Let $$A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$$ and $$B = 7A^{20} - 20A^7 + 2I$$, where $$I$$ is an identity matrix of order $$3 \times 3$$. If $$B = [b_{ij}]$$, then $$b_{13}$$ is equal to ___.

Write $$A = I + N$$ where $$N = A - I = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$. Since $$N$$ is strictly upper triangular, it is nilpotent: $$N^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ and $$N^3 = 0$$.

By the binomial theorem for commuting matrices, $$A^n = (I+N)^n = I + nN + \binom{n}{2}N^2$$ (since all higher powers of $$N$$ vanish).

The $$(1,3)$$ entry (top-right) of $$I$$ is 0, of $$N$$ is 0, and of $$N^2$$ is 1. So the $$(1,3)$$ entry of $$A^n$$ is $$\binom{n}{2} = \frac{n(n-1)}{2}$$.

Computing: the $$(1,3)$$ entry of $$A^{20}$$ is $$\binom{20}{2} = 190$$, and of $$A^7$$ is $$\binom{7}{2} = 21$$. The $$(1,3)$$ entry of $$I$$ is 0.

Therefore: $$b_{13} = 7 \cdot 190 - 20 \cdot 21 + 2 \cdot 0 = 1330 - 420 = 910$$