Determinant of elementary permutation matrix Why is the determinant of an elementary permutation matrix equal to $-1$?
I am brand new to determinants and I've tried expanding it and using cofactor expansion, but it's messy and complicated. I would prefer if someone could show me using expansion, but alternative methods are welcome. Thanks.
 A: Not every permutation matrix has determinant $-1$, but the elementary matrices which are permutation matrices (corresponding to interchanges of two rows) have determinant $-1$.  The easy way to see this is that (1) the identity matrix has determinant $1$, and (2) interchanging two rows or columns of a matrix multiplies its determinant by $-1$.
A: Recall that if we interchange two rows (or two coulumns) of a matrix $A$ then its determinant change the sign. 
The permutation matrix is obtained from the identity   matrix  (with determinant $1$) by interchanging their rows so its determinant is $\pm 1$.
A: The elementary permutation matrix that swaps rows $i$ and $j$ is
$$\mathrm E_{ij} := \mathrm I - \mathrm{e}_i \mathrm{e}_i^\top - \mathrm{e}_j \mathrm{e}_j^\top + \mathrm{e}_i \mathrm{e}_j^\top + \mathrm{e}_j \mathrm{e}_i^\top = \mathrm I - \left( \mathrm{e}_i - \mathrm{e}_j \right) \left( \mathrm{e}_i - \mathrm{e}_j \right)^\top$$
Using the Weinstein-Aronszajn determinant identity,
$$\det \left( \mathrm E_{ij} \right) = \det \left( \mathrm I - \left( \mathrm{e}_i - \mathrm{e}_j \right) \left( \mathrm{e}_i - \mathrm{e}_j \right)^\top \right) = 1 - \left( \mathrm{e}_i - \mathrm{e}_j \right)^\top \left( \mathrm{e}_i - \mathrm{e}_j \right) = 1 - \underbrace{\| \mathrm{e}_i - \mathrm{e}_j \|_2^2}_{=2} = \color{blue}{-1}$$
