Geometric Interpretation of Determinant of an Inverse Matrix

Question

The $\mathbf{A}$ be an $n\times n$ full rank matrix. Then, the (signed) volume enclosed by the rows (or columns) of $\mathbf{A}$ is equal to $\det(\mathbf{A})$. My question is, what is a geometric interpretation of $\det(\mathbf{A}^{-1})$? What would be the parallelotope defined by $\mathbf{A}^{-1}$ and in which way is it related to the parallelotope defined by $\mathbf{A}$?

Answer 1

The determinant is multiplicative, i.e. $|(AA^{-1})| = |1| = |A||A^{-1}|$ and as such $|A^{-1}| = |A|^{-1}$ which shows the intuitively obvious fact that if the determinant is the signed area of the image when transformed by $A$ of the unit cube with sides given by the basis vectors of the space. Clearly if $A$ grows this cube by a certain amount, then $A^{-1}$ must shrink it by the same amount, as when they are applied together they should give the unit cube back.

Answer 2

It is no different from a multiplicative/divisive bijective relationship.

If $ V_2 = 3\; V_1$ ,then $ V_1 = \frac 13 \; V_2. $ Under action of the determinant volume expands or contracts by the same factor.