Let $A, B$ be two matrices. Let $\det(A)=4$, $\det(B)=2/3$. Find $\det(AB^{-1})$.
I already found that $\det(B^{-1})= 3/2$ as a determinant to the power of $-1$ equals $1$ over the determinant. I was just using one of the theorems in my text book that stated that
$$\det(B^{-1}) = \frac{1}{\det(B)}$$
it is known that $\det(XY) = \det(X)\dot\det(Y)$ and $\det(X^{-1}) = \frac{1}{\det(X)}$.
So, your question is $\det(AB^{-1}) = \frac{\det(A)}{\det(B)}=6$