%\input ../../../../Papers/def.tex
%\twelvepoint
\def\RR{R}
\def\pmatrix#1{\left[\matrix{#1}\right]}
\def\ttul{\_}
\def\hrulefill\relax
{\hbox to \hsize{% Seal of NCSU Seal
\hbox to 0cm{\hskip -0.3cm\vbox to 1.4cm{\vbox to 2.7cm{\vfill%
%\special{psfile=/afs/eos.ncsu.edu/users/k/kaltofen/Postscript/ncsu-mac-my.ps}
%\special{psfile=/home/erich/Postscript/seal.ps}
}\vss}\hss}\hss
%\Bigrm
North Carolina State University\hss}
\medskip\centerline{%\tenpoint
Department of Mathematics}
%\centerline{\tenpoint College of Physical and Mathematical Sciences}
}
\vtop{%
\hbox{\bf MA-305 Elem Linear Algebra}
\hbox{{\sl First Midsemester Quiz---Solution}}}
\rightline{Spring 1997}
\vskip 2cm
\par\noindent{\sl Your Name:} \hbox to 5cm{\hrulefill}
\hfill\break For purpose of anonymous grading, please do {\bf not} write your name on
the subsequent pages.
\vskip 0.5cm
This examination consists of 5 questions, each question counting for the
given number of points, adding to a total of {\bf 20 points}.
Please write your answers in the spaces indicated, or
below the questions (using the back of the sheets if necessary).
%If you need more space, there is an extra page at the end of the exam.
You are allowed to consult {\bf two} 8.5in $\times$ 11in sheets
with notes, but {\bf not} your book or your class notes.
If you get stuck on a problem, it may be advisable to go to another
problem and come back to that one later.
\medskip
You will have {\bf 75 minutes} to do this test.
\par
\rightline{Good luck!}
\vskip 1cm
$$\matrix{%
\hbox{Problem 1}\hfill&\cr
\noalign{\bigskip}
\hfill 2&\cr
\noalign{\bigskip}
\hfill 3&\cr
\noalign{\bigskip}
\hfill 4&\cr
\noalign{\bigskip}
\hfill 5&\cr
&\cr
\noalign{\bigskip}
\hfill Total&\cr}$$
\vskip 2cm\noindent
If you are taking the exam later, please sign the following statement:
\bigskip\noindent
{\sl I,% \vrule height0.4pt width4cm depth 0.0pt,
\hbox to 4cm{\hss}
affirm that I have no knowledge of the contents of this exam.}
\bigskip
%
\rightline{\hbox to 5cm{\hrulefill}}
\rightline{\hbox to 5cm{\hss Signature\hss}}
\vfill\eject
%
%
\noindent{\bf Problem 1} (7 points, 1 point for part (a), 1.5 points for each other part):
Please answer the following questions.

\item{(a)}Please name two well-known mathematicians
who have made contributions to the field of linear algebra.
\vfill
\docLink{sol1_1a.tex}{Solution}
%%%Carl Friedrich Gauss, (M.-E.) Camille Jordan, Leonardo of Pisa Fibonacci
\vfill
\item{(b)}Is the following identity true for all pairs of invertible
matrices $A,B\in \RR^{n\times n}$?
$$ \left( (AB)^{-1} \right)^T = (A^{-1})^T (B^{-1})^T $$
Please explain your answer.
\vfill
\docLink{sol1_1b.tex}{Solution}
%%% Yes. $((AB)^{-1})^T = (B^{-1} A^{-1})^T = (A^{-1})^T (B^{-1})^T$
\vfill

\item{(c)}Please compute the explicit value of the vector
$\left[\matrix{0&1\cr 1&1\cr}\right]^4 \cdot \left[\matrix{0\cr 1\cr}\right]$.
\vfill
\docLink{sol1_1c.tex}{Solution}
%%%$\left[\matrix{3\cr 5\cr}\right]$
\vfill

\item{(d)}The set of all $n\times n$ matrices over $\RR$
forms a commutative group with matrix addition as its binary operation.
What does this mean?
\vfill
\docLink{sol1_1d.tex}{Solution}
%%%For all $A,B,C\in \RR^{n\times n}$ we have
%%%$$\eqalign{%
%%%&A + (B + C) = (A + B) + C\hbox{\quad ($+$ is associative)}\cr
%%%&A + 0^{n\times n} = 0^{n\times n} + A\hbox{\quad ($0^{n\times n}$ is
%%%an additive unit element)}\cr
%%%&A + (-A) = (-A) + A = 0^{n\times n}\hbox{\quad ($-A$ is an additive
%%%inverse)}\cr
%%%&A + B = B + A\hbox{\quad ($+$ is commutative)}\cr
%%%}$$
\vfill

\item{(e)}The name of the Maple procedure {\tt rref} in the {\tt linalg} package
is an acronym of which sequence of words?
\vfill
\docLink{sol1_1e.tex}{Solution}
%%%reduced row echelon form
\eject
%
%
\noindent{\bf Problem 2} (4 points):
The Gaussian elimination process has produced the following augmented
matrix
$$
  \left[\matrix{-3 & 2 & 1 & 4 & | & 6 \cr
                0 &  2 & 1 & 2 & | & -4 \cr
                0 &  0 & 2 & 1 & | & 2 \cr}\right]
$$
where the first four columns correspond to the variable $x$, $y$, $z$, and $w$.
Please give the solution to this system in terms of linear forms in
the free variables.
\vfill
$$\eqalign{
w &{}=w\cr
z &{}=1 - {1\over 2} w\cr
y &{}=-2 - {1\over 2}z - w = -2 - {1\over 2} + {1\over 4}w - w\cr
  &{}= -{5\over 2} - {3\over 4} w\cr
x &{}=-2 + {2\over 3}y + {1\over 3}z + {4\over 3}w= -2 + (-{5\over 3} - {1\over 2} w) + ({1\over 3} - {1\over 6}w)
   + {4\over 3} w\cr
  &{}= -{10\over 3} + {2\over 3}w\cr
}$$
\vfill
\eject
%
%
\noindent{\bf Problem 3} (5 points, 1 point for part (a) and (b), 1.5 points for
part (c) and (d)):
Consider the following matrix with
symbolic real parameters $\alpha$, $\beta$, and $\gamma$:
$$
  A = \pmatrix{ \alpha &  \beta  & 0      & 0      \cr
                \gamma &  \alpha & \beta  & 0      \cr
                0      &  \gamma & \alpha & \beta  \cr
                0      &  0      & \gamma & \alpha \cr}.
$$
Give necessary {\bf and sufficient} conditions that the
parameters $\alpha$, $\beta$, and $\gamma$ must satisfy
(simultaneous equalities and inequalities---there may be more than one) such that
\item{(a)}$A$ is upper triangular
\vfill
$\gamma = 0$
\vfill
\item{(b)}$A$ is diagonal {\bf and invertible}
\vfill
$\gamma = 0$ and $\beta = 0$ and $\alpha \not= 0$
\vfill
\item{(c)}$A$ is in row echelon form
\vfill
$\gamma=0$
\vfill
\item{(d)}$A$ is in reduced row echelon form
\vfill
$\gamma=0$ and [($\alpha = 1$ and $\beta = 0$) or ($\alpha = 0$
and $\beta = 1$)]
\eject
%
%
\noindent
{\bf Problem 4} (4 points): The following is product of 4 matrices with 3 rows
and 3 columns.  The matrices contain the variables $a$, $b$, $c$, $d$, $e$, and $f$.
$$
  \pmatrix{ 1 & 0 & 0\cr 
            0 & 1 & 0\cr
            0 & db-e & 1\cr} \cdot
  \pmatrix{ 1 & 0 & 0\cr
            0 & 1 & 0\cr
            -d & 0 & 1\cr}\cdot
  \pmatrix{ 0 & 1 & 0\cr
            1 & 0 & 0\cr
            0 & 0 & 1\cr} \cdot
  \pmatrix{ 0 & 1 & a\cr 
            1 & b & c\cr
            d & e & f\cr}.
$$
Give a 3 by 3 matrix that is the product multiplied out;  some entries
in the resulting matrix depend on the variables. Hint: perform the
product from right to left using properties of elementary matrices.
\vfill
Using properties of elementary matrices, we get
$$\eqalign{
&\pmatrix{ 0 & 1 & 0\cr
            1 & 0 & 0\cr
            0 & 0 & 1\cr} \cdot
  \pmatrix{ 0 & 1 & a\cr
            1 & b & c\cr
            d & e & f\cr} =
\pmatrix{ 1&b&c\cr 0&1&a\cr d&e&f\cr}\cr
&   \pmatrix{ 1 & 0 & 0\cr
            0 & 1 & 0\cr
            -d & 0 & 1\cr}\cdot
\pmatrix{ 1&b&c\cr 0&1&a\cr d&e&f\cr} =
\pmatrix{ 1&b&c\cr 0&1&a\cr 0&e-db&f-dc\cr}\cr
& \pmatrix{ 1 & 0 & 0\cr
            0 & 1 & 0\cr
            0 & db-e & 1\cr} \cdot
\pmatrix{ 1&b&c\cr 0&1&a\cr 0&e-db&f-dc\cr} =
\pmatrix{ 1&b&c\cr 0&1&a\cr 0&0&f-dc + a(db-a)\cr}\cr
}$$
\vfill
\noindent {\bf Bonus question} (1 point):
Give a sequence of Maple expressions that compute
the above product automatically.
\vfill\noindent{\tt
with(refpkg);\hfill\break
A := matrix(3,3,[0,1,a, 1,b,c, d,e,f]);\hfill\break
E1 := E\ttul I(3,1,3); E2:=E\ttul III(3,1,3,-d); E3:=E\ttul III(3,2,3,d*b-e);\hfill\break
evalm( E3 \&* E2 \&* E1 \&* A);
}
\vfill
\bye