TITLE: determinant of permutation matrix
QUESTION [7 upvotes]: 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?

REPLY [10 votes]: Perhaps you can elaborate on what exactly is confusing to you.
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
$$P = E_t\cdots E_2E_1 \implies \det(P) = \prod^t_{i=1}\det(E_i)=(-1)^t$$