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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.)

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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 $\det(I - (-A^{-1}B)^s)$.

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