Find the inverse,by elementary row operations (if possible),of the following matrix: $\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]$

(D) Suppose $A = \left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]$.
To find the inverse using elementary row operations,we use the relation $A = IA$.
$\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] A$
Apply $R_2 \rightarrow R_2 + 5R_1$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 22\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 5 & 1\end{array}\right] A$
Apply $R_2 \rightarrow \frac{1}{22} R_2$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Apply $R_1 \rightarrow R_1 - 3R_2$:
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 - 3(\frac{5}{22}) & 0 - 3(\frac{1}{22}) \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}\frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Thus,$A^{-1} = \frac{1}{22} \left[\begin{array}{cc}7 & -3 \\ 5 & 1\end{array}\right]$.