The determinant behaves multiplicatively with respect to the usual matrix product
$$
\det(AB) = \det(A)\det(B),
$$
and also with respect to the Kronecker (or tensor) product of square matrices
$$
\det(A\otimes B) = \det(A)^q \det(B)^p,
$$
when $A$ and $B$ are $p\times p$ and $q \times q$ matrices, respectively.
Are there other natural types of matrix products under which the determinant behaves multiplicatively? To be completely precise, the property I need is that the determinant of the product is $0$ if and only if the determinant of at least one of its factors is $0$.
Best Answer
Direct summation (taking a $p \times p$ matrix $A$ and a $q \times q$ matrix $B$ and returning a block-diagonal $(p+q) \times (p+q)$ matrix $A \oplus B := \begin{pmatrix} A & 0 \\\ 0 & B \end{pmatrix}$) also works:
$$\det(A \oplus B) = \det(A) \det(B).$$
One can debate whether this operation deserves to be called a "matrix product", though (for instance, it is not distributive over addition).
EDIT: Another (somewhat trivial) example is the reversed multiplication operation $(A, B) \mapsto BA$. More generally, if there was a linear automorphism $T$ on $Mat_n$ that preserved the singular variety $\{ A \in Mat_n: \det A = 0 \}$, one could conjugate the usual matrix multiplication operation by $T$. In the above example, $T$ is the transpose operation $T: A \mapsto A^t$. As another example, one could let $T$ be a left multiplication operator $A \mapsto SA$ for some invertible $S$, in which case the matrix multiplication operation becomes $(A, B) \mapsto ASB$, which also seems to work. One can combine the two and obtain another operation $(A, B) \mapsto BSA$. I'm not sure if these are the only examples that can be constructed by this method.