Determinant properties when determinant is given and $n \times n$ is known

Question

$A$ and $B$ are $4\times 4$ matrices $(n=4)$ and $\det(A)= -2$, $\det(B)=3$.

What is $\det (A^3 \cdot \det(B^{-1}))$?

I'm not sure if the dot product comes into play in this exercise or if I follow the $\det(AB)= \det(A)\det(B)$ property. I understand that $\det(B^{-1}) = \frac{1}{\det(B)}$, but I'm confused overall.

Answer 1

You have to find the determinant of $\frac{1}{3}A^3$.

This is just $(\frac{1}{3})^4$ multiplied by the determinant of $A^3$ and is therefore $\frac{-8}{81}$.

Answer 2

It $k$ is a scalar and $M$ is a matrix, avoid writing $M\cdot k$: write $kM$ or $k\cdot M$ instead.

Said that, if $M$ is a $n\times n$ matrix, $\det (kM)=k^n\det M$. Note that if you multiply a single row or column of a $n\times n$ matrix by a scalar, the determinat of the matrix becomes multiplied by this sclalar. So if you multiply every column (or row) of the matrix, the determinant becomes multiplied by $k$ $n$ times, that is, by $k^n$.

So, $$\det(\det(B^{-1})\cdot A^3)=\det(B^{-1})^4\det(A)^3=-\frac{8}{81}$$