Question

Let p be an odd prime number & \(T_p\) be the following set of \(2 \times 2\) matrices \[T_p= \left\{ A=\left[\begin{matrix}a & b \\ c & a\end{matrix}\right]; a,b,c \epsilon \left\{ 0,1,2,.....,p-1 \right\} \right\}\]

Answer 1

then the number of A in \(T_p\) such that A is either symmetric or skew symmetric or both , & det(A) is divisible by p is
1). \((p-1)^2\)
2). 2(p-1)
3).\((p-1)^2+1\)
4). 2p-1

Answer 2

All skew symmetric matrices would be a=0 and c=p-b so there will be \(p\) many skew symmetric matrices (although the 0 matrix is also symmetric so don't double count it when counting the symmetric matrices) and also we have an exception with p=2, since that's the only case where we can have b+c=p with b=c so we actually have another matrix that is both symmetric and skew symmetric simultaneously.

In fact, we can stop for a moment to appreciate p=2 since it is dead simple, there is quite clearly only 4 matrices that are symmetric and skew symmetric (out of 8 possible matrices to choose from)

oooh also I just realized something about this determinant condition for skew symmetric matrices is the same as

\[a^2 +b^2 \equiv 0 \pmod p\]
So that's cute and related to that Fermat two square theorem or whoever did it.

Anyways I was gonna use that p=2 case to try to rule out some of the answer choices if possible but I'm kinda distracted and tired

Answer 3

Oh right, so it can be shown for p=2 the determinant condition is the same for both, a+b=0 and so that rules out 2 of the 4 meaning the expression for p=2 needs to be 2. That rules out options 1 and 4. Haha.