Inverse, Product of PermutationsDate: 04/27/2002 at 09:37:54 From: Meena Varma Subject: The inverse and product of permutations Hello Dr.Math, I know how to calculate the signum of permutations, but I don't understand how to calculate the inverse or the product of permutations. Could you please solve for: p1 = |1 2 3 4 5 6| and p2 = |1 2 3 4 5 6| |2 3 1 5 4 6| |1 2 3 6 4 5| Thanking you in anticipation. Regards, Meena Date: 04/27/2002 at 10:42:06 From: Doctor Mitteldorf Subject: Re: The inverse and product of permutations Dear Meena, The inverse of a permutation is the permutation that "undoes" it. If p1 takes 1 into 2, then the inverse of p1 must take 2 into 1. You can write the inverse of the permutation you've listed as p1 by putting the top line on the bottom and the bottom line on top. Then, for convention's sake, you should rearrange the columns to be in standard order, starting with |1 2 3 4 5 6| |3 1 etc. - you fill it in. To calculate the product you must follow the chain of events: ______________ / \ p1 takes 1->2 and p2 takes 2->2, so the product takes 1->2. p1 takes 2->3 and p2 takes 3->3, so the product takes 2->3. p1 takes 3->1 and p2 takes 1->1, so the product takes 3->1. p1 takes 4->5 and p2 takes 5->4, so the product takes 4->4. ...etc. (Notice that the product p1*p2 is not the same as p2*p1. The product that we've calculated above is the one that is conventionally written p2*p1, with p1 acting first, then p2.) - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 04/27/2002 at 11:04:39 From: Meena Varma Subject: The inverse and product of permutations Dear Dr.Mitteldorf, That was an excellent explanation and a very expeditious reply! Now I can do my university level Linear Algebra homework question :) This was the first question I ever asked here and I'm so delighted with the result! Thank you so much! Regards, Meena |
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