Any hints how to calculate this determinant? The result should be 0 (it's a determinant of distance matrix of a even length cycle).
$$\begin{vmatrix} 0 & 1 & 2 & 3 & \dots & k-1 & k & k-1 & \dots & 2 & 1\\ 1 & 0 & 1 & 2 & \dots & k-2 & k-1 & k & \dots & 3 & 2\\ 2 & 1 & 0 & 1 & \dots & k-3 & k-2 & k-1 & \dots & 4 & 3\\ \vdots & \vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots & \vdots \\ k-1 & k-2 & k-3 & \dots & \dots & 0 & \dots & \dots & \dots & k-1 & k \\ k & k-1 & k-2 & \dots & \dots & \dots & 0 & \dots & \dots & k-2 & k-1 \\ k-1 & k & k-1 & \dots & \dots & \dots & \dots & 0 & \dots & k-3 & k-2 \\ \vdots & & & & & & & & & \vdots \\ 1 & 2 & 3 & 4 & \dots & k & k-1 & k-2 & \dots & 1 & 0 \end{vmatrix} $$
If we add the first row to the $(k+1)$-th row and add the second row to the $(k+2)$-th row then those two rows will be equal and so we can conclude the determinant is zero.