TITLE: Determinant of the inverse matrix
QUESTION [5 upvotes]: I'm seeking for a proof of the following:

Let $A$ be an invertible matrix. Then the determinant of $A^{-1}$ equals:
  $$\left|A^{-1}\right|=|A|^{-1} $$

I don't know where to begin the proof.
Any suggestions?

REPLY [2 votes]: By interpreting the determinant as the (signed) ratio between the hypervolume of $f_A(\Gamma)$ and the hypervolume of $\Gamma$, where $\Gamma$ is a symplex associated with the canonical base and $f_A$ is the linear map associated with $A$, the claim is trivial, since:
$$ \left(f_A\right)^{-1} = f_{A^{-1}}. $$
This is just the classic "measure-theoretic" proof of Binet's theorem.