A more efficient way to find $A^{-1}$
Given an $n\times n$ matrix $A$, form the $n\times 2n$ matrix
\[ \def\m#1{\left[
\begin{array}{@{} c|c {}@} % it does autodetection
#1
\end{array}
\right]}\m{A&I_n}\]
and use EROs to put this matrix into RREF. One of two things can happen:
Either you get a row of the form $[0~0~\dots~0~|~*~*~\dots~*]$ which starts with $n$ zeros. You can then conclude that $A$ is not invertible.
Or you end up with a matrix of the form $\m{I_n&B}$ for some $n\times n$ matrix $B$. You can then conclude that $A$ is invertible, and $A^{-1}=B$.
Examples
Consider $A=\def\mat#1{\begin{matrix}#1\end{matrix}}\left[\mat{1&3\\2&6}\right]$. \begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&6}&\mat{1&0\\0&1}}
\def\go#1#2{\m{\mat{#1}&\mat{#2}}}
\def\ar#1{\\[6pt]\xrightarrow{#1}&}
\ar{R2\to R2-2R1}\go{1&3\\0&0}{1&0\\-2&1}
\end{align*} Conclusion: $A$ is not invertible.
Consider $A=\left[\mat{1&3\\2&7}\right]$.\begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&7}&\mat{1&0\\0&1}}
\ar{R2\to R2-2R1}\go{1&3\\0&1}{1&0\\-2&1}
\ar{R1\to R1-3R1}\go{1&0\\0&1}{7&-3\\-2&1}
\end{align*} Conclusion: $A$ is invertible and $A^{-1}=\left[\mat{7&-3\\-2&1}\right]$.
Consider $A=\left[\mat{3&1&0\\-2&-4&3\\5&4&-2}\right]$.\begin{align*}\m{A&I_3}&=\go{3&1&0\\-2&-4&3\\5&4&-2}{1&0&0\\0&1&0\\0&0&1}
\ar{R1\to R1+R2}
\go{1&-3&3\\-2&-4&3\\5&4&-2}{1&1&0\\0&1&0\\0&0&1}
\ar{R2\to R2+2R1,\ R3\to R3-5R1}
\go{1&-3&3\\0&-10&9\\0&19&-17}{1&1&0\\2&3&0\\-5&-5&1}
\ar{R3\leftrightarrow R2}
\go{1&-3&3\\0&19&-17\\0&-10&9}{1&1&0\\-5&-5&1\\2&3&0}
\ar{R2\to R2+2R3}
\go{1&-3&3\\0&-1&1\\0&-10&9}{1&1&0\\-1&1&1\\2&3&0}
\ar{R1\to R1+3R2,\ R3\to R3-10R2}
\go{1&0&0\\0&-1&1\\0&0&-1}{4&-2&3\\-1&1&1\\12&-7&-10}
\ar{R2\to R2+R3}
\go{1&0&0\\0&-1&0\\0&0&-1}{4&-2&3\\11&-6&-9\\12&-7&-10}
\ar{R2\to -R2,\ R3\to -R3}
\go{1&0&0\\0&1&0\\0&0&1}{4&-2&3\\-11&6&9\\-12&7&10}
\end{align*} Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.