Help Determinant Binary Matrix I was messing around with some matrices and found the following result. 

Let $A_n$ be the $(2n) \times (2n)$ matrix consisting of elements $$a_{ij} = \begin{cases} 1 & \text{if } (i,j) \leq (n,n) \text{ and } i \neq j \\ 1 & \text{if } (i,j) > (n,n) \text{ and } i \neq j \\ 0 & \text{otherwise}.  \end{cases} $$
   Then, the determinant of $A_n$ is given by $$\text{det}(A_n) = (n-1)^2.$$

Example: $$A_2 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, A_3 = \begin{pmatrix} 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{pmatrix},$$ with det$(A_2)$ and det$(A_3)$ being $1$ and $4$ respectively. I was wondering if anybody could prove this statement for me.

The matrix is diagonal by blocks. so the determinant is equal to the product of the determinants of each block.
So we just have to prove that:
$\det\begin{pmatrix}0 & 1 & \dots & 1 \\ 1 & 0 & \dots & 1 \\ & &\vdots \\ 1 & 1 & \dots & 0\\ \end{pmatrix}=\pm(n-1)$
This is done in various ways here, in fact it is equal to $(-1)^n(n-1)$
