Tools of linear algebra work with any field ( https://secure.wikimedia.org/wikipedia/en/wiki/Field_(mathematics) ), including finite ones like integers mod 3. The problem is effectively Gaussian Elimination.

But we can't do transformations with antiequations such as with equations. For example, let we have a system: x1 != 0 (mod 3) x2 != 2 (mod 3). It has a solution (2; 0) If we add the antiequations, we'll get this one: x1+x2 != 2 (mod 3), but it's wrong for solution (2; 0) of the system! Am I right?

I think that your advice is absolutely right! Let for example det(Aij)<>0. Then Answer is 2^k,because (A*X)i in {0,1,2}\{bi} If det(Aij)==0 we can use gauss method to make standard worm of the matrix A.