Determinant of matrix with zeros on diagonal and square identity matricies on non-diagonal.

Question

I am attempting to prove that $\begin{vmatrix}0 & I_{m}\\I_{m} & 0\end{vmatrix} = (-1)^m$. I have seen that if I work out the problem for a m = 2 I get the matrix $\begin{vmatrix}0&0&1&0 \\ 0 &0&0&1 \\ 1 &0&0&0 \\ 0 &1&0&0\end{vmatrix}$ This matrix requires 2*m elementary row operations to be transformed into the identity matrix, meaning the determinant would be equal to $(-1)^{2m}$. This does not match $(-1)^{m}$ if m is odd, my result and the desired result would be different. Am I missing a more obvious way to prove this? My work would not even be a real proof so I am guessing that I missed something obvious.

Answer 1

Actually, it takes $m$ elementary row operations to get the given matrix to the identity: swap columns $i$ and $m+i$ for $1\le i\le m$. So the determinant is indeed $(-1)^m$.