It's either every choice of a, or no choice of a. Convert this into a matrix-vector equation Ax=b(a). I write b(a) to denote a vector b that depends on the choice of a. Then you're asking for values of a such that the preimage of b(a) under the linear transformation A is a single vector. But that happens (no matter what b(a) is) if and only if A is injective, i.e. when the matrix A is non-singular. If A is singular (zero determinant), then you may have no solution or infinitely many solutions, depending on a, but never a unique solution.