Find the inverse of the matrix (if it exists): $\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$

Let $A = \left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$.
We know that the inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,which exists if $|A| \neq 0$.
First,calculate the determinant $|A|$:
$|A| = \left|\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right| = (2 \times 3) - (4 \times -2) = 6 + 8 = 14$.
Since $|A| = 14 \neq 0$,the inverse $A^{-1}$ exists.
Next,calculate the adjoint of $A$ (adj $A$):
For a $2 \times 2$ matrix $\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$,the adjoint is $\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Thus,$\text{adj}(A) = \left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right]$.
Finally,calculate $A^{-1}$:
$A^{-1} = \frac{1}{14} \left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right] = \left[\begin{array}{cc}\frac{3}{14} & \frac{2}{14} \\ -\frac{4}{14} & \frac{2}{14}\end{array}\right] = \left[\begin{array}{cc}\frac{3}{14} & \frac{1}{7} \\ -\frac{2}{7} & \frac{1}{7}\end{array}\right]$.