How to calculate the determinant

Question

$$\begin{vmatrix} x & y & a & a & \dots & a & a & a\\ y & x & y & a & \dots & a & a & a\\ a & y & x & y & \dots & a & a & a\\ a & a & y & x & \dots & a & a & a\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ a & a & a & a& \dots & x & y & a\\ a & a & a & a & \dots & y & x & y\\ a & a & a & a & \dots & a & y & x \end{vmatrix}$$

Answer 1

I'm not sure that this determinant has a nice closed form. However, you might find the following approach to be helpful.

We can write the matrix $A$ of interest as a rank-1 update of a tridiagonal Toeplitz matrix. Write $A = T + axx^\top$, where $$ T = \pmatrix{x-a & y-a & 0\\y-a & x-a & y-a\\ &y-a & \ddots & \ddots\\ &&\ddots&&y-a\\ &&&y-a&x-a}, \quad x = \pmatrix{1\\1\\\vdots\\ 1}. $$ Per the linked wikipedia page, we can express the determinant of $T$ as the product of its eigenvalues, which is to say that $$ \det(T) = \prod_{k=1}^n \left(x-a + |y-a| \cos\left(\frac{\pi k}{n+1} \right) \right). $$ Alternatively, the determinant of $T$ can be described by a recurrence, as explained in this post and this post.

From there, $\det(A)$ can be computed with the help of the matrix determinant lemma. In particular, we have $$ \det(A) = (1 + a\cdot x^\top T^{-1}x)\det(T). $$ Notably, $T^{-1}$ has the closed form that is described here (among other places). $x^\top T^{-1}x$ is the sum of all entries of $T^{-1}$.