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\input amssym.def
\input amssym
\newcommand{\reals}{{\Bbb R}}

\newcounter{xprobno}
\newcommand{\xprob}[1]{\addtocounter{xprobno}{1}
   \par\medskip\noindent\thexprobno . (#1 points) }
\newcommand{\by }{\times}
\newcommand{\norm}[1]{||#1||}
\newcommand{\complex }{\Bbb C}
\newcommand{\field }{\Bbb F}

\begin{document}

\begin{flushright}
  A. Miller\\
  M340 Exam 1\\
  February 27, 1991
\end{flushright}


{\bf There is a total of 100 points.
Put your answers on separate sheets of paper.
Keep this exam,  hand in the answers and all work sheets.
}

\xprob{16} Let $A\in\field^{m\by n}$, $B\in\field^{n\by p}$,
and $C\in \field^{p\by r}$.
Prove that $A(BC)=(AB)C$.

\xprob{16} Find the inverse of the following matrix:
$$
\left[
\begin{array}{rrr}
         1   &     2   &     0   \\
         1   &     3   &     1    \\
        -1   &    -1   &     2
\end{array}
\right]
$$

\xprob{16} Find a $4\by 4$ elementary matrix $E$ such that
for any $4\by n$ matrix $A$, $EA$ is the matrix obtained from
$A$ by replacing the third row of $A$ by the third row of $A$ plus
$5$ times the fourth row of $A$.  What is $E^{-1}$?

\xprob{18} Find a matrix  $R$ in rref such that for some
invertible matrix $M$,  $MA=R$.  Find two solutions to
the equation $AX=Z$ where $Z$ is the $3\by 1$ zero matrix.

$$
A=
\left[
\begin{array}{rrrrr}
         1    &     2    &    -2    &    -3    &     2\\
        -1    &    -2    &     3    &     4    &     0\\
         3    &     6    &    -3    &    -6    &    -3
\end{array}
\right]
$$

\xprob{18} Let $A\in \field^{n\by n}$ and suppose that $A$ is
not invertible.   Prove that there is a $Y\in \field^{n\by 1}$
for which there does not exist $X\in \field^{n\by 1}$.
such that $AX=Y$.

\xprob{16} Let $T:V\rightarrow W$ is a mapping and
$X,Y,Z\subseteq V$.  Define $T(Z)$.
Prove that
$T(X\cap Y)\subseteq T(X)\cap T(Y)$.
Give an example of $T,X,Y$ such that
$T(X\cap Y)\not= T(X)\cap T(Y)$.

\newpage

\begin{flushright}
  A. Miller\\
  M340 Exam 1 answers\\
  February 27, 1991
\end{flushright}


1. see example 2.20 page 20.


$$2.
\hspace{.15in}
\left[
\begin{array}{rrr}
          7  &  -4  &   2    \\
         -3  &   2  &  -1    \\
          2  &  -1  &   1
\end{array}
\right]
\hspace{.25in}
3.
\hspace{.15in}
E=
\left[
\begin{array}{rrrr}
          1  &  0  &   0 & 0    \\
          0  &  1  &   0 & 0    \\
          0  &  0  &   1 & 5    \\
          0  &  0  &   0 & 1
\end{array}
\right]
\hspace{.15in}
%
E^{-1}=
\left[
\begin{array}{rrrr}
          1  &  0  &   0 & 0    \\
          0  &  1  &   0 & 0    \\
          0  &  0  &   1 & -5    \\
          0  &  0  &   0 & 1
\end{array}
\right]
$$

$$
4.
\hspace{.25in}
R=
\left[
\begin{array}{rrrrr}
          1 &    2  &   0  &  -1 &    0 \\
          0 &    0  &   1  &   1 &    0 \\
          0 &    0  &   0  &   0 &    1
\end{array}
\right]
\hspace{.5in}
X_1=
\left[
\begin{array}{r}
          0  \\
          0  \\
          0  \\
          0  \\
          0  \\
\end{array}
\right]
%
\hspace{.5in}
X_2=
\left[
\begin{array}{r}
         -2  \\
          1  \\
          0  \\
          0  \\
          0  \\
\end{array}
\right]
$$

5. see Thm 4.11 page 70-71.

6.
Define $T(Z)=\{T(z) : z\in Z\}$, or (in English) $T(Z)$ is the
set of all $T(z)$ for $z$ an element of $Z$.

THM. $T(X\cap Y)\subseteq T(X)\cap T(Y)$.

proof. Let $z\in T(X\cap Y)$. Then
for some $x\in X\cap Y$ we must have that $T(x)=z$.  Since
$x\in X\cap Y$ we have that both $x\in X$ and $x\in Y$.  So
that $z=T(x)\in T(X)$ and $z=T(x)\in T(Y)$ and therefore
$z\in T(X)\cap T(Y)$.

Example.  Let $T:\reals\rightarrow\reals$ be defined by
$T(x)=x^2$.  Let $X=\{0,1\}$ and $Y=\{0,-1\}$.   Then
$T(X\cap Y)=\{0\}\not=\{0,1\}=T(X)\cap T(Y)$.

\end{document}