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Prove this property of the matrix

Source: 2019 Jozsef Wildt International Math Competition

May 19, 2020
linear algebramatrixdeterminantcomplex numbers

Problem Statement

We consider a natural number nn, n≥2n \geq 2 and the matrices
\begin{tabular}{cc} A=(123⋯nn12⋯n−1n−1n1⋯n−2⋯⋯⋯⋯⋯234⋯1)A= \begin{pmatrix} 1 & 2 & 3 & \cdots & n\\ n & 1 & 2 & \cdots & n - 1\\ n - 1 & n & 1 & \cdots & n - 2\\ \cdots & \cdots & \cdots & \cdots & \cdots\\2 & 3 & 4 & \cdots & 1 \end{pmatrix}
\end{tabular}
Show thatϵndet(In−A2n)+ϵn−1det(ϵIn−A2n)+ϵn−2det(ϵ2In−A2n)+⋯+det(ϵnIn−A2n)\epsilon^ndet\left(I_n-A^{2n}\right)+\epsilon^{n-1}det\left(\epsilon I_n-A^{2n}\right)+\epsilon^{n-2}det\left(\epsilon^2 I_n-A^{2n}\right)+\cdots +det\left(\epsilon^n I_n-A^{2n}\right) =n(−1)n−1[nn(n+1)2]2n2−4n(1+(n+1)2n(2n+(−1)n(2nn)))=n(-1)^{n-1}\left[\frac{n^n(n+1)}{2}\right]^{2n^2-4n}\left(1+(n+1)^{2n}\left(2n+(-1)^n{{2n}\choose{n}}\right)\right)where ϵ∈C\R\epsilon \in \mathbb{C}\backslash \mathbb{R}, ϵn+1=1\epsilon^{n+1}=1
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