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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?

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up vote 1 down vote accepted

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 $$P = E_t\cdots E_2E_1 \implies \det(P) = \prod^t_{i=1}\det(E_i)=(-1)^t$$

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"An elementary row switch matrix has determinant −1" How can this be shown in a general case for n x n matrix? –  Savch Oct 26 '12 at 18:20
    
There are numerous proofs available in many introductory linear algebra textbooks. It depends on how you are characterizing the determinant. If you're using Laplace expansion, I would go for an inductive argument. If you're using Leibniz formula then I would prove that it is an alternating form. If you simply define it as an alternating multi-linear form, well then that's that. –  EuYu Oct 26 '12 at 18:23
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