Let A be a $N\times N$ real symmetric matrix. I define the $N^2\times N^2$ matrix $\Sigma$ such that: $$\Sigma_{n(i-1)+j, n(k-1)+l}=A_{j l} \delta_{i k}+\delta_{i l} \delta_{j k} b \quad \forall i, j, k, l \in \mathbb{N} \cap[1, n]$$
My question is, is there a way to efficiently find the determinant of $\Sigma$? I already know the eigenvalues and thus the determinant of $A$. If $b=0$ then we have a diagonal block matrix and the following holds: $|\Sigma|=|A|^N$. Of course, I would be interested in the case where $b\neq 0$.
In the case $N=2$:
$$\Sigma=\left( \begin{array}{cccc} A_{1,1}+b & A_{1,2} & 0 & 0 \\ A_{2,1} & A_{2,2} & b & 0 \\ 0 & b & A_{1,1} & A_{1,2} \\ 0 & 0 & A_{2,1} & A_{2,2}+b \\ \end{array} \right)$$
In the case $N=3$, the matrix $\Sigma$ looks like this: $$\Sigma=\left( \begin{array}{ccccccccc} A_{1,1}+b & A_{1,2} & A_{1,3} & 0 & 0 & 0 & 0 & 0 & 0 \\ A_{2,1} & A_{2,2} & A_{2,3} & b & 0 & 0 & 0 & 0 & 0 \\ A_{3,1} & A_{3,2} & A_{3,3} & 0 & 0 & 0 & b & 0 & 0 \\ 0 & b & 0 & A_{1,1} & A_{1,2} & A_{1,3} & 0 & 0 & 0 \\ 0 & 0 & 0 & A_{2,1} & A_{2,2}+b & A_{2,3} & 0 & 0 & 0 \\ 0 & 0 & 0 & A_{3,1} & A_{3,2} & A_{3,3} & 0 & b & 0 \\ 0 & 0 & b & 0 & 0 & 0 & A_{1,1} & A_{1,2} & A_{1,3} \\ 0 & 0 & 0 & 0 & 0 & b & A_{2,1} & A_{2,2} & A_{2,3} \\ 0 & 0 & 0 & 0 & 0 & 0 & A_{3,1} & A_{3,2} & A_{3,3}+b \\ \end{array} \right)$$
Playing around, I found that:
In the case $N=2$:
$|\Sigma|=|A+bI|(|A|-|bI|)$
In the case $N=3$:
$|\Sigma|=|A+bI|\left(|A|^2-b^2\operatorname{Tr}(A)|A|-|bI|^2+b^4\operatorname{Tr}(\Lambda^2A)\right)$
Where $\operatorname{Tr}\left(\Lambda^2A\right)$ is the trace of the $k$th exterior power of A, taken from this formula: $$ |A-bI|=-\sum_{k=0}^{n} b^{n-k}(-1)^{k} \operatorname{tr}\left(\Lambda^{k} A\right) $$ I would hope that there is a relatively simple formula for general $N$.
Any ideas or remarks are always appreciated, thank you.