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{\bf Question}

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

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{\bf Answer}

Elimination method to find inverse of A.

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


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

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

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

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

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


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

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

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

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

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

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


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