Does anyone have a geometrical intuition about how this works? Maybe just for 3d case? How does condensation compute the volume of a parallelepiped from the individual areas of parallelogram? And what does the division do?
[a b c]
[d e f]
[g h i]
For condensation purposes, we view our 3x3 matrix as a 2x2 matrix instead:
[A B]
[D E]
We then take the determinant AE - DB of our 2x2 matrix. If we look back up at the original 3x3 scalar matrix, the only element common to all four of these matrices-of-convenience is e, and that's what we use to "normalize" the determinant back down. The determinant of the original matrix is (|A||E| - |D||B|)/e.
The product of two 2x2 determinants has four terms, so the sum we compute has eight instead of six, but two of them, which just happen to be the two terms with no factor of e, cancel out.
Here's a hand-wavy version in a sort of pseudo-notation: if the matrix is M = (a,b,c), then the first step of condensation forms (a∧b, b∧c). Confusingly, you drop the last row and column of the minor matrix, the one that wraps around to the start. Then you compute (ab ∧ bc) / b, which if you squint looks like it cancels the b out and leaves (a∧b∧c).
I guess if you didn't know anything about vectors and their multiplication (or lack of it) you might assume that (abc) = (ab * bc / b), and the way this algorithm works is basically "oh actually multiplication does kinda work on vectors, if you're really careful to not erase any information".
But there's clearly a missing 'piece' that explains exactly what is going on. I'm pretty sure there is a way to make it rigorous but I haven't thought about in a while. (I have an ongoing silly project to make sense of all the weird formulas of exterior algebra, cf https://alexkritchevsky.com/2018/10/09/exterior-2.html ... but I haven't worked on it in a while and haven't looked closely at Dodgson Condensation yet.)
Regarding the division aspect, the video doesn't explicitly mention it. However, the concept of subtracting the product of certain elements from the product of other elements in the determinant formula can be understood as a form of division operation. This subtraction or difference in scaling factors is what contributes to the computation of the determinant and reflects the volume change or scaling of the parallelepiped.
It is true that it doesn't answer all the questions one might have about the problem, but it does answer some of them and is likely the video mentioned earlier, which unfortunately wasn't linked by either of you. So, I linked to it so we'd have it here.
Yes, we all already know that. It is explicitly present as assumed background knowledge in the original question that ivan_ah asked:
>>> How does condensation compute the volume of a parallelepiped from the individual areas of parallelogram?
The question is how the area of the parallelogram defined by the two vectors
[e f]
[h i]
[a b c]
[d e f]
[g h i]
and the 3b1b video not only doesn't address this, or mention that condensation is a thing you can do, it actively discourages you from calculating determinants by any method.