Let \(A, B \in \mathbb{R}^{3\times 3}\).
Suppose that \(\det(A) = 2\) and \(\det(B) = 4\).
What is \(\det(2A^\mathsf{T}B^{-1})\)?
The answer is 4.
Since \(A\) and \(B\) are \(3 \times 3\) matrices,
the product \(A^\mathsf{T}B^{-1}\) is also \(3\times 3\).
Hence,
\begin{eqnarray}
\det(2A^\mathsf{T}B^{-1}) & = & 2^3 \det(A^\mathsf{T}B^{-1}) \\
& = & 8 \det(A^\mathsf{T})\det(B^{-1}) \\
& = & 8 \det(A)/\det(B) \\
& = & 2\cdot 8/4 = 4.
\end{eqnarray}