Let $$A=\left[\begin{array}{lll} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{array}\right], a, b \in \mathbb{R}$$. If for some

$$n \in \mathbb{N}, A^{n}=\left[\begin{array}{ccc} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{array}\right] $$ then $n+a+b$ is equal to ____________.

<p>$$A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] + \left[ {\matrix{ 0 & a & a \cr 0 & 0 & b \cr 0 & 0 & 0 \cr } } \right] = I + B$$</p> <p>$${B^2} = \left[ {\matrix{ 0 & a & a \cr 0 & 0 & b \cr 0 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & a & a \cr 0 & 0 & b \cr 0 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 0 & {ab} \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$</p> <p>${B^3} = 0$</p> <p>$\therefore$ $${A^n} = {(1 + B)^n} = {}^n{C_0}I + {}^n{C_1}B + {}^n{C_2}{B^2} + {}^n{C_3}{B^3} + \,\,....$$</p> <p>$$ = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] + \left[ {\matrix{ 0 & {na} & {na} \cr 0 & 0 & {nb} \cr 0 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & 0 & {{{n(n - 1)ab} \over 2}} \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$</p> <p>$$ = \left[ {\matrix{ 1 & {na} & {na + {{n(n - 1)} \over 2}ab} \cr 0 & 1 & {nb} \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {48} & {2160} \cr 0 & 1 & {48} \cr 0 & 0 & 1 \cr } } \right]$$</p> <p>On comparing we get $na = 48$, $nb = 96$ and</p> <p>$na + {{n(n - 1)} \over 2}ab = 2160$</p> <p>$\Rightarrow a = 4,n = 12$ and $b = 8$</p> <p>$n + a + b = 24$</p>