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

JEE MainJEE Main 2022Matrices

Solution:

$ A^2 = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{bmatrix} $
$ =\begin{bmatrix} 1 & 2 a & 2 a+a b \\ 0 & 1 & 2 b \\ 0 & 0 & 1\end{bmatrix} $
$ A^2 A = \begin{bmatrix} 0 & 1 & 2 b \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 0 & 1 & b \\ 0 & 0 & 1\end{bmatrix} $
$ A ^3 = \begin{bmatrix} 1 & 3 a & 3 a+3 a b \\ 0 & 1 & 3 b \\ 0 & 0 & 1\end{bmatrix}$
$A ^4= \begin{bmatrix} 1 & 4 a & 4 a +6 ab \\ 0 & 1 & 4 b \\ 0 & 1 & 1\end{bmatrix} $
$ A ^{ n }= \begin{bmatrix} 1 & \text { na } & \frac{\left( n ^2- n \right)}{2} ab + na \\ 0 & 1 & nb \\ 0 & 0 & 1\end{bmatrix} $
$ na =48, nb =96$
$ na +\frac{ nab }{2}( n -1)=2160$
$ 48+24 b ( n -1)=2160 $
$ 48+24 \times 96-24 b =2160 $
$ b =8 and a =4, n =12 $
$ n + a + b =24$