Question

If ${\left[\begin{array}{ccc}1& 2& a\\ 0& 1& 4\\ 0& 0& 1\end{array}\right]}^n=\left[\begin{array}{ccc}1& 18& 2007\\ 0& 1& 36\\ 0& 0& 1\end{array}\right]$, then find $\frac{a+n}{100}.$

Answer 1

${[\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}0& 2& a\\ 0& 0& 4\\ 0& 0& 0\end{array}\right]]}^n=\left[\begin{array}{ccc}1& 18& 2007\\ 0& 1& 36\\ 0& 0& 1\end{array}\right]$
I A
$\Rightarrow (I+A{)}^n=\left[\begin{array}{ccc}1& 18& 2007\\ 0& 1& 36\\ 0& 0& 1\end{array}\right]$
$\Rightarrow I+nA+\frac{n(n-1)}{2}{A}^2=\left[\begin{array}{ccc}1& 18& 2007\\ 0& 1& 36\\ 0& 0& 1\end{array}\right]$
$(A^3,A^4.......isanullmatrix)$
On solving it, we get $n=9$ and $a=191$