Answer
See below
Work Step by Step
Given $A=\begin{bmatrix}
a_{11} &a_{12}\\a_{21} & a_{22}
\end{bmatrix}$
To find the inverse of $A$, we obtain augmented matrix:
$\begin{bmatrix}
a_{11} &a_{12} | 1&0\\a_{21} & a_{22}| 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
0 &\frac{a_{12} }{a_{11} }| \frac{1}{a_{11}}&0\\a_{21} & a_{22}| 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
0 &\frac{a_{12} }{a_{11} }| \frac{1}{a_{11}}&0\\ 0 &\frac{ a_{11} a_{22}- a_{21} a_{12}}{ a_{11}}| -\frac{ a_{21}}{ a_{11}} & 1
\end{bmatrix} \approx \begin{bmatrix}
0 &\frac{a_{12} }{a_{11} }| \frac{1}{a_{11}}&0\\ 0 &1| -\frac{ a_{21}}{a_{11}a_{22}-a_{12}a_{21}} & \frac{a_{11} }{a_{11} a_{22}- a_{21} a_{12}}
\end{bmatrix} \approx \begin{bmatrix}
1 & 0| \frac{a_{22} }{a_{11} a_{22}- a_{21} a_{12}} & \frac{-a_{12} }{a_{11} a_{22}- a_{21} a_{12}}\\ 0 &1| -\frac{ a_{21}}{a_{11}a_{22}-a_{12}a_{21}} & \frac{a_{11} }{a_{11} a_{22}- a_{21} a_{12}}
\end{bmatrix}$
Hence, $A^{-1}= \begin{bmatrix}
\frac{a_{22} }{a_{11} a_{22}- a_{21} a_{12}} & \frac{-a_{12} }{a_{11} a_{22}- a_{21} a_{12}}\\-\frac{ a_{21}}{a_{11}a_{22}-a_{12}a_{21}} & \frac{a_{11} }{a_{11} a_{22}- a_{21} a_{12}}
\end{bmatrix}= \begin{bmatrix}
\frac{a_{22} }{\Delta} & \frac{-a_{12} }{\Delta}\\-\frac{ a_{21}}{\Delta} & \frac{a_{11} }{\Delta}
\end{bmatrix}=\frac{1}{\Delta}\begin{bmatrix}
a_{22} & -a_{12}\\ -a_{21} & a_{11}
\end{bmatrix}$
where $\Delta \ne0$
$1.M_1(\frac{1}{a_{11}})\\
2.A_{12}(-a_{21})\\
3.M_2(\frac{a_{11} }{a_{11} a_{22}- a_{21} a_{12}})\\
4. A_{21}(\frac{-a_{12}}{a_{11}})$
