<s>[INST] Determinant of a special skew-symmetric matrix Simple calculation show that:
$$
\begin{align}
\det(A_2)=\begin{vmatrix} 
   0& 1  \\
   -1& 0  
\end{vmatrix}&=1\\
\det(A_4)=\begin{vmatrix} 
   0& 1 &1 &1 \\
   -1& 0 &1&1\\
   -1& -1& 0&1\\
   -1& -1& -1&0 
\end{vmatrix}&=1
\end{align}
$$
Here is my question:

Is it true that $\det(A_{2n})=1$ for all $n\in{\mathbb Z_+}$?

With MAPLE, I tried some large $n$. And I guess it is true. But temporarily I have no idea how to show it. 
 [/INST] Let $A_{2n}$ the $2n\times 2n$ matrix defined by $(A_{2n})_{i,j} = \begin{cases}
1&\mbox{if }i<j\\
0&\mbox{if }i=j\\
-1&\mbox{if } i>j
\end{cases}$.
We can show the result ($\det(A_{2n}) = 1$) by induction. The case $n=1$ and $n=2$ has been solved. Now, we can write $A_{2n+2}=\begin{pmatrix}A_{2n}&B\\-B^t &C\end{pmatrix}$ where $B$ is a $2n\times 2$ matrix whose entries are all $1$ and $C=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Consider the line $l_{2n+2}$: we can transform it, computing $l_{2n+2}\leftarrow l_{2n+2}-l_{2n+1}$. It becomes $(0,\ldots,0,-1,-1)^t$ (the $2n$ first components are $0$). After that we make $c_{2n+2}\leftarrow c_{2n+2}-c_{2n+1}$: we get $(0,\ldots,0,1,0)$. Now we conclude, expanding by the last column and after the last row (the first expansion gives $(-1)^{2n-1+2n}=-1$ and the second $(-1)^{2n-1+2n-1}\cdot (-1)$), hence we get $\det A_{2n+2}=(-1)\cdot(-1)\cdot\det A_{2n} =\det A_{2n}$.  
</s>