Determinant of the $n \times n$ exchange matrix

Question

I have to find the determinant of an $n \times n$ exchange matrix. After applying values to $n$, I found out the determinant has a periodicity of $4$ and its formula is $(-1)^{n(n-1)/2}$. Literature revision confirms this. However, I don't know how to reach this conclusion formally neither how to prove it.

Answer 1

Let $J_n$ denote the exchange matrix of size $n$. It suffices to show the following:

• $\det(J_1) = 1$
• $\det(J_2) = -1$
• For $n \geq 3$, $\det(J_n) = -\det(J_{n-2})$.

The first two statements are easy to prove directly. For the third statement, proceed as follows: $$ \begin{align} \det(J_n) &= \det \pmatrix{0&J_{n-2}\\J_2 & 0} \\ & = \det\pmatrix{J_2 & 0\\0&J_{n-2}} \\ & = \det(J_2)\det(J_{n-2}) = -\det(J_{n-2}). \end{align} $$