How to prove distributive property of a determinant?

Question

How to prove that $|A\cdot B| = |A|\cdot|B|$ where A and B are square matrices of the same size?

P.S.: This proof is not mentioned in my textbook, nor was I able to find it on the web.

Answer 1

Here's a proof that avoids heavy calculations with triple indexed matrix entries (at the cost of being quite abstract): Let $V$ be an $n$-dimensional space over a field $k$. One can show that the vector space of alternating multilinear $n$-forms (that is, maps $V^n\to k$ that are linear in each component and such that the result is zero if two input components are equal) is one-dimensional. The determinant of $n\times n$-matrices is such an alternating multilinear $n$-form (in the $n$ columns of matrices) and is uniquely determined within this one-dimensional space by the fact that $\det I_n=1$ (in fact, this can be used as definition of $\det$). For any matrix $A$, the map $X\mapsto \det(AX)$ is also an alternating multilinear $n$-form, hence is a scalar multiple of $\det$, i.e. $\det(AX)=\alpha\det(X)$ for some $\alpha$ depending on $A$. By comparing values for $X=I_n$, we conclude that $\alpha=\det(A)$.