Question

Let $S=\left\{\left(\begin{array}{cc}-1& a\\ 0& b\end{array}\right);a,b\in \{1,2,3,\dots 100\}\right\}$ and let $T_n=\left\{A\in S:A^{n(n+1)}=I\right\}$. Then the number of elements in $\bigcap _{n=1}^{100}{T}_n$ is _____

Answer 1

Answer: 100

$A=\left[\begin{array}{cc}-1& a\\ 0& b\end{array}\right]$
$A^2=\left[\begin{array}{cc}-1& a\\ 0& b\end{array}\right]\left[\begin{array}{cc}-1& a\\ 0& b\end{array}\right]$
$=\left[\begin{array}{cc}1& -a+ab\\ 0& b^2\end{array}\right]$
$\therefore T_n=\left\{A\in S;A^{n(n+1)}=I\right\}$
$\therefore$ b must be equal to $1$
$\therefore$ In this case $A^2$ will become identity matrix and a can take any value from $1$ to $100$
$\therefore$ Total number of common element will be $100$ .