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{\bf Question}

Use the elimination method to find the inverse of the matrix $$
{\bf A}=\left(\begin{array}{ccc}1&amp;2&amp;-1 \\ 1&amp; -1&amp; 1 \\-1 &amp; 3 &amp; 2
\end{array}\right)$$
Check that the answer obeys the relations ${\bf AA}^{-1} = {\bf
A}^{-1}{\bf A} = {\bf I}$

\vspace{.25in}

{\bf Answer}

Elimination method to find inverse of A.

$\displaystyle \left(\begin{array}{ccc}1&amp;2&amp;-1 \\ 1&amp; -1&amp; 1 \\-1 &amp; 3
&amp; 2
\end{array}\right)\left(\begin{array}{ccc}1&amp;0&amp;0 \\ 0&amp; 1&amp; 0
\\0 &amp; 0 &amp; 1\end{array}\right) \begin{array}{l} \\ {\rm row}\  2
\rightarrow {\rm row}\  2 - {\rm row}\  1 \\ {\rm row}\  3
\rightarrow {\rm row}\  3 + {\rm row}\  1
\end{array}$


$\displaystyle \rightarrow\left(\begin{array}{ccc}1&amp;2&amp;-1 \\ 0&amp; -3&amp;
2 \\0 &amp; 5 &amp; 1\end{array}\right)\left(\begin{array}{ccc}1&amp;0&amp;0 \\
-1&amp; 1&amp; 0
\\1 &amp; 0 &amp; 1\end{array}\right) \begin{array}{l} {\rm row}\  1 \rightarrow
{\rm row}\  1 + \frac{2}{3} {\rm row}\  2\\  \\ {\rm row}\  3
\rightarrow {\rm row}\  3 + \frac{5}{3} {\rm row}\  1 \end{array}$

$\displaystyle \rightarrow\left(\begin{array}{ccc}1&amp;0&amp;\frac{1}{3}
\\ 0&amp; -3&amp; 2
\\0 &amp; 0 &amp; \frac{13}{3}\end{array}\right)\left(\begin{array}{ccc}
\frac{1}{3}&amp;\frac{2}{3}&amp;0 \\ -1&amp;1&amp; 0
\\-\frac{2}{3} &amp; \frac{5}{3} &amp; 1\end{array}\right) \begin{array}{l} {\rm row}\  1 \rightarrow
{\rm row}\  1 - \frac{1}{13} {\rm row}\  3\\  {\rm row}\  2
\rightarrow {\rm row}\  2 -\frac{6}{13} {\rm row}\  3 \\ \\
\end{array}$

$\displaystyle \rightarrow\left(\begin{array}{ccc}1&amp;0&amp;0 \\ 0&amp; -3&amp;
0
\\0 &amp; 0 &amp; \frac{13}{3}\end{array}\right)\left(\begin{array}{ccc}
\frac{5}{13}&amp;\frac{7}{13}&amp;-\frac{1}{13} \\
-\frac{9}{13}&amp;\frac{3}{13}&amp; -\frac{6}{13}
\\-\frac{2}{3} &amp; \frac{5}{3} &amp; 1\end{array}\right) \begin{array}{l} \\{\rm row}\  2 \rightarrow
- \frac{1}{3}{\rm row}\  2 \\ {\rm row}\  3 \rightarrow
\frac{3}{13}{\rm row}\  3
\end{array}$

$\displaystyle \rightarrow\left(\begin{array}{ccc}1&amp;0&amp;0 \\ 0&amp; 1&amp; 0
\\0 &amp; 0 &amp; 1\end{array}\right)\left(\begin{array}{ccc}
\frac{5}{13}&amp;\frac{7}{13}&amp;-\frac{1}{13} \\
-\frac{3}{13}&amp;-\frac{1}{13}&amp; \frac{2}{13}
\\-\frac{2}{13} &amp; \frac{5}{13} &amp; \frac{3}{13}\end{array}\right)$

To check that this is the inverse check that $AA^{-1} = I$ and
$A^{-1}A = I$


$\left(\begin{array}{ccc}1&amp;2&amp;-1 \\ 1&amp; -1&amp; 1 \\-1 &amp; 3 &amp; 2
\end{array}\right)\left(\begin{array}{ccc} \frac{5}{13}&amp;\frac{7}{13}&amp;-\frac{1}{13}
\\ -\frac{3}{13}&amp;-\frac{1}{13}&amp; \frac{2}{13}
\\-\frac{2}{13} &amp; \frac{5}{13} &amp; \frac{3}{13}\end{array}\right)$

$ = \frac{1}{13} \left(\begin{array}{ccc} 5+6+2&amp;7-2-5 &amp;-1+4-3 \\
5-3-2 &amp; 7+1+5 &amp; -1-2+3 \\ -5+9-4 &amp; -7-3+10 &amp; 1+6+6
\end{array}\right) = I$

$\left(\begin{array}{ccc}1&amp;2&amp;-1 \\ 1&amp; -1&amp; 1 \\-1 &amp; 3 &amp; 2
\end{array}\right)\left(\begin{array}{ccc} \frac{5}{13}&amp;\frac{7}{13}&amp;-\frac{1}{13}
\\ -\frac{3}{13}&amp;-\frac{1}{13}&amp; \frac{2}{13}
\\-\frac{2}{13} &amp; \frac{5}{13} &amp; \frac{3}{13}\end{array}\right)$

$ = \frac{1}{13}\left(\begin{array}{ccc} 5+7+1 &amp; 10-7-3 &amp;-5+7-2 \\
3-1-2 &amp; 6+1+6 &amp; -3-1+4 \\ -2+5-3 &amp; -4-5+9 &amp; 2+5+6
\end{array}\right) = I$

Hence $A^{-1} = \left(\begin{array}{ccc}
\frac{5}{13}&amp;\frac{7}{13}&amp;-\frac{1}{13}
\\ -\frac{3}{13}&amp;-\frac{1}{13}&amp; \frac{2}{13}
\\-\frac{2}{13} &amp; \frac{5}{13} &amp; \frac{3}{13}\end{array}\right)$

and $AA^{-1} = A^{-1}A = I$ as required.


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