Pseudo determinant of product of two square matrices Let $A$ and $B$ be square symmetric matrices. $A$ is singular and $B$ is non-singular. Is there a way to decompose:

$Det(AB)$ in terms of $Det(A)$ and $det(B)$.

$Det(.)$ refers to the pseudo determinant and $det(.)$ refers to the usual determinant of a square non-singular matrix.

No. For example, consider
\begin{align*}
A_1 &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \\
A_2 &= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\
B &= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}.
\end{align*}
Then, $\operatorname{Det} A_1 = \operatorname{Det} A_2 = 1$, but $\operatorname{Det}(A_1 B) = 1$ and $\operatorname{Det} (A_2 B) = 2$. That is, $\operatorname{Det}(AB)$ is not purely a function of $\operatorname{Det} A$ and $\det B$.
