As it sounds like you're aware, if a 3x3 matrix is considered as three vectors that form a parallelogram, the 2x2 minors form three vectors in the bivector spac

As it sounds like you're aware, if a 3x3 matrix is considered as three vectors that form a parallelogram, the 2x2 minors form three vectors in the bivector space ∧^2 R^2 that represent the area vectors for the three faces of that parallelogram. 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 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.)