Determinant of a large symmetric block matrix Consider a given matrix $Q \in \text{Mat}_N(\mathbb{R})$, which is invertible, and $n \geq 1$. I am looking for the determinant of the symmetric block matrix $I_n(Q)$ of total size $nN \times nN$:
$$I_n(Q) = \begin{pmatrix}Id_N & Q  & \cdots & Q \\
Q^\intercal & Id_N & \cdots & Q \\
\vdots & \vdots & \ddots & \vdots \\
Q^\intercal & Q^\intercal & \cdots & Id_N \\
\end{pmatrix}$$
If $Q$ is symmetric, I know that $\det I_n(Q) = \det((Id_N-Q)^{n-1}) \det(Id_N+(n-1)Q)$, and I also have a formula for $I_n(Q)^{-1}$, but I couldn't derive a form for the non-symmetric case. Any help would be appreciated ! Thanks :)
 A: Here is a partial answer. Let us consider the case where the number of block rows is $n=4$ first. The other cases are similar. We shall perform blockwise elementary row/column operations on the block matrix
$$
\pmatrix{
I&Q&Q&Q\\
Q^T&I&Q&Q\\
Q^T&Q^T&I&Q\\
Q^T&Q^T&Q^T&I}.
$$
First, for $i=n,n-1,\ldots$ down to $2$, subtract the $i$-th block row by the one above it. We get
$$
\pmatrix{
I&Q&Q&Q\\
Q^T-I&I-Q&0&0\\
0&Q^T-I&I-Q&0\\
0&0&Q^T-I&I-Q}.
$$
Suppose $1$ is not an eigenvalue of $Q$. Let
$$
M=(I-Q)^{-1}(Q^T-I).
$$
Then for $j=n-1,n-2,\ldots$ down to $1$, perform the block column operation $C_j\leftarrow C_j-C_{j+1}M$ successively. When $n=4$, we will obtain, in three steps, the following matrices:
\begin{aligned}
&\pmatrix{
I&Q&Q-QM&Q\\
Q^T-I&I-Q&0&0\\
0&Q^T-I&I-Q&0\\
0&0&0&I-Q},\\
&\pmatrix{
I&Q-QM+QM^2&Q-QM&Q\\
Q^T-I&I-Q&0&0\\
0&0&I-Q&0\\
0&0&0&I-Q},\\
&\pmatrix{
I-QM+QM^2-QM^3&Q-QM+QM^2&Q-QM&Q\\
0&I-Q&0&0\\
0&0&I-Q&0\\
0&0&0&I-Q}.
\end{aligned}
Hence at the end we get a block upper triangular matrix. In general, if $1$ is not an eigenvalue of $Q$, the determinant of the original block matrix is given by
$$
\det\left(I+Q\left[(-M)+(-M)^2+..+(-M)^{n-1}\right]\right)\det(I-Q)^{n-1}.\tag{1}
$$
Here we do not replace the square bracket term by $[(-M)-(-M)^n](I+M)^{-1}$ because $-1$ may be an eigenvalue of $M$ (in particular, this always occurs if the size $N$ of the matrix $Q$ is odd).
Since the determinant of the original block matrix is a polynomial in the entries of $Q$, if a generic formula exists, we don't expect it to contain any matrix inverse term. Unfortunately I cannot manage to cancel out the inverse term $(I-Q)^{-1}$ inside the powers of $M$. Therefore formula $(1)$ cannot be further simplified at the moment and the requirement that $1$ is not an eigenvalue of $Q$ cannot be dropped yet.
