If the system of equations 4x − 3y = 3; 3x − 5y = 7 has unique solution, then

The given system can be written as AX=B,
where
$\mathrm{A}=\left[\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-3\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-5\end{array}\right],\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right],\mathrm{B}=\left[\begin{array}{l}3\\ 7\end{array}\right]$
Here, $\left|\mathrm{A}\right|=\left|\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-3\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-5\end{array}\right|=4(-5)-3(-3)$
$\text{=−20+9=−11≠0}$
Thus, A is non-singular. Therefore, its inverse exists.
Therefore, the given system has a unique solution given by X=A^-1 B.
$\begin{array}{l}\therefore \mathrm{adj}\left(\mathrm{A}\right)=\left[\begin{array}{ll}-5& 3\\ -3& 4\end{array}\right]\\ \therefore {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}(\mathrm{adj}\mathrm{A})=\frac{1}{-11}\left[\begin{array}{ll}-5& 3\\ -3& 4\end{array}\right]\end{array}$
Now, $\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}=\frac{1}{-11}\left[\begin{array}{l}-5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\\ -3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\end{array}\right]\left[\begin{array}{l}3\\ 7\end{array}\right]$
$\begin{array}{l}=\frac{1}{-11}\left[\begin{array}{r}-15+21\\ -9+28\end{array}\right]\\ =\frac{1}{-11}\left[\begin{array}{r}6\\ 19\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left[\begin{array}{r}6\\ -\frac{19}{11}\end{array}\right]\\ \therefore \mathrm{x}=-\frac{6}{11}\text{ and }\mathrm{y}=-\frac{19}{11}\end{array}$