\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Describe in geometrical terms the transformations defined by the
following matrices.  What effect do these transformations have on
\begin{description}
\item[(i)] The square with vertices$(\pm1, \pm1),$
\item[(ii)] the unit circle?
\item
\item[(a)] $\ds \left(\begin{array}{cc} 4 & -6 \\ 6  & 4
\end{array}\right)$
\item[(b)] $\ds \left(\begin{array}{cc} 4 & 6 \\ -6  & 4
\end{array}\right)$
\item[(c)] $\ds \left(\begin{array}{cc} 1 & 2 \\ 2  & 4
\end{array}\right)$
\item[(d)] $\ds \left(\begin{array}{cc} \ds \frac{1}{\sqrt 2}
& \ds -\frac{1}{\sqrt 2} \\ \ds \frac{1}{\sqrt2}  & \ds
\frac{1}{\sqrt 2}
\end{array}\right)$
\item[(e)] $\ds \left(\begin{array}{cc} 2 & 0 \\ 1  & 1
\end{array}\right)$
\end{description}
\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] $\left| \begin{array}{cc} 4 & -6 \\ 6 & 4\end{array}
\right| = 52$

$\left( \begin{array}{cc} 4 & -6 \\ 6 & 4\end{array} \right) =
\sqrt{52}\left( \begin{array}{cc} \ds \frac{4}{\sqrt{52}} & \ds
\frac{-6}{\sqrt{52}} \\ \ds \frac{6}{\sqrt{52}} & \ds
\frac{4}{\sqrt{52}}\end{array} \right) = 52$

So the matrix performs a magnification by a factor $\sqrt{52}$ and
a rotation anticlockwise through $\cos^{-1} \frac{4}{\sqrt{52}}$

$\left( \begin{array}{cc} 4 & -6 \\ 6 & 4\end{array} \right)\left(
\begin{array}{cccc} 1 & 1 & -1 & -1  \\ 1 & -1 & 1 & -1
\end{array} \right) = \left( \begin{array}{cccc} -2 & 10 & -10 & 2
\\ 10 & 2 & -2 & -10 \end{array} \right)$

$x^2+y^2 = 1 \rightarrow X^2 + Y^2 = 52$

${}$

\item[(b)] $\left( \begin{array}{cc} 4 & 6 \\ -6 & 4\end{array} \right) =
\sqrt{52}\left( \begin{array}{cc} \ds \frac{4}{\sqrt{52}} & \ds
\frac{6}{\sqrt{52}} \\ \ds \frac{-6}{\sqrt{52}} & \ds
\frac{4}{\sqrt{52}}\end{array} \right) = 52$

So the matrix performs a magnification by a factor $\sqrt{52}$ and
a rotation clockwise through $\cos^{-1} \frac{4}{\sqrt{52}}$

${}$

\item[(c)] $\left( \begin{array}{c} X \\ Y \end{array} \right) =
\left( \begin{array}{cc} 1 & 2 \\ 3 &4 \end{array} \right) \left(
\begin{array}{c} x \\ Y \end{array} \right) = \left(
\begin{array}{c} x+2y \\ 2x + 4y  \end{array} \right)$ So ${\bf
R}^2 \rightarrow Y = 2X$

The inverse image of $(k, 2k)$ is the line $x+2y = k$.  For the
square the extremities of the image are $(\pm3, \pm6)$

\begin{center}
$\begin{array}{c}
\epsfig{file=cn-15-1.eps, width=45mm}
\end{array}
\ \ \ 
\begin{array}{l}
\textrm{So the extremities are}\\
\ds \left ( \pm \frac{5}{\sqrt{5}}, \pm \frac{10}{\sqrt{5}} \right ) =
\left (\pm \sqrt{5} , \pm 2 \sqrt{5} \right )
\end{array}$
\end{center}

${}$

\item[(d)] Rotation through $45^\circ$ anitclockwise

${}$

\item[(e)] $\left( \begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}
\right) \left( \begin{array}{cccc} 1 & 1 & -1 & -1 \\ 1 & -1 & 1 &
-1 \end{array} \right) = \left( \begin{array}{cccc} 2 & 1 & -1 &
-1 \\ 2 & 0 & 0 & -2 \end{array} \right)$

Gives a magnification and shear.

$x = \frac{1}{2}X \hspace{.3in} y = -\frac{1}{2} X + Y$

$x^2 + y^2 = 1 \rightarrow \frac{1}{2}X^2 - XY + Y^2 = 1$ -
ellipse.
\end{description}

\end{document}