# determinant of permutation matrix

It's a well known fact that $\det(P)=(-1)^t$, where $t$ is the number of row exchanges in the $PA=LU$ decomposition. Can somebody point me to a (semi) formal proof as why it is so?

-

An elementary row switch matrix has determinant $-1$. A permutation matrix is just a product of such elementary matrices, so every row switch introduces a factor of $-1$. If you have $t$ row switches, then $$P = E_t\cdots E_2E_1 \implies \det(P) = \prod^t_{i=1}\det(E_i)=(-1)^t$$