Let $$R_1$$ and $$R_2$$ be two relations defined on $$\mathbb{R}$$ by $$a R_1 b \Leftrightarrow ab \geq 0$$ and $$a R_2 b \Leftrightarrow a \geq b$$, then

We can write $$A = I + N$$ where $$N = \begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}$$.

Since $$N$$ is strictly upper triangular, it is nilpotent: $$N^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ and $$N^3 = O$$.

Because $$I$$ and $$N$$ commute, by the binomial theorem: $$A^n = (I + N)^n = I + nN + \binom{n}{2}N^2$$ for all $$n \geq 0$$.

The $$(1,3)$$ entry of $$A^n$$ is $$0 + n \cdot 0 + \binom{n}{2} \cdot 1 = \dfrac{n(n-1)}{2}$$.

Now $$B = 7A^{20} - 20A^7 + 2I$$, so $$b_{13} = 7 \cdot \dfrac{20 \cdot 19}{2} - 20 \cdot \dfrac{7 \cdot 6}{2} + 2 \cdot 0 = 7 \cdot 190 - 20 \cdot 21 = 1330 - 420 = 910$$.

The correct answer is Option A: $$910$$.