Let $A=\left[\begin{array}{lll}1 & a & a \ 0 & 1 & b \ 0 & 0 & 1\end{array} ight], a, b \in R$. I

Question

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

Professor PC Thomas Classes · Answer

(24) : $A=\left[\begin{array}{lll}1 & a & a \ 0 & 1 & b \ 0 & 0 & 1\end{array}ight]$\ $\Rightarrow \quad A^{2}=\left[\begin{array}{ccc}1 & 2 a & a+a b+a \ 0 & 1 & 2 b \ 0 & 0 & 1\end{array}ight]$\ $A^{3}=\left[\begin{array}{ccc}1 & 3 a & 3 a+3 a b \ 0 & 1 & 3 b \ 0 & 0 & 1\end{array}ight]$\ $A^{4}=\left[\begin{array}{ccc}1 & 4 a & 4 a+6 a b \ 0 & 1 & 4 b \ 0 & 0 & 1\end{array}ight]$\ $	herefore \quad$ We can say $n a=48, n b=96$,\ $n a+\frac{n(n-1)}{2} a b=2160$\ By solving these equations, we get $n=12$,\ $a=4$ and $b=8$\ $\Rightarrow n+a+b=12+4+8=24$\