The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det(A) divisible by p is:

For a matrix A to be symmetric, A=AT, which is possible if b=c. Hence we get |A|=a2-b2. We must have a2-b2=kp ⇒ (a+b)(a-b)=kp ⇒ either a-b or a+b is a multiple of p. a-b will be divisible by p only when a=b (since a and b can take values from 0 to p); hence number of such matrices is p and when a+b=multiple of p ⇒ a, b can take p values. (the pairs (1,p-1),(2,p-2),(3,p-3) and so on) ∴ Total number of matrices=p+p=2p. If a matrix A is skew symmetric then A=-AT, which, in this case, is possible only if a=0 and b=0. That gives a null matrix which has already been counted once in skew symmetric case. Hence, total number of matrix is still 2p