Non brute force proof of multiplicative property of determinants

Question

Does anybody have a "non brute" force way to prove the following for non-singular matrices $A, B$:

$$\det(AB) = \det(A) \det(B)$$

Answer 1

It's extremely simple using geometric algebra.

In geometric algebra we define the determinant geometrically by $F(I) = \det(F)I$, where $F$ is an extension of a linear function called an outermorphism and $I$ is the unit pseudoscalar. Essentially this just says that the unit $n$-cube in $n$-dimensional space is scaled by a factor of $\det(F)$ by the linear transformation $F$.

Remember that if $F(x) = Ax$ for matrix $A$ and $G(x) = Bx$ for matrix $B$, then $(F\circ G)(x) = ABx$. Also keep in mind that $\det(F) = \det(A)$ -- that is, the determinant of a linear transformation is equal to the determinant of the matrix representing that linear transformation wrt any basis. Therefore to prove $\det(AB) = \det(A)\det(B)$ it suffices (and in fact is more general) to prove $\det(F\circ G) = \det(F)\det(G)$.

So... Proof:

$$\begin{align}\det(F\circ G)I &=(F\circ G)(I) \\ &= F(G(I)) \\ &=F(\det(G)I) \\ &=\det(G)F(I)& \text{(because $\det(G)$ is a scalar)} \\ &=\det(G)\det(F)I\end{align}$$ $$\implies \require{enclose}\enclose{box}{\det(F\circ G)=\det(G)\det(F)}\ \ \ \ \Box$$