$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right]$

.:

Let us take A = $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right]$

We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}&3\\2&2&3\\3&{ - 2}&2\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A$

Applying ${R_1} \to {R_1} - {R_2}$,

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 5}&0\\2&2&3\\2&{ - 2}&2\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&0\\0&1&0\\0&0&1\end{array}} \right]A$

Applying ${R_3} \to {R_3} - {R_2}$,

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 5}&0\\2&2&3\\1&{ - 4}&{ - 1}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&0\\0&1&0\\0&{ - 1}&1\end{array}} \right]A$

Applying ${R_2} \to {R_2} - 2{R_3}$,

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 5}&0\\0&{10}&5\\1&{ - 4}&{ - 1}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&0\\0&3&{ - 2}\\0&{ - 1}&1\end{array}} \right]A$

Applying ${R_1} \leftrightarrow {R_3},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 4}&{ - 1}\\0&{10}&5\\0&{ - 5}&0\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&1\\0&3&{ - 2}\\1&{ - 1}&0\end{array}} \right]A$

Applying ${R_3} \leftrightarrow {R_2}$,

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 4}&{ - 1}\\0&{ - 5}&0\\0&{10}&5\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&1\\1&{ - 1}&0\\0&3&{ - 2}\end{array}} \right]A$

Applying ${R_3} \to {R_3} + 2{R_2},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 4}&{ - 1}\\0&{ - 5}&0\\0&0&5\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&1\\1&{ - 1}&0\\2&1&{ - 2}\end{array}} \right]A$

Applying ${R_2} \to - {R_2},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 4}&{ - 1}\\0&5&0\\0&0&5\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&1\\{ - 1}&1&0\\2&1&{ - 2}\end{array}} \right]A$

Applying ${R_2} \to \cfrac{1}{5}{R_2}$and ${R_3} \to \cfrac{1}{5}$ ${R_3}$,

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 4}&{ - 1}\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&1\\{\cfrac{{ - 1}}{5}}&{\cfrac{1}{5}}&0\\{\cfrac{2}{5}}&{\cfrac{1}{5}}&{\cfrac{{ - 2}}{5}}\end{array}} \right]A$

Applying ${R_1} \to {R_1} + 4{R_2},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{ - 1}\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \cfrac{4}{5}}&{ - \cfrac{1}{5}}&1\\{ - \cfrac{1}{5}}&{\cfrac{1}{5}}&0\\{\cfrac{2}{5}}&{\cfrac{1}{5}}&{\cfrac{{ - 2}}{5}}\end{array}} \right]A$

Applying ${R_1} \to {R_1} + {R_3}$,

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cfrac{{ - 2}}{5}}&0&{\cfrac{3}{5}}\\{\cfrac{{ - 1}}{5}}&{\cfrac{1}{5}}&0\\{\cfrac{2}{5}}&{\cfrac{1}{5}}&{\cfrac{{ - 2}}{5}}\end{array}} \right]$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cfrac{{ - 2}}{5}}&0&{\cfrac{3}{5}}\\{\cfrac{{ - 1}}{5}}&{\cfrac{1}{5}}&0\\{\cfrac{2}{5}}&{\cfrac{1}{5}}&{\cfrac{{ - 2}}{5}}\end{array}} \right]$