Let $A$ be a $0/1$ matrix in $\mathbb Z^{n\times n}$ such that $I+A$ is invertible $\bmod 3$. Consider $Q=(I-A)(I+A)^{-1}$.
Let $Det(M)$ and $Per(M)$ be determinant and permanent respectively of matrix $M$.
If $Det(Q)\bmod 3$ and $Per(Q)\bmod 3$ are known can we compute $Det(A)\bmod3$ and $Per(A)\bmod3$ directly faster without computing the determinant and permanent values using the usual techniques?
$A_1 = \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$ and $A_2 = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ both yield a corresponding $Q$ with determinant and permanent $0$, but $0 = \det A_1 = \operatorname{perm} A_1 \neq \operatorname{perm} A_2 = \det A_2 = 1$.
If we go to $n=3$ we can also exclude the possibility of determining $\operatorname{perm} A$ purely from $(\det Q, \operatorname{perm} Q, \det A)$:
$A_3 = \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}$ and $A_4 = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}$ both have determinant $0$ and yield a corresponding $Q$ with determinant and permanent $0$, but $\operatorname{perm} A_3 = 0 \neq 2 = \operatorname{perm} A_4$.
