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QUESTION
\begin{description}

\item[(i)]
Evaluate the determinant of the matrix
$$\textbf{A}=\left(\begin{array}{ccc}1&amp;2&amp;0\\2&amp;3&amp;-1\\\beta&amp;5&amp;1\end{array}\right),$$
where $\beta$ is a constant.

\item[(ii)]
For what values of $\beta$ does $\textbf{A}^{-1}$ exist?

\item[(iii)]
Determine $\textbf{A}^{-1}$ when $\beta=3$, and verify your
answer.

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(i)]

\begin{eqnarray*}
\det
A&amp;=&amp;\det\left(\begin{array}{ccc}1&amp;2&amp;0\\2&amp;3&amp;-1\\\beta&amp;5&amp;1\end{array}\right)\\
&amp;=&amp;1(3(1)-(-1)5)-2(2(1)-(-1)\beta)+0\\ &amp;=&amp;3+5-2(2+\beta)\\
&amp;=&amp;8-4-2\beta=4-2\beta
\end{eqnarray*}


\item[(ii)]
$A^{-1}$ exists when $\det A\neq0$ i.e. $\beta\neq2$.

\item[(iii)]
When $\beta=3,\
A=\left(\begin{array}{ccc}1&amp;2&amp;0\\2&amp;3&amp;-1\\3&amp;5&amp;1\end{array}\right)$

By Gaussian elimination

$\left(\begin{array}{ccc|ccc}1&amp;2&amp;0&amp;1&amp;0&amp;0\\
2&amp;3&amp;-1&amp;0&amp;1&amp;0\\3&amp;5&amp;1&amp;0&amp;0&amp;1\end{array}\right)\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;2&amp;0&amp;1&amp;0&amp;0\\
0&amp;-1&amp;-1&amp;-2&amp;1&amp;0\\0&amp;-1&amp;1&amp;-3&amp;0&amp;1\end{array}\right)
\ \ (r_2'=r_2-2r_1, r_3'=r_3-3r_1)\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;2&amp;0&amp;1&amp;0&amp;0\\
0&amp;1&amp;1&amp;2&amp;-1&amp;0\\0&amp;-1&amp;1&amp;-3&amp;0&amp;1\end{array}\right)
\ \ (r_2'=-r_2)
\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;0&amp;-2&amp;-3&amp;2&amp;0\\
0&amp;1&amp;1&amp;2&amp;-1&amp;0\\0&amp;0&amp;2&amp;-1&amp;-1&amp;1\end{array}\right)
\ \ (r_1'=r_1-2r_2,\ r_3'=r_3+r_2)
\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;0&amp;-2&amp;-3&amp;2&amp;0\\
0&amp;1&amp;1&amp;2&amp;-1&amp;0\\0&amp;0&amp;1&amp;-\frac{1}{2}&amp;-\frac{1}{2}&amp;\frac{1}{2}\end{array}\right)
\ \ (r_3'=\frac{1}{2}r_3)\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;0&amp;0&amp;-4&amp;1&amp;1\\
0&amp;1&amp;0&amp;2\frac{1}{2}&amp;-\frac{1}{2}&amp;-\frac{1}{2}\\
0&amp;0&amp;1&amp;-\frac{1}{2}&amp;-\frac{1}{2}&amp;\frac{1}{2}\end{array}\right)
\ \ (r_1'=r_1+2r_3,\ r_2'=r_2-r_3)
$

Therefore

$A^{-1}=\left(\begin{array}{ccc}-4&amp;1&amp;1\\\frac{5}{2}&amp;-\frac{1}{2}&amp;-\frac{1}{2}\\
-\frac{1}{2}&amp;-\frac{1}{2}&amp;\frac{1}{2}\end{array}\right)$

\begin{eqnarray*}
AA^{-1}&amp;=&amp;\left(\begin{array}{ccc}1&amp;2&amp;0\\2&amp;3&amp;-1\\3&amp;5&amp;1\end{array}\right)
\left(\begin{array}{ccc}-4&amp;1&amp;1\\\frac{5}{2}&amp;-\frac{1}{2}&amp;-\frac{1}{2}\\
-\frac{1}{2}&amp;-\frac{1}{2}&amp;\frac{1}{2}\end{array}\right)\\
&amp;=&amp;\left(\begin{array}{ccc}-4+5+0&amp;1-1&amp;1-1+0\\
-8+\frac{15}{2}+\frac{1}{2}&amp;2-\frac{3}{2}+\frac{1}{2}&amp;2-\frac{3}{2}-\frac{1}{2}\\
-12+\frac{25}{2}-\frac{1}{2}&amp;3-\frac{5}{2}-\frac{1}{2}&amp;3-\frac{5}{2}+\frac{1}{2}
\end{array}\right)\\
&amp;=&amp;\left(\begin{array}{ccc}1&amp;0&amp;0\\0&amp;1&amp;0\\0&amp;0&amp;1\end{array}\right)
\end{eqnarray*}

\begin{eqnarray*}
A^{-1}A&amp;=&amp;\left(\begin{array}{ccc}-4&amp;1&amp;1\\\frac{5}{2}&amp;-\frac{1}{2}&amp;-\frac{1}{2}\\
-\frac{1}{2}&amp;-\frac{1}{2}&amp;\frac{1}{2}\end{array}\right)
\left(\begin{array}{ccc}1&amp;2&amp;0\\2&amp;3&amp;-1\\3&amp;5&amp;1\end{array}\right)\\
&amp;=&amp;\left(\begin{array}{ccc}-4+2+3&amp;-8+3+5&amp;-1+1\\
\frac{5}{2}-1-\frac{3}{2}&amp;5-\frac{3}{2}-\frac{5}{2}&amp;\frac{1}{2}-\frac{1}{2}\\
-\frac{1}{2}-1+\frac{3}{2}&amp;-1-\frac{3}{2}+\frac{5}{2}&amp;\frac{1}{2}+\frac{1}{2}
\end{array}\right)\\
&amp;=&amp;\left(\begin{array}{ccc}1&amp;0&amp;0\\0&amp;1&amp;0\\0&amp;0&amp;1\end{array}\right)\\
\end{eqnarray*}

\end{description}



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