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(A) Let $A = I + B$,where $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & a & a \ 0 & 0 & b \ 0

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$24$

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(A) Let $A = I + B$,where $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & a & a \ 0 & 0 & b \ 0 & 0 & 0 \end{bmatrix}$.Calculating $B^2 = \begin{bmatrix} 0 & 0 & ab \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$ and $B^3 = 0$.Using the Binomial Theorem,$A^n = (I + B)^n = I + nB + \frac{n(n-1)}{2} B^2$.$A^n = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & na & na \ 0 & 0 & nb \ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & \frac{n(n-1)ab}{2} \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & na & na + \frac{n(n-1)ab}{2} \ 0 & 1 & nb \ 0 & 0 & 1 \end{bmatrix}$.Comparing with the given matrix $\begin{bmatrix} 1 & 48 & 2160 \ 0 & 1 & 96 \ 0 & 0 & 1 \end{bmatrix}$:$na = 48$,$nb = 96$,and $na + \frac{n(n-1)ab}{2} = 2160$.From $na = 48$ and $nb = 96$,we get $b = 2a$.Substituting into the third equation: $48 + \frac{n(n-1)a(2a)}{2} = 2160 \Rightarrow 48 + n(n-1)a^2 = 2160 \Rightarrow n(n-1)a^2 = 2112$.Since $na = 48$,$a = \frac{48}{n}$. Substituting this: $n(n-1)(\frac{48}{n})^2 = 2112 \Rightarrow (n-1) \frac{2304}{n} = 2112 \Rightarrow 2304n - 2304 = 2112n \Rightarrow 192n = 2304 \Rightarrow n = 12$.Then $a = \frac{48}{12} = 4$ and $b = \frac{96}{12} = 8$.Thus,$n + a + b = 12 + 4 + 8 = 24$.

Let $A = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix}$,where $a, b \in \mathbb{R}$. If for some $n \in \mathbb{N}$,$A^n = \begin{bmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{bmatrix}$,then $n + a + b$ is equal to:

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