Determinant of $\mathbb{F}_2$ square matrices

Question

Let $A_1,A_2,A_3$ and $B$ be in $\mathbb F_2^{n\times n}$. Is $$\mathsf{Det}(A_{1}A_2A_{3}+B)=\mathsf{Det}(A_{1}A_3A_{2}+B)=\mathsf{Det}(A_{3}A_2A_{1}+B)=\mathsf{Det}(A_{2}A_3A_{1}+B)=\mathsf{Det}(A_{2}A_1A_{3}+B)=\mathsf{Det}(A_{3}A_1A_{2}+B)?$$

Answer 1

Here is a quick counterexample. Let $E_{ij}$ be the elementary matrix with exactly one $1$ entry at position $(i,j)$, else zero entries. Then $E_{ij}E_{jk}=E_{ik}$. So in dimension $n=2$ $$ \begin{aligned} E_{12}E_{21} + E_{22} &= E_{11} + E_{22} = I &\text{ has determinant }1\ ,\\ E_{21}E_{12} + E_{22} &= E_{22} + E_{22} = 0 &\text{ has determinant }0\ . \end{aligned} $$ (In higher dimensions add diagonally an identity block.)