Let A=left[begin{array}{lll}1 & a & a 0 & 1 & b 0 & 0 & 1end{array}right], a

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3 and 4 .Determinants and Matrices

hard

Let $A=\left[\begin{array}{lll}1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{array}\right], a, b \in R$. If for some $n \in N$, $A ^{ n }=\left[\begin{array}{ccc}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{array}\right]$ then $n + a + b$ is equal to $\dots\dots$

$24$

$23$

$22$

$21$

(JEE MAIN-2022)

Solution

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

$B^{2}=\left[\begin{array}{lll}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{array}\right]+\left[\begin{array}{lll}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}0 & 0 & a b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

$B^{3}=0$

$\therefore A^{n}=(1+B)^{n}={ }^{n} C_{0} I+{ }^{n} C_{1} B+{ }^{n} C_{2} B^{2}+{ }^{n} C_{3} B^{3}+$

$=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]+\left[\begin{array}{ccc} 0 & n a & n a \\ 0 & 0 & n b \\ 0 & 0 & 0 \end{array}\right]+\left[\begin{array}{ccc} 0 & 0 & \frac{n(n-1) a b}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

$=\left[\begin{array}{ccc} 1 & n a & n a+\frac{n(n-1)}{2} a b \\ 0 & 1 & n b \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 48 & 2160 \\ 0 & 1 & 48 \\ 0 & 0 & 1 \end{array}\right]$

On comparing we get $n a=48, n b=96$ and

$n a+\frac{n(n-1)}{2} a b=2160$

$\Rightarrow a=4, n=12 \text { and } b=8$

$\quad n+a+b=24$

Standard 12

Mathematics