Permutation and combination in matrices

Let p be an odd prime number and T be the following set of 2X2 matrices,

T={A=matrix such that a11=a, a12=b, a21=c, a22=a where aij denotes the term in ith row and jth column

a,b,c belongs to {0,1,2,........,p-1}

THen number of A in T such that trace of A is not divisible by p but det(A) is divisible by p is?

trace of matrix isthe sum of its diagnol entries.

Re: Permutation and combination in matrices

Let be any of these described matricies, .

Then (since is odd).

And .

Since via the nonzero trace, clearly , and .

From that it's obvious to identify the given set of usable integers with , the integer remainders modulo p.

Let , and denote the multiplicative inverse of by .

So we want the cardinality of .

But amounts to simply .

Thus .

Thus the function , is surjective.

But looking at the first row shows it's obviously injective. Hence it's bijective.

Hence .

Thus there are disitinct matricies your problem described, exactly one such for each top row made by choosing any values in {1, 2, ..., p-1}.

Re: Permutation and combination in matrices

Thanks. But can you give a simpler proof. I am not pro at number theory.

Re: Permutation and combination in matrices

There's nothing advanced there. Maybe some of the notation I used you weren't familiar with? (those funky twists in front of thosee "mod p" mean "congruent to". It's tempting to write "32 = 7 mod 5", but I suppose that mathematicians loathe to write, even when it makes sense, that 32 = 7, just on general principles. It's just easier to write congruence rather than equals, because otherwise they have to write it as "[32] = [7] in ", which is a pain. All it means is "modulo p" which means "remainders when dividing by p".

I filled in almost all the details - that's the only real reason it seems long. To someone who I already knew was comfortable with math, my answer might've gone like this (or much much less - just writing down the form of the answer): (Note that this isn't a demonstration so much as an explanation):

If represents those matricies, then obviously .

You can quickly see that as required.

However, since isn't zero and , . Obviously can't be zero, since its inverse is required.

Because of the top row, each choice of pairs gives a different matrix in .

All the matricies of are obviously included.

Thus .