TITLE: given the inverse of a matrix, is there an efficient way to find the determinant?
QUESTION [13 upvotes]: Suppose one has the inverse $A^{-1}$ of an $N\times N$ non-singular matrix $A$.   Is there an ''efficient'' way to obtain $\det{A}$?  
With ''efficient'' I mean anything that has a better scaling than the standard $\mathcal{O}(N^3)$.  Naively, one can argue that $\det{A}$ is already implicitly incorporated in the inverse via the adjugate $A^{-1}=\displaystyle\frac{1}{\det{A}}\operatorname{adj}A$, so all the work has "already been done", and one only needs an efficient trick to distill the determinant. 
E.g.: for $N=2$, we have $A=\left(\begin{array}{cc} a & b \\ c & d\end{array}\right)$ and $A^{-1}=\displaystyle\frac{1}{\det{A}}\left(\begin{array}{cc} d & -b \\ -c & a\end{array}\right)$.  The determinant can efficiently be obtained from $\displaystyle\frac{A_{11}}{[A^{-1}]_{22}}=\det{A}$ in $\mathcal{O}(1)$.

REPLY [3 votes]: This is strictly harder than a problem we already don't know how to solve: quickly taking the determinant of an orthogonal matrix.
If $Q$ is an orthogonal matrix, then $Q Q^{\mathsf T} = I$. In particular, if you're given $Q$, you are also automatically given its inverse. The determinant $\det(Q)$ is guaranteed to be $\pm 1$; however, determining which one it is appears to be just as hard as taking any other determinant.
If we could, in general, efficiently compute $\det(A)$ given both $A$ and $A^{-1}$, then in particular, for an orthogonal matrix $Q$, we could efficiently compute $\det(Q)$.