Yes, the matrix is in rref.
Singular matrices have some non-zero vector they map to the zero-vector (via matrix multiplication). Non-singular matrices only map the zero-vector to the zero-vector (they have a trivial null space).
The matrix you have is singular because:
A(2,-1,0,0) = (0,0,0)
(other definitions of singular exist).
In this particular case, it is easy to tell that A will be singular, because rank(A) = 3, whereas dim(dom(A)) = 4. You can't fit all of a 4-dimensional space inside a 3-dimensional one without zeroing out at least one line.