Find the determinant and the inverse of A = 5, 2; 7, 3…

$A^{-1} = \dfrac{\text{adj}\,A}{|A|}$ exists exactly when $|A| \ne 0$, and $A\,\text{adj}\,A = |A|I$. For $2 \times 2$, swap the diagonal and negate the other two entries. Also $(AB)^{-1} = B^{-1}A^{-1}$, $|A^{-1}| = \dfrac{1}{|A|}$ and $|\text{adj}\,A| = |A|^{n-1}$.