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QUESTION

\begin{description}

\item[(i)]
Evaluate the determinant of the matrix

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

where $\beta$ is a constant.

\item[(ii)]
For what value of $\beta$ does $\textbf{A}^{-1}$ not exist?

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

\end{description}



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ANSWER

\begin{description}

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

so
$\det\textbf{A}=1(3\beta-(-1)5)-2(2\beta-(-1)2)+0=3\beta+5-4\beta-4=1-\beta$,

\item[(ii)]
$\mathbf{A}^{-1}$ does not exist if det$\mathbf{A}=0$,
i.e.$\beta=1$

\item[(iii)]
Using 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\\2&amp;5&amp;3&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;3&amp;-2&amp;0&amp;1\end{array}\right),\
r_2'=r_2-2r_1,\ r_3'=r_3-2r_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;3&amp;-2&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;1&amp;1&amp;2&amp;-1&amp;0\\0&amp;0&amp;2&amp;-4&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;-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;-7&amp;3&amp;1\\0&amp;1&amp;0&amp;4&amp;-\frac{3}{2}&amp;-\frac{1}{2}
\\0&amp;0&amp;1&amp;-2&amp;\frac{1}{2}&amp;\frac{1}{2}\end{array}\right),\
r_1'=r_1+2r_3,\ r_2'=r_2-r_3 $

Hence
$\mathbf{A}^{-1}=\left(\begin{array}{ccc}-7&amp;3&amp;1\\4&amp;-\frac{3}{2}&amp;-\frac{1}{2}
\\-2&amp;\frac{1}{2}&amp;\frac{1}{2}\end{array}\right)$

To verify that this is correct
\begin{eqnarray*}
\mathbf{AA^{-1}}&amp;=&amp;
\left(\begin{array}{ccc}1&amp;2&amp;0\\2&amp;3&amp;-1\\2&amp;5&amp;3\end{array}\right)
\left(\begin{array}{ccc}-7&amp;3&amp;1\\4&amp;-\frac{3}{2}&amp;-\frac{1}{2}\\
-2&amp;\frac{1}{2}&amp;\frac{1}{2}\end{array}\right)\\
&amp;=&amp;\left(\begin{array}{ccc}-7+8+0&amp;3-3+0&amp;1-1+0\\
-14+12+2&amp;6-\frac{9}{2}-\frac{1}{2}&amp;2-\frac{3}{2}-\frac{1}{2}\\
-14+20-6&amp;6-\frac{15}{2}+\frac{3}{2}&amp;2-\frac{5}{2}+\frac{3}{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)
\\ \mathbf{A^{-1}A}&amp;=&amp;
\left(\begin{array}{ccc}-7&amp;3&amp;1\\4&amp;-\frac{3}{2}&amp;-\frac{1}{2}\\
-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\\2&amp;5&amp;3\end{array}\right)\\
&amp;=&amp;\left(\begin{array}{ccc}-7+6+2&amp;-14+9+5&amp;0-3+3\\
4-3-1&amp;8-\frac{9}{2}-\frac{5}{2}&amp;0+\frac{3}{2}-\frac{3}{2}\\
-2+1+1&amp;-4+\frac{3}{2}+\frac{5}{2}&amp;0-\frac{1}{2}+\frac{3}{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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