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Statement: If $P_1$ is an odd and $P_2$ be an even permutation matrices of size $n$ then, prove that $\det(2P_1-2P_2)=0$

My approach: I wrote the term as $2^n \det(P_1-P_2)=0$. Now we need to show that the matrix obtained from $P_1-P_2$ will have at least of its pivot $0$. But I am unable to prove the existence of a zero pivot.

Kindly help me with this proof. If there is any other way to approach the proof apart from the one I mentioned, please do share.

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    $\begingroup$ You might want to have look here $\endgroup$ Commented Oct 2, 2020 at 8:16

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Hint: what happens when you multiply P1 and P2 by the column vector $(1,1,\dots,1)^T$?

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  • $\begingroup$ Got it! Thanks a lot! $\endgroup$ Commented Oct 2, 2020 at 15:23

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