If $$A = \frac{1}{2}\begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}$$ then,

We have $$A = \frac{1}{2}\begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}$$. Recognizing that this is a rotation matrix, we write $$ A = \begin{bmatrix} \cos(-\pi/3) & -\sin(-\pi/3) \\ \sin(-\pi/3) & \cos(-\pi/3) \end{bmatrix} = \begin{bmatrix} \cos(\pi/3) & \sin(\pi/3) \\ -\sin(\pi/3) & \cos(\pi/3) \end{bmatrix}, $$ which represents a rotation by $$-\pi/3$$.

Since $$A^n$$ corresponds to a rotation by $$-n\pi/3$$, it follows that $$A^6$$ is the rotation by $$-2\pi$$, which yields the identity matrix $$I$$.

Therefore, $$ A^{30} = (A^6)^5 = I^5 = I, \quad A^{25} = A^{24} \cdot A = (A^6)^4 \cdot A = I \cdot A = A. $$ Hence, $$ A^{30} + A^{25} - A = I + A - A = I. $$ The correct answer is Option C: $$A^{30} + A^{25} - A = I$$.