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

.:

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

We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&3&{ - 2}\\{ - 3}&0&{ - 5}\\2&5&0\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A$

Applying ${R_2} \to {R_2} + 3{R_1},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&3&{ - 2}\\0&9&{ - 11}\\2&5&0\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\3&1&0\\0&0&1\end{array}} \right]A$

Applying ${R_3} \to {R_3} - 2{R_1},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&3&{ - 2}\\0&9&{ - 11}\\0&{ - 1}&4\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\3&1&0\\{ - 2}&0&1\end{array}} \right]A$

Applying ${R_1} \to {R_1} + 3{R_3},$

wet get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{10}\\0&9&{ - 11}\\0&{ - 1}&4\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&0&3\\3&1&0\\{ - 2}&0&1\end{array}} \right]A$

Interchanging ${R_2}$and ${R_3}$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{10}\\0&{ - 1}&4\\0&9&{ - 11}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&0&3\\{ - 2}&0&1\\3&1&0\end{array}} \right]A$

Applying ${R_3} \to {R_3} + 9{R_2},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{10}\\0&{ - 1}&4\\0&0&{25}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&0&3\\{ - 2}&0&1\\{ - 15}&1&9\end{array}} \right]A$

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

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{10}\\0&1&{ - 4}\\0&0&{25}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&0&3\\2&0&{ - 1}\\{ - 15}&1&9\end{array}} \right]A$

Applying ${R_3} \to \cfrac{1}{{25}}{R_3},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&{10}\\0&1&{ - 4}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&0&3\\2&0&{ - 1}\\{\cfrac{{ - 15}}{{25}}}&{\cfrac{1}{{25}}}&{\cfrac{9}{{25}}}\end{array}} \right]A$

Applying ${R_1} \to {R_1} - 10{R_3}$,

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&{ - 4}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{\cfrac{{ - 2}}{5}}&{\cfrac{{ - 3}}{5}}\\2&0&{ - 1}\\{\cfrac{{ - 3}}{5}}&{\cfrac{1}{{25}}}&{\cfrac{9}{{25}}}\end{array}} \right]A$

Applying ${R_2} \to {R_2} + 4{R_3},$

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{\cfrac{{ - 2}}{5}}&{\cfrac{{ - 3}}{5}}\\{\cfrac{{ - 2}}{5}}&{\cfrac{4}{{25}}}&{\cfrac{{11}}{{25}}}\\{\cfrac{{ - 3}}{5}}&{\cfrac{1}{{25}}}&{\cfrac{9}{{25}}}\end{array}} \right]A$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{\cfrac{{ - 2}}{5}}&{\cfrac{{ - 3}}{5}}\\{\cfrac{{ - 2}}{5}}&{\cfrac{4}{{25}}}&{\cfrac{{11}}{{25}}}\\{\cfrac{{ - 3}}{5}}&{\cfrac{1}{{25}}}&{\cfrac{9}{{25}}}\end{array}} \right]$