29th Putnam 1968 Problem B5 Let F be the field with p elements. Let S be the set of 2 x 2 matrices over F with trace 1 and determinant 0. Find |S|. Solution
: p 2 + p. Let the matrix be a b c d If we take any a not 0 or 1, then d is fixed as 1 - a and is also not 0. Hence bc is fixed as a(1 - a) and is non-zero. So we can take any b not 0 and d is then fixed. Altogether that gives us (p - 2)(p - 1) possibilities. If we take a as 0 or 1, the d is fixed as 1 - a and bc must be 0. That gives 2p-1 possibilities for bc, or 4p-2 possibilities in all. So in total we have (p - 2)(p - 1) + 4p - 2 = p 2 + p. 29th Putnam 1968 © John Scholes jscholes@kalva.demon.co.uk 14 Jan 2002