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

\begin{document}

\begin{flushright}
  A. Miller\\
  M340 Exam 2\\
  April 17, 1991
\end{flushright}


\xprob{20} Let $A\in\field^{m\by n}$ be a matrix which determines
a 1-1 map $A:\field^{n\by 1}\rightarrow \field^{m\by 1}$.  Prove that
$A$ has a left inverse.

\xprob{20} Let $A\in\field^{m\by n}$.
\par a. Define the column space of $A$, ${\cal C}(A)$.
\par b. Prove that ${\cal C}(A)$ is a subspace
of $\field^{m\by 1}$.


\xprob{15} Let $\sigma\in S_4$ be the permutation defined
by $\sigma(1)=4,\sigma(2)=3,\sigma(3)=1,\sigma(4)=2$.
\par a. What is sgn$(\sigma)$?
\par b. Write $\sigma$ as a product of transpositions.

\xprob{10} a. What is the (2,3)-cofactor of $A$, ie cof$_{2,3}(A)$?
\par b. What is entry$_{3,1}($adj$(A))$?

$$
A=
\left[
\begin{array}{rrr}
         1    &     2    &    -2 \\
        -1    &    -1    &     5 \\
         3    &     2    &    -4
\end{array}
\right]
$$

\xprob{10} Use Cramer's rule to solve the equations $AX=B$ where
$$
A=
\left[
\begin{array}{rr}
         1    &     2  \\
        -1    &     1  \\
\end{array}
\right]
\hspace{1in}
B=
\left[
\begin{array}{r}
         3  \\
         2  \\
\end{array}
\right]
$$

\xprob{25} (see minimat output).   Find a basis matrix for
\par a. $\cal{N}(A)$
\par b. $\cal{N}(R)$
\par c. $\cal{N}(D)$
\par d. $\cal{C}(A)$
\par e. $\cal{C}(R)$
\par f. $\cal{C}(D)$
\par g. $\cal{N}(A^T)$
\par h. $\cal{C}(A^T)$

\newpage

\begin{verbatim}
 A =
 [        -3.000    -4.000     9.000
          -5.000    -7.000    15.000
          -5.000    -7.000    15.000
          -3.000    -4.000     9.000     ] 

 #> [R M]=rrefm(A)

 R = 
 [        1.000    0.000   -3.000
          0.000    1.000    0.000
          0.000    0.000    0.000
          0.000    0.000    0.000     ] 

 M = 
 [        0.000    0.000    4.000   -7.000
          0.000    0.000   -3.000    5.000
          1.000    0.000   -0.000   -1.000
          0.000    1.000   -1.000   -0.000     ] 

 #> [DT NT]=rrefm(R'), D=DT', N=NT'

 DT = 
 [        1.000    0.000    0.000    0.000
          0.000    1.000    0.000    0.000
          0.000    0.000    0.000    0.000     ] 

 NT = 
 [        0.000    0.000   -0.333
          0.000    1.000    0.000
          1.000    0.000    0.333     ]

 D = 
 [        1.000    0.000    0.000
          0.000    1.000    0.000
          0.000    0.000    0.000
          0.000    0.000    0.000     ] 

 N = 
 [        0.000    0.000    1.000
          0.000    1.000    0.000
         -0.333    0.000    0.333     ] 

 #> invN=inv(N), invM=inv(M)

 invN = 
 [        1.000    0.000   -3.000
          0.000    1.000    0.000
          1.000    0.000    0.000     ] 

 invM = 
 [       -3.000   -4.000    1.000    0.000
         -5.000   -7.000    0.000    1.000
         -5.000   -7.000    0.000    0.000
         -3.000   -4.000    0.000    0.000     ]
\end{verbatim}


\newpage

\begin{flushright}
  A. Miller\\
  M340 Exam 2 answers\\
  April 17, 1991
\end{flushright}


1. By row reduction and column reduction we can find invertible
matrices $M\in \field^{m\by m}$ and $N\in \field^{n\by n}$ such
that $MAN=D$ where
$$
D=\left[
\begin{array}{rr}
          I_r & 0    \\
           0 &  0    \\
\end{array}
\right]
$$
Since $A,M,$ and $N$ are all 1-1,  their composition, $D$, is 1-1.
This means that $D$ must have a trivial null space, or equivalently
it cannot have any zero columns.
$$
D=\left[
\begin{array}{r}
          I_n   \\
           0    \\
\end{array}
\right]
$$
Noting that $D^TD=I_n$ and $A=M^{-1}DN^{-1}$, if we let $B=ND^TM$ then
$$BA=(ND^TM)(M^{-1}DN^{-1})=ND^T(MM^{-1})DN^{-1}=ND^TIDN^{-1}=$$
$$ND^TDN^{-1}=NIN^{-1}=NN^{-1}=I$$
so $BA=I$ and $B$ is a left inverse of $A$.

2. a. ${\cal C}(A)=\{Y\in \field^{m\by 1}:\exists X\in \field^{n\by 1}\;\;
AX=Y\}$ or ${\cal C}(A)=\{AX: X\in \field^{n\by 1}\}$.
\par b. Suppose $Y_1,Y_2\in {\cal C}(A)$ and $c_1,c_2\in \field$.
Then for some $X_1,X_2\in \field^{n\by 2}$ we have
$AX_1=Y_1$ and $AX_2=Y_2$.
$$A(c_1X_1+c_2X_2)=Ac_1X_1+Ac_2X_2=c_1AX_1+c_2AX_2=c_1Y_1+c_2Y_2$$
and so $c_1Y_1+c_2Y_2\in {\cal C}(A)$.

3. a. $-1$  \hspace{.24in} b. $<3,4> <2,4> <1,3>$ or $<2,3><3,4><1,3>$ or
$<1,4><2,3><3,4>$ where
$<i,j>$ is the transposition which swaps $i$ and $j$.

4. a. $(-1)^{2+3}$det$\left[
\begin{array}{rr}
           1 & 2   \\
           3 & 2     \\
\end{array}
\right]=4$

b.entry$_{3,1}($adj$(A))$=cof$_{1,3}(A)=(-1)^{1+3}$
det$\left[
\begin{array}{rr}
         -1 & -1   \\
           3 & 2     \\
\end{array}
\right]=1$

5. $$x_1={det \left[
\begin{array}{rr}
         3    &     2  \\
         2   &     1  \\
\end{array}
\right]
\over
det\left[
\begin{array}{rr}
         1    &     2  \\
        -1    &     1  \\
\end{array}
\right] }=-1/3
\hspace{1in}
x_2={det \left[
\begin{array}{rr}
         1    &     3  \\
        -1    &     2  \\
\end{array}
\right]
\over
det\left[
\begin{array}{rr}
         1    &     2  \\
        -1    &     1  \\
\end{array}
\right] }=5/3
$$

6.
\par a. $\cal{N}(A)$  same as b or last column of $N$.
\par b. $\cal{N}(R)$  $\left[
\begin{array}{r}
         3      \\
         0      \\
         1
\end{array}
\right]$
\par c. $\cal{N}(D)$   $\left[
\begin{array}{r}
         0      \\
         0      \\
         1
\end{array}
\right] $
\par d. $\cal{C}(A)$   First two columns of $A$.
\par e. $\cal{C}(R)$   First two columns of $R$.
\par f. $\cal{C}(D)$   First two columns of $D$.
\par g. $\cal{N}(A^T)$ Last two columns of $M^T$.
\par h. $\cal{C}(A^T)$ First two columns of $(N^{-1})^T$.

For g. and h. use: $MAN=D$ implies $N^TA^TM^T=D^T$

\bigskip
{\bf \large
Final Exam: Tuesday, May 14, 12:25-2:25, B127 (Here).
}

\end{document}