Question

Let $A=\left[\begin{array}{ccc}1& a& a\\ 0& 1& b\\ 0& 0& 1\end{array}\right],a,b\in R$. 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 _________

Answer 1

Answer: 24

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