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QUESTION
\begin{description}

\item[(i)]
Evaluate the determinant of the matrix
$$\textbf{A}=\left(\begin{array}{ccc}1&2&0\\2&3&-1\\\beta&5&1\end{array}\right),$$
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&=&\det\left(\begin{array}{ccc}1&2&0\\2&3&-1\\\beta&5&1\end{array}\right)\\
&=&1(3(1)-(-1)5)-2(2(1)-(-1)\beta)+0\\ &=&3+5-2(2+\beta)\\
&=&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 $\beta=3,\
A=\left(\begin{array}{ccc}1&2&0\\2&3&-1\\3&5&1\end{array}\right)$

By Gaussian elimination

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

Therefore

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

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

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

\end{description}



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