00:01
In this problem, we have to show that if we keep multiplying a given matrix a by some permutation p, the first row of matrix a comes back to its original position.
00:18
So, to prove this, consider let us multiply the given matrix a by permutation p.
00:27
We get p -a is equal to p -a.
00:32
Now, multiplying again with permutation p, we get p times p a is equal to p square a, and p times p times p a is equal to p cube a.
01:02
And if we multiply the given matrix a with the permutation pn times, we get p times p times up to p times a is equal to pn times a.
01:39
Since we know that when a permutation matrix is multiplied with itself, the resultant matrix is also a permutation matrix, thus, p power 1, p squared, p cube, up to p raised to the power n are all permutation matrices.
02:10
If p is a permutation matrix of order n, then total possible permutations of p will be n factorial...