Find inverse, by elementary row operations (if possible),

(i) Let $A=\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]$

Now,

$\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$

$\left[\begin{array}{cc}1 & 3 \\ 0 & 22\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 5 & 1\end{array}\right] A \quad\left[\because R_{2} \rightarrow R_{2}+5 R_{1}\right]$

$\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 5 / 22 & 1 / 22\end{array}\right] A \quad\left[\because R_{2} \rightarrow \frac{1}{22} R_{2}\right]$

$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 / 22 & -3 / 22 \\ 5 / 22 & 1 / 22\end{array}\right] A \quad\left[\because R_{1} \rightarrow R_{1}-3 R_{2}\right]$

$\mathbf{I}=\mathbf{B} \mathbf{A}$, where $\mathbf{B}$ is the inverse of $\mathbf{A} .$

Hence,

$B=\frac{1}{22}\left[\begin{array}{ll}7 & -3 \\ 5 & -1\end{array}\right]$

(ii) Let $A=\left[\begin{array}{cc}1 & -3 \\ -2 & 6\end{array}\right]$

Now, $\left[\begin{array}{cc}1 & -3 \\ -2 & 6\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$

$\left[\begin{array}{cc}1 & -3 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] A$ $\left[\because R_{2} \rightarrow R_{2}+2 R_{1}\right]$

As we obtain all the zeroes in a row of the matrix A on the L.H.S., A-1 does not exist.

$\left[\begin{array}{cc}x y & 4 \\ z+6 & x+y\end{array}\right]=\left[\begin{array}{ll}8 & w \\ 0 & 6\end{array}\right]$