For $n\in\Bbb{N}$, define three matrices $A\_n(x,y), B\_n$ and $M\_n$ as follows:


(a) the $n\times n$ tridiagonal matrix $A\_n(x,y)$ with main diagonal all $y$'s, superdiagonal all $x$'s and subdiagonal all $-x$'s. For example,
$$ A\_4(x,y)=\begin{pmatrix} y&x&0&0\\-x&y&x&0\\0&-x&y&x
\\0&0&-x&y\end{pmatrix}. $$


(b) the $n\times n$ antidigonal matrix $B\_n$ consisting of all $1$'s. For example,
$$B\_4=\begin{pmatrix} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}.$$


(c) the $n^2\times n^2$ block-matrix $M\_n=A\_n(B\_n,A\_n(1,1))$ or using the Kronecker product $M\_n=A\_n(1,0)\otimes B\_n+I\_n\otimes A\_n(1,1)$.



> 
> **Question.** What is the determinant of $M\_n$?
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**UPDATE.** For even indices, I conjecture that



> 
> $$\det(M\_{2n})=\prod\_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2.$$
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This would confirm what Philipp Lampe's "perfect square" claim.



Block matrices and their determinants
Flip the order of the Kronecker products to get $M'=A\_n(I\_n,I\_n)+B\_n\otimes T\_n$, where $T\_n=A\_n(1,0)$. Note that $\det M=\det M'$. Since all blocks are polynomial in $A$, they commute, and therefore the determinant of $M'$ is $\det(f(T\_n))$, where $f(x)=\det(A\_n(1,1)+xB\_n)$. That is, $f(x)=\det(B\_n)\det(x+A\_n(1,1)B\_n)$ is $(-1)^{n\choose 2}$ times the characteristic polynomial of $H=-A\_n(1,1)B\_n$.


Let $t\_n(x)$ be the characteristic polynomial of $T\_n$. By repeated cofactor expansion on the first row, $t\_n(x)=xt\_{n-1}(x)+t\_{n-2}(x)$. The initial conditions then imply that the $t\_n$ is the $(n+1)^{th}$ Fibonacci polynomial. The roots of $t\_n$ are $2i\cos(k\pi/(n+1))$ for $k=1,\dots,n$.


The eigenvalues of $H$ are worked out in ["The eigenvalues of some anti-tridiagonal Hankel matrices"](https://journalskuwait.org/kjs/index.php/KJS/article/view/2372). When $n$ is odd they are $1$ and $\pm\sqrt{3+2\cos(\frac{2k\pi}{n+1})}$ for $k=1,\dots,\frac{n-1}{2}$. When $n$ is even they are $\pm\sqrt{1+4\cos^2(\frac{(2k+1)\pi}{n+1})}$ for $k=0,\dots,\frac{n}{2}-1$.


By a quick diagonalization argument $\det(f(T\_n))$ is the resultant of $f$ and $t\_n$. This plus some trig gives
$$
\det(M\_{2n})=\prod\_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2
$$
and
$$
\det(M\_{2n-1})=\prod\_{j=1}^{n-1}\left[1+4\cos^2\left(\frac{j\pi}{2n}\right)\right]^2\prod\_{k=1}^{n-1}\left[1+4\cos^2\left(\frac{j\pi}{2n}\right)+4\cos^2\left(\frac{k\pi}{2n}\right)\right]^2.
$$


