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Find the determinant $\Delta_n$

$A_n = \begin{bmatrix} 0 & 1 & 0 &\dots &\dots&0\\ -1 & 0 &1 & 0&&\vdots\\ 0&-1 & 0 &1 &\ddots&\vdots\\ \vdots& & & &\ddots&0\\ \vdots& & \ddots &\ddots &\ddots&1\\ 0 & \dots & \dots &0&-1&0 \end{bmatrix} \in M_n(\mathbb{K})$

After doing some tests I conclude $\Delta_n$ is $0$ if $n$ is odd and $1$ if $n$ is even. How can I prove it formally? Any hint?

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    $\begingroup$ Do the Laplace expansion along the first row and column to get $\Delta_n=\Delta_{n-2}$. Then $\Delta_1$ and $\Delta_2$ are obvious. $\endgroup$ – A.Γ. Jan 9 at 21:40
  • $\begingroup$ An alternate approach for the case of odd $n$: $A$ is skew-symmetric. So, if $n$ is odd, we have $$ \det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A) = -\det(A) $$ That is, we have $\det(A) = -\det(A)$ which means that $\det(A) = 0$. $\endgroup$ – Omnomnomnom Jan 9 at 22:43
  • $\begingroup$ You could also use more general results about tridiagonal Toeplitz matrices $\endgroup$ – Omnomnomnom Jan 9 at 22:46
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Hint:

Expanding by the last row, prove that $$\Delta_n=\begin{vmatrix}0&1&0&\dots&0&0&0 \\ -1&0&1&\dots&0&0&0 \\ 0&-1& 0 & \cdots&0&0&0\\ \vdots&&& \vdots &&&\vdots \\0&0&0&\cdots&0&1&0 \\0&0&0&\dots&-1&0&0 \\0&0&0&\dots&0&-1&1\end{vmatrix}=\Delta_{n-2} \;\text{ (expanding by the last column)}$$

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