Let $$A = \begin{pmatrix} 4 & -2 \\ \alpha & \beta \end{pmatrix}$$. If $$A^2 + \gamma A + 18I = O$$, then $$\det(A)$$ is equal to

Consider the $$2 \times 2$$ matrix $$A = \begin{pmatrix} 4 & -2 \\ \alpha & \beta \end{pmatrix}$$. Its trace is $$\text{tr}(A) = 4 + \beta$$ and its determinant is $$\det(A) = 4\beta - (-2)\alpha = 4\beta + 2\alpha$$. The characteristic equation is given by $$\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \implies \lambda^2 - (4+\beta)\lambda + (4\beta + 2\alpha) = 0$$.

By the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation, we have $$A^2 - (4+\beta)A + (4\beta + 2\alpha)I = O$$. Since we are also given $$A^2 + \gamma A + 18I = O$$, comparing coefficients of $$A$$ and $$I$$ yields $$\gamma = -(4+\beta)$$ and $$18 = 4\beta + 2\alpha = \det(A)$$.

Therefore, $$\det(A) = \boxed{18}$$, and the answer is Option B.