*** I have notes on this ***

Why does the cross product compute area,
and why can you compute it via a determinant?
B/c the cross product -is- (adjunct to) the determinant!
(or rather, to a volume form)
The determinant is the natural map End(V) -> K,
given by End(V) -> End(\Lambda^n V) = K
  (the top exterior power is a 1-dimensional vector space,
   whose endomorphism are exactly scalar multiplication,
   hence naturally identified.
   Properly, K -> End(V) via scalar multiplication,
   and here it's epi, so an isomorphism).
A volume form is an element of (V^n)^*
(given n vectors, compute the volume);
given an identification End(V) -~-> V^n,
the determinant gives you a volume form,
and a choice of basis identifies End(V) with V^n
[this is the formal way of saying:
 "A linear map is freely determined by where it sends a basis".]
[I should be careful: End(V) -~-> V^n,
 but not V^{\otimes n}; the map (V^n) -> K is multilinear,
 not linear.]

So a volume form is a multilinear map
V^n -> K
The cross product comes from adjunction on one (say, the last) coordinate:
V^{n-1} -> V^*
...together with identifying V^* with V (via an inner product,
or more generally non-degenerate form):
V^{n-1} -> V^* -~-> V

Hey, weird:
in 3D, the determinant gives you
V x V -> V
&
V -> V (x) V
(I think)
...so a Hopf structure?


?? this is basically -affine-
  (we're using a 3D determinant to compute 2D area;
   are we basically seeing it as a 2D plane in 3D space?)
You can do the same in higher dimensions, to compute the area of a parellelipiped.
Note that you get signed area, accordingly as whether you agree or disagree with the given orientation.

I do not think you can use this to compute general Grammians.