<s>[INST] Find the determinant of the following general matrix Let $A_r$ and $B_r$ be the $r\times r$ matrix blocks  
$A_r=\left(
\begin{array}{A}
 1-t & t^2 & 0 & 0 & 0 & \cdots & 0 \\
 t^2 & 0 & 0 & 0 & 0 & \cdots & 0 \\
 0 & 0 & t^2 & 0 & 0 & \cdots & 0 \\
 0 & 0 & 0 & t^2 & 0 & \cdots & 0 \\
 0 & 0 & 0 & 0 & t^2 & \cdots & 0 \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & 0 \\
 0 & 0 & 0 & 0 & 0   & \cdots & t^2 \\ 
 \end{array} 
\right)$,  
$B_r=\left(
\begin{array}{B}
 0 & t^2 & 0 & 0 & 0 & \cdots & 0 & -1 \\
 -1 & 0 & 0 & 0 & 0 & \cdots & 0 & 1-t \\
 0 & -1 & 0 & 0 & 0 & \cdots & 0 & 1-t \\
 0 & 0 & -1 & 0 & 0 & \cdots & 0 & 1-t \\
 0 & 0 & 0 & -1 & 0 & \cdots & 0 & 1-t \\
 \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & 1-t \\
 0 & 0 & 0 & 0 & 0   & \ddots & 0  & 1-t \\
 0 & 0 & 0 & 0 & 0   & \cdots & -1 & 1-t \\
 \end{array} 
\right) 
$.
Find the determinant of the following $rs\times rs$ matrix $C$:
$$C = \left(
\begin{array}{C}
 A_r & B_r & 0 & 0 & 0 & \cdots & 0 & 0 \\
 0 & A_r & B_r & 0 & 0 & \cdots & 0 & 0 \\
 0 & 0 & A_r & B_r & 0 & \cdots & 0 & 0 \\
 0 & 0 & 0 & A_r & B_r & \cdots & 0 & 0 \\
 0 & 0 & 0 & 0 & A_r & \ddots & 0 & 0 \\
 \vdots & \vdots & \vdots & \vdots & \vdots   & \ddots & \ddots & \vdots \\
 0 & 0 & 0 & 0 & 0   & \cdots & A_r & B_r \\
 B_r & 0 & 0 & 0 & 0   & \cdots & 0 & A_r \\
 \end{array}
\right) $$
Following user1551's answer, I find the matrix $A^{-1}B$ to be
$$
 -\frac{1}{t^4} \left(
\begin{array}{D}
 t^2   & 0      & 0      & 0 & 0 & \cdots & 0 & -t^2 \\
 -(1-t)   & -t^4   & 0      & 0 & 0 & \cdots & 0 & 1-t+t^2 \\
 0      & t^2      & 0      & 0 & 0 & \cdots & 0 & t^2(1-t) \\
 0      & 0      & t^2      & 0 & 0 & \cdots & 0 & t^2(1-t) \\
 0      & 0      & 0      & t^2 & 0 & \cdots & 0 & t^2(1-t) \\
 \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & 0 & 0   & \ddots & 0 & t^2(1-t) \\
 0 & 0 & 0 & 0 & 0   & \cdots & t^2 & t^2(1-t) \\
 \end{array}
\right) 
$$
But then I am stuck at calculating $(-A^{-1}B)^s$...
(By the way, from the SciLab calculations I have done, I guess that the final result of the determinant depends on $\mu$, $D$ and $N$ where $r$ = $\mu D$ and $s$ = $\mu N$ and $D$ and $N$ are relatively prime.)
 [/INST] Here is a partial answer. For convenience, let us drop the subscript $r$. Write $C=\pmatrix{A&R\\ S&T}$, where $T$ is $r(s-1)\times r(s-1)$. Note that
$$
T^{-1} =
\left[\begin{array}{rrrrr}
A^{-1}&-A^{-1}BA^{-1}&A^{-1}BA^{-1}BA^{-1}&\cdots&(-1)^{s-2}(A^{-1}B)^{s-2}A^{-1}\\
&A^{-1}&-A^{-1}BA^{-1}&\ddots&\ddots\\
&&\ddots&\ddots&\ddots\\
&&A^{-1}&-A^{-1}BA^{-1}&A^{-1}BA^{-1}BA^{-1}\\
&&&A^{-1}&-A^{-1}BA^{-1}\\
&&&&A^{-1}
\end{array}\right].
$$
Therefore, using Schur complement, we get
\begin{align*}
\det(C) &= \det(T)\det(A-RT^{-1}S)\\
&= \det(A)^{s-1} \det(A - (-1)^{s-2}B(A^{-1}B)^{s-2}A^{-1}B)\\
&= \det(A)^s \det(I - (-1)^{s-2}A^{-1}B(A^{-1}B)^{s-2}A^{-1}B)\\
&= (-t^{2r})^s \det(I - (-A^{-1}B)^s).
\end{align*}
So, the question boils down to finding $\det(I - (-A^{-1}B)^s)$.
</s>