There's probably a slick group-theoretic way of doing this, but I would just use a simple counting strategy.

There are elements in , which I'm thinking of as the set of all 2×2 matrices over . Count the number of those for which the determinantisequal to 0:

there is one matrix with all four components zero (the zero element),

there are 8 matrices with three components zero,

there are 8 matrices with two zero components (in the same row),

there are 8 matrices with two zero components (in the same column).

There are no matrices with zero determinant and just one zero. Finally, there are 8 matrices with zero determinant and no zero components (a, b and c can all be 1 or 2, and then d is determined by the condition ad–bc=0).

That gives a total of 33 matrices with determinant 0, hence 81 – 33 = 48 matrices with nonzero determinant.