal to both $v_1$ and
$v_2$.





% #> rref(a)
%
% ans =
% [        0    1   -1    0    1
%          0    0    0    1   -3
%          0    0    0    0    0     ] 
%

\end{document}