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\newcommand\ds{\displaystyle}
\begin{document}

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QUESTION

Show that the system of equations

\begin{eqnarray*}
x-3y+2z&amp;=&amp;5\\3x+y-z&amp;=&amp;\beta\\\alpha x+2y-3z&amp;=&amp;15
\end{eqnarray*}

has a unique solution, whatever the value of $\beta$, provided
that $\alpha\neq16$.

If $\alpha=16$ find the value of $\beta$ for which the system of
equations is consistent and obtain the general solution in this
case.

\bigskip

ANSWER
\begin{eqnarray*}
x-3y+2z&amp;=&amp;5\\ 3x+y-z&amp;=&amp;\beta\\ \alpha x+2y-3z&amp;=&amp;15
\end{eqnarray*}

\begin{eqnarray*}
\left|\begin{array}{ccc}1&amp;-3&amp;2\\3&amp;1&amp;-1\\\alpha&amp;2&amp;-3\end{array}\right|
&amp;=&amp;1(-3-(-2))+(-3)(-\alpha-(-9))+2(6-\alpha)\\
&amp;=&amp;1(-1)-3(9-\alpha)+2(6-\alpha)\\&amp;=&amp;-1-27+3\alpha+12-2\alpha\\
&amp;=&amp;-16+\alpha.\end{eqnarray*}

$-16+\alpha=0$ when $\alpha=16$.  Hence there is a unique solution
when $\det A\neq0$, i.e. $\alpha\neq16$.

When $\alpha=16$,

\begin{eqnarray*}
x-3y+2z&amp;=&amp;5\\3x+y-z&amp;=&amp;\beta\\16x+2y-3z&amp;=&amp;1
\end{eqnarray*}

$\left(\begin{array}{ccc}1&amp;-3&amp;2\\3&amp;1&amp;-1\\16&amp;2&amp;-3\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)
=\left(\begin{array}{c}5\\\beta\\15\end{array}\right)\\
\left(\begin{array}{ccc}1&amp;-3&amp;2\\0&amp;10&amp;-7\\0&amp;50&amp;-35\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}5\\\beta-15\\-65\end{array}\right)$ $r_
2-3r_1, r_3-16r_1$\\
$\left(\begin{array}{ccc}1&amp;-3&amp;2\\0&amp;10&amp;-7\\0&amp;0&amp;0\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}5\\\beta-15\\10-5\beta\end{array}\right)$\\
Consistent only when $10-5\beta=0$, i.e. $\beta=2$\\ The reduced
system is
\begin{eqnarray*}
x-3y+2z&amp;=&amp;5\\ 10y-7z&amp;=&amp;-13
\end{eqnarray*}

$\ds z=c,\Rightarrow y=\frac{7c-13}{10},\Rightarrow
x=5+3y-2z=5+\frac{21c-39}{10}-2c=\frac{c+11}{10}$\\ The solution
is $\left(\begin{array}{c}x\\y\\z\end{array}\right)
=\left(\begin{array}{c}\frac{c+11}{10}\\\frac{7c-13}{10}\\c\end{array}\right)$





\end{document}
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