JEE Main 2022 — Matrices Question with Solution

Given,

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

Now on rearranging we get,

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

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

And $B^3=0$

Now by using binomial expansion we get,$A^n={\left(1+B\right)}^n={}^{n}C_{0}I+{}^{n}C_{1}B+{}^{n}C_2{B}^2+{}^{n}C_3{B}^3+...$

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

$\left(\text{as} B^3 \text{and higher power will become zero}\right)$

$=\left[\begin{array}{ccc}1& na& na+\frac{n\left(n-1\right)}{2}ab\\ 0& 1& nb\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 48& 2160\\ 0& 1& 96\\ 0& 0& 1\end{array}\right]$

On comparing we get $na=48, nb=96$ and $na+\frac{n\left(n-1\right)}{2}ab=2160$

On solving above equations we get,

$\Rightarrow a=4,n=12$ and $b=8$

So, $n+a+b=4+12+8=24$