Determinant of an anti-diagonal block matrix composed of identity matrices I'm having a little trouble solving this problem :
What is the determinant of the matrix ${A= \begin{bmatrix}{0_{p,n}} && {I_p} \\ {I_n} && {0_{n,p}}\end{bmatrix} }$ ?
This page says $\det (A)=(-1)^{np} $.
But I don't really see how to get there.
The way I see it,  by switching the columns of A, you can get to the (n+p) identity  matrix ${ I_{n+p} }$ .
In that case, since switching the columns of a matrix inverts the sign of it's determinant, then $\det(A)=(-1)^ q $ , with q the number of column switches needed to obtain the $(n+p)$ identity matrix $I_{n+p}$ .
That would mean you can always get to $I_{n+p}$ with $n \cdot p$ switches.
Now, I don't really know how to show this.
Any help would be appreciated, thanks.

There are $n$ columns in the first column block. Each has to swap with the $p$ remaining columns to get the whole thing in the form $I_{n+p}$.
