JEE Main 2022 — Mathematics Algebra

Given,

$$A=[\begin{matrix}1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{matrix}]$$A=[100​a10​ab1​]

Now on rearranging we get,

$$A=[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]+[\begin{matrix}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{matrix}]=I+B$$A=[100​010​001​]+[000​a00​ab0​]=I+B

Now finding ${B}^{2}=[\begin{matrix}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{matrix}][\begin{matrix}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{matrix}]=[\begin{matrix}0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}]$B2=[000​a00​ab0​][000​a00​ab0​]=[000​000​ab00​]

And ${B}^{3}=0$B3=0

Now by using binomial expansion we get,${A}^{n}={(1+B)}^{n}=C0nI+C1nB+C2n{B}^{2}+C3n{B}^{3}+...$An=(1+B)n=C0nI+C1nB+C2nB2+C3nB3+...

$$=[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]+[\begin{matrix}0 & na & na \\ 0 & 0 & nb \\ 0 & 0 & 0\end{matrix}]+[\begin{matrix}0 & 0 & \frac{n(n-1)ab}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}]$$=[100​010​001​]+[000​na00​nanb0​]+[000​000​2n(n−1)ab​00​]

$$(\text{as}{B}^{3}\text{and higher power will become zero})$$(asB3and higher power will become zero)

$$=[\begin{matrix}1 & na & na+\frac{n(n-1)}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1\end{matrix}]=[\begin{matrix}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{matrix}]$$=[100​na10​na+2n(n−1)​abnb1​]=[100​4810​2160961​]

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

On solving above equations we get,

$\Rightarrow a=4,n=12$⇒a=4,n=12 and $b=8$b=8

So, $n+a+b=4+12+8=24$n+a+b=4+12+8=24