Solution of coupled non-linear matrix difference equations encountered in calculating the determinat of a block partitioned matrix. The determinant of the following $n N \times n N$ block partitioned complex symmetric matrix ($N \times N$ blocks)
$$\begin{bmatrix}\mathbb{A} & \mathbb{B} &\cdots &\mathbb{B} \\
\mathbb{B}_{}^{T} & \ddots &\ddots &\vdots \\
\vdots & \ddots & \ddots & \mathbb{B}\\
\mathbb{B}_{}^{T}& \cdots & \mathbb{B}_{}^{T}& \mathbb{A}\end{bmatrix}$$
is given as 
$$\prod_{i=1}^{N}\det\left[\mathbb{A}_{i}^{}\right].$$
Here $\mathbb{A}_{}^{}$, $\mathbb{B}_{}^{}$ are $n \times n$ complex matrices with $\mathbb{A}_{}^{T}=\mathbb{A}_{}^{}$.
$\mathbb{A}_{i}^{}$ are $n \times n$ matrices defined by the following recurrence equations (difference equations)
$$\mathbb{A}_{k}^{}=\mathbb{A}_{k-1}^{}-\mathbb{B}_{k-1}^{T}\mathbb{A}_{k-1}^{-1}\mathbb{B}_{k-1}^{}$$
$$\mathbb{B}_{k}^{}=\mathbb{B}_{k-1}^{}-\mathbb{B}_{k-1}^{T}\mathbb{A}_{k-1}^{-1}\mathbb{B}_{k-1}^{}$$
with the initial conditions
$$\mathbb{A}_{1}^{}=\mathbb{A}_{}^{}$$
$$\mathbb{B}_{1}^{}=\mathbb{B}_{}^{}$$
Is there a method to find the solutions of the above recurrence equations? 
Is there a closed form expression for the determinant of the above block partitioned matrix?
$\textbf{Progress achieved:}$
After i have posted this question, i could make some progress as follows:
(i) If $\mathbb{A}_{}^{T}=\mathbb{A}_{}^{}$, then $\mathbb{A}_{k}^{T}=\mathbb{A}_{k}^{}$.
(ii) $\mathbb{B}_{k}^{T}-\mathbb{B}_{k}^{}=\mathbb{B}_{}^{T}-\mathbb{B}_{}^{}$.
(iii) $\mathbb{A}_{k}^{}-\mathbb{B}_{k}^{}=\mathbb{A}_{}^{}-\mathbb{B}_{}^{}$.
(iv) Using (iii), the above two recurrence equations can be decoupled and one only needs to solve for the following
$$\mathbb{A}_{k}^{}=\left[\mathbb{S}_{}^{T}+\mathbb{S}_{}^{}\right]-\mathbb{S}_{}^{T}\mathbb{A}_{k-1}^{-1}\mathbb{S}_{}^{}$$ 
where $\mathbb{S}_{}^{}=\left(\mathbb{A}_{}^{}-\mathbb{B}_{}^{}\right)$ and the initial condition is $\mathbb{A}_{1}^{}=\mathbb{A}_{}^{}$.
This recurrence equation seems to be related to discrete time algebraic riccatti equation (someone please confirm this). If this is true, can analytical solution (not a simple iterative solution in terms of repeated fractions but some closed form solution) be achieved for the above recurrence equation?
 A: Here's some semi-justified black magic. Perhaps it helps arrive at an answer
$$A_{k}^{}=\left[S^{T} + S \right] - S^{T} A_{k-1}^{-1} S$$
$$\mathbb{1} = A_{k}^{-1} \left[S^{T} + S \right] - A_{k}^{-1} S^{T} A_{k-1}^{-1} S$$
Let $Z_k = A_k^{-1}S^T$ and $C = (S^T)^{-1} S$, then
$$\mathbb{1} = Z_k \left[\mathbb{1} + C \right] - Z_k Z_{k-1} C$$
$$\mathbb{1} = Z_k \left[\mathbb{1} - (Z_{k-1} - \mathbb{1})C \right]$$
Subtract the bracket from both sides
$$(Z_{k-1} - \mathbb{1})C =  (Z_k - \mathbb{1}) \left[\mathbb{1} - (Z_{k-1} - \mathbb{1})C \right]$$
Let $Y_k = Z_k - \mathbb{1}$, then
$$Y_{k-1}C = Y_k(\mathbb{1} - Y_{k-1}C)$$
$$Y_{k-1} = Y_k(C^{-1} - Y_{k-1})$$
$$Y_{k}^{-1} = (C^{-1}Y_{k-1}^{-1} - \mathbb{1})$$
Finally with $W_k = Y_k^{-1}$ and $B = C^{-1}$ we get
$$W_k =B W_{k-1} - \mathbb{1} = B^k W_0 - \sum_{i=0}^{k-1} B^i$$
Substituting back the expressions for $W_k$ and $B$ via $A_k$ and $S$ we get an analytical solution for $A_k$.
Of course, it all depends on the assumption that a bunch of matrices are invertible, but relaxing these assumptions would produce a lot of cases to consider. Hopefully this is sufficient to get you going
