<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\begin{document}

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QUESTION

\begin{description}

\item[(a)]
Find the eigenvalues of the matrix

$$\left(\begin{array}{cc}1&amp;3\\3&amp;1\end{array}\right)$$ and
determine the corresponding eigenvectors.

\item[(b)]
Writing $z=x+jy$ find the locus of the point $z$ in the Argand
diagram which satisfies the equation
$\ds\left|\frac{z-2j}{z+1}\right|=1$.\\ Illustrate your result on
an Argand diagram, and explain briefly how the locus is
geometrically related to the points $2j$ and $-1$.

\end{description}



\bigskip

ANSWER

\begin{description}

\item[(a)]
$A=\left(\begin{array}{cc}1&amp;3\\3&amp;1\end{array}\right)$. Eigenvalues
satisfy $\det(A-\lambda I)=0$
\begin{eqnarray*}
\det(A-\lambda
I)&amp;=&amp;\det\left(\begin{array}{cc}1-\lambda&amp;3\\3&amp;1-\lambda\end{array}\right)\\
&amp;=&amp;(1-\lambda)^2-3^2\\ &amp;=&amp;(1-\lambda-3)(1-\lambda+3)\\
&amp;=&amp;(-\lambda-2)(4-\lambda)\\ &amp;=&amp;(\lambda-4)(\lambda+2)\\ &amp;=&amp;0\
{\rm if\ } \lambda=4,-2.
\end{eqnarray*}



${\lambda=4}:\\ (A-4I)\textbf{X}=0\\
\left(\begin{array}{cc}-3&amp;3\\3&amp;-3\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\end{array}\right)=
\left(\begin{array}{c}0\\0\end{array}\right)\\
\left.\begin{array}{r}-3x_1+3x_2=0\\3x_1-3x_2=0\end{array}\right\}\
x_1=x_2=C$

Therefore the eigenvector is
$C\left(\begin{array}{c}1\\1\end{array}\right)$

${\lambda=-2:}\\ (A-(-2)I)\textbf{X}=0,\ (A+2I)\textbf{X}=0\\
\left(\begin{array}{cc}3&amp;3\\3&amp;3\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\end{array}\right)=
\left(\begin{array}{c}0\\0\end{array}\right)\\
\left.\begin{array}{r}3x_1+3x_2=0\\3x_1+3x_2=0\end{array}\right\}\
x_2=-x_1, \, \Rightarrow x_1=D,\ x_2=-D$

Therefore the eigenvector is
$D\left(\begin{array}{c}1\\-1\end{array}\right)$


\item[(b)]

$\ds\left|\frac{z-2j}{z+1}\right|=1$, i.e.
$\ds\frac{|z-2j|}{|z+1|}=1$, or $|z-2j|=|z+1|$\\ Using $z=x+jy,\
|x+jy-2j|=|x+jy+1|$, or $|x+j(y-2)|=|x+jy+1|$\\
i.e.$\ds\{x^2+(y-2)^2\}^\frac{1}{2}=\{(x+1)^2+y^2\}^\frac{1}{2}$\\
Squaring both sides gives $x^2+(y-2)^2=(x+1)^2+y^2$\\ i.e.
$x^2+y^2-4y+4=x^2+2x+1+y^2,\ 4y=-2x+3$\\ i.e. $\ds
y=-\frac{1}{2}x+\frac{3}{4}$.

$|z-2j|=$ distance of $z$ from $2j$, $|z+1|=$ distance of $z$ from
$-1$.

The distances are the same along the straight line shown.

\setlength{\unitlength}{0.5in}

\begin{picture}(6,5)

\put(0,2){\line(1,0){6}}

\put(3,0){\vector(0,1){5}}

\put(2,2){\line(1,2){1}}

\put(.5,4){\line(2,-1){5}}

\put(5.7,1.5){$y=-\frac{x}{2}+\frac{3}{4}$}

\put(2,1.7){$-1$}

\put(4,1.7){1}

\put(5,2.1){2}

\put(3.1,3){1}

\put(3.1,4){2}

\end{picture}


\end{description}




\end{document}
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