Is there a geometric proof that the determinant of a 3x3 matrix is invariant under switching rows and columns?

Solution 1:

If you're willing to posit that the determinant is linear, here's something for the $2 \times 2$ case: consider $\left(\begin{smallmatrix} a & b \\ 1 & 0 \end{smallmatrix}\right)$. Rowwise, that gives us a parallelogram with vertices $(a,b)$, $(1,0)$, and $(a+1,b)$. (And the origin, of course, but I won't bother writing that.) That's pretty clearly a parallelogram with base 1 and height $b$.

After transposing, $\left(\begin{smallmatrix} a & 1 \\ b & 0 \end{smallmatrix}\right)$ gives a parallelogram with vertices $(a,1)$, $(b,0)$, and $(a+b,1)$; now the height is 1 and the base is $b$.

Geometrically, you can slide the top of one of those until you get a rectangle, reflect about $y=x$, and slide the top again (all area-preserving transformations) to turn one into the other.

Scaling the $(1,0)$ row/column obviously just scales the resulting parallelograms. The $\left(\begin{smallmatrix} a & b \\ 0 & 1 \end{smallmatrix}\right)$ case works out the same. So if the linearity of the determinant is okay, you've got the $2 \times 2$ case.

For $3 \times 3$, my guess is that it suffices to do $\left(\begin{smallmatrix} a & b & c \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{smallmatrix}\right)$.

But this is still a bit unsatisfying. I'm guessing the OP wants a really concrete operation that turns one parallelepiped into the other so that the equality of volumes is forcefully obvious.