Let A = [ cases 1 & a & a 0 & 1 & b 0 & 0 & 1 cases ], a, b ∈ ℝ ....

Given,

A = \left[\begin{cases}1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{cases}\right]

Now on rearranging we get,

A = \left[\begin{cases}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}\right] + \left[\begin{cases}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{cases}\right] = I + B

Now finding {B}^{2} = \left[\begin{cases}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{cases}\right]\left[\begin{cases}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{cases}\right] = \left[\begin{cases}0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{cases}\right]

And ${B}^{3} = 0$

Now by using binomial expansion we get,${A}^{n} = {\left(1 + B\right)}^{n} = C0 n I + C1 n B + C2 n {B}^{2} + C3 n {B}^{3} + . . .$

= \left[\begin{cases}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}\right] + \left[\begin{cases}0 & na & na \\ 0 & 0 & nb \\ 0 & 0 & 0\end{cases}\right] + \left[\begin{cases}0 & 0 & \dfrac{n\left(n - 1\right)ab}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{cases}\right]

\left(\text{as}{B}^{3}\text{and higher power will become zero}\right)$</p><p>$= \left[$\begin{cases}1 & na & na + \dfrac{n\left(n - 1\right)}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1\end{cases}$\right] = \left[$\begin{cases}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>On comparing we get$na = 48, nb = 96$and$na + \dfrac{n\left(n - 1\right)}{2}ab = 2160$</p><p>On solving above equations we get,</p><p>$⇒ a = 4, n = 12$and$b = 8$</p><p>So,$n + a + b = 4 + 12 + 8 = 24