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QUESTION

\begin{description}

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

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

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}



\bigskip

ANSWER

\begin{description}

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

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&2&0&1&0&0\\2&3&-1&0&1&0\\2&5&3&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&3&-2&0&1\end{array}\right),\
r_2'=r_2-2r_1,\ r_3'=r_3-2r_1
\\
\rightarrow
\left(\begin{array}{ccc|ccc}1&2&0&1&0&0\\0&1&1&2&-1&0\\0&1&3&-2&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&1&1&2&-1&0\\0&0&2&-4&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&-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&-7&3&1\\0&1&0&4&-\frac{3}{2}&-\frac{1}{2}
\\0&0&1&-2&\frac{1}{2}&\frac{1}{2}\end{array}\right),\
r_1'=r_1+2r_3,\ r_2'=r_2-r_3 $

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

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