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

.:

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

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

Interchanging ${R_1}$and ${R_2}$,

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

Applying ${R_1} \to {R_1} - 2{R_2},$

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

Applying ${R_2} \to {R_2} - 2{R_1},$

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

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

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

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

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

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

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

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

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

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

we get
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]A$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]$