Consider the following product:
$$\langle A | B \rangle=\begin{bmatrix} a_{00} & a_{01} | b_{00} & b_{01} \\ a_{10} & a_{11} | b_{10} & b_{11} \\ \end{bmatrix} = a_{00} b_{11} - b_{01} a_{10} - a_{01} b_{10} + b_{00} a_{11} $$
this product has interesting properties, one of them is that:
$$ \det(A) = 1/2 \langle A | A \rangle $$
But I think the most appealing property is that allows a relationship of distributivity of determinants:
$$ \det(A + B) = \langle A + B | A + B \rangle = \langle A A \rangle + 2 \langle A B \rangle + \langle B B \rangle = 2 \det(A) + 2 \det(B) + 2 \langle A B \rangle $$
Written in alternative form:
$$ \langle A B \rangle = \frac{\det(A + B)}{2} - \det(A) - \det(B) $$
Can this product be generalized to higher than $2 \times 2$ matrices, such that both interesting properties are preserved?
Disclaimer: I was about to ask this question in Physics.SE instead, since algebraic bra-ket know-how typically lies on a kind of gray area of ownership in the boundaries between physics history and mathematics traditional training, nonetheless and against my best judgment I decided to post the question here
The "distributivity" propriety you noticed is simply because the determinant of a 2x2 matrix is a quadratic form. Basically the map $\widetilde{\det}$ defined as :
$$\widetilde{\det}(A,B)=\frac{1}{2}\left(\det(A+B)-\det(A)-\det(B)\right)$$ is bilinear symmetric. You can easily check it matches your bilinear form. It may be called the polar form of $\det$. It follows that :
$$\widetilde{\det}(A,A)=\det(A)$$ $$\det(A+B)=\widetilde{\det}(A+B,A+B)=\det(A)+\widetilde{\det}(A,B)+\widetilde{\det}(B,A)+\det(B)$$
I'm not sure it generalizes in a helpful way. The distributivity looks nice, but $\widetilde{\det}$ is defined from the determinant anyways and, as far as I know, it wouldn't really simplify expressions involving the determinant of a sum of a matrix.