Let $$A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix}$$. If $$A^3 = 4A^2 - A - 21I$$, where $$I$$ is the identity matrix of order $$3 \times 3$$, then $$2a + 3b$$ is equal to

We have $$A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix}$$ and $$A^3 = 4A^2 - A - 21I$$, which implies $$A^3 - 4A^2 + A + 21I = 0$$, so by the Cayley-Hamilton theorem $$A$$ satisfies its own characteristic equation.

$$p(\lambda) = \lambda^3 - 4\lambda^2 + \lambda + 21 = 0$$

The trace of $$A$$ is $$2 + 3 + b = 5 + b$$, and since the coefficient of $$\lambda^2$$ in the characteristic polynomial is the negative of the trace, we have $$-(5 + b) = -4$$ which gives $$5 + b = 4$$ and hence $$b = -1$$.

The sum of the cofactors of the diagonal entries equals the coefficient of $$\lambda$$ in the characteristic polynomial. The cofactor of $$a_{11}$$ is $$3b - 5 = -3 - 5 = -8$$, of $$a_{22}$$ is $$2b - 0 = -2$$, and of $$a_{33}$$ is $$6 - a$$, so their sum is $$-8 - 2 + 6 - a = -4 - a$$ which must equal $$1$$. Therefore $$-4 - a = 1$$ and hence $$a = -5$$.

To verify, the determinant of $$A$$, which equals the negative of the constant term of the characteristic polynomial, is $$2(3b - 5) - a(b) + 0 = 2(-8) -(-5)(-1) = -16 - 5 = -21$$. Since the constant term is $$-\det(A) = 21$$, the calculation is consistent.

Finally, $$2a + 3b = 2(-5) + 3(-1) = -10 - 3 = -13$$, so the correct answer is Option B: $$-13$$.