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{\bf Question}

\begin{description}
\item[(a)] Solve the following system of equations
\begin{eqnarray*} x + 2y + 2z & = & 11 \\ 2x - y + z & = & 3 \\
-4x + 7y + z & = & 13 \end{eqnarray*} Give a geometrical
interpretation.
\item[(b)] Find {\underline {all}} the 2x2 matrices $A$ with the
property that $$A^2 = \left(\begin{array}{cc} 4 & 0 \\ 0 & 1
\end{array} \right)$$ Hence find all the solutions $X$ of the
equation $$X^2 + 2X = \left( \begin{array}{cc} 3 & 0 \\ 0 & 0
\end{array}\right)$$ where X is a 2x2 matrix.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] $\left.\begin{array}{ccc}1 & 2 & 2 \\ 2 & -1 & 1 \\ -4
& 7 & 1 \end{array} \right | \begin{array}{c} 11 \\ 3 \\ 13
\end{array} \rightarrow \left. \begin{array}{ccc} 1 & 2 & 2 \\ 0 &
-5 & -3 \\ 0 & 15 & 9 \end{array} \right| \begin{array}{l}
11\\-19\\57 \leftarrow {\rm This\ is\ }-3 \times R2 \end{array}$

$\begin{array}{rcl} {\rm So \ \ } x + 2y + 2z & = & 11 \\ 5y + 3z
& = & 19 \end{array}$

Let $z=t$ then $\ds y = \frac{19-3t}{5}$

Thus $\ds x = 11 - 2t - 2\left(\frac{19 -3t}{5} \right) = \frac{55
- 10t -38 +6t}{5} = \frac{17-4t}{5}$

So $\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{c} \frac{17}{5} \\ \frac{19}{5} \\ 0
\end{array} \right) + t \left( \begin{array}{c} \frac{-4}{5} \\
\frac{-3}{5} \\ 1 \end{array} \right) $

This system represents three plans meeting in a common line, whose
equation is the solution.
\item[(b)] $\left( \begin{array}{cc} a & b \\ c & d \end{array}
\right)\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)
= \left( \begin{array}{cc} 4 & 0 \\ 0 & 1 \end{array} \right)$

Thus \begin{eqnarray} a^2 +bc & = & 4 \\ b(a+d) & = & 0 \\ c(a+d)
& - & 0 \\ d^2 + bc & = & 1 \end{eqnarray}

(1) and (4)  imply that $a^2 \not= d^2$ So $a+d \not=0$.  Thus
$b=c=0$ and $ a = \pm 2$ and $d = \pm 1$

So there are 4 possibilities: $$\left( \begin{array}{cc} 2 & 0 \\
0 & 1 \end{array} \right), \hspace{.2in} \left( \begin{array}{cc}
-2 & 0 \\ 0 & 1 \end{array} \right), \hspace{.2in} \left(
\begin{array}{cc} 2 & 0 \\ 0 & -1 \end{array} \right), \hspace{.2in}
\left( \begin{array}{cc} -2 & 0 \\ 0 & -1 \end{array} \right).$$

${}$

$X^2 + 2X = \left( \begin{array}{cc} 3 & 0 \\ 0 & 0
\end{array}\right)$

if and only if $X^2 + 2X +I =  \left( \begin{array}{cc} 4 & 0 \\ 0
& 1 \end{array}\right)$

Thus $(X +I)^2 = \left( \begin{array}{cc} 4 & 0 \\ 0 & 1
\end{array}\right)$

$X+I = $ any of the 4 above.

So $X = \left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array}
\right), \hspace{.1in} \left( \begin{array}{cc} -2 & 0 \\ 0 & 1
\end{array} \right), \hspace{.1in} \left(
\begin{array}{cc} 2 & 0 \\ 0 & -1 \end{array} \right), \hspace{.1in}
\left( \begin{array}{cc} -2 & 0 \\ 0 & -1 \end{array} \right).$

\end{description}
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