Let S = - 1 a 0 b ; a , b ∈ 1 , 2 , 3 , … 100 and let T n = A ∈ S : A n n + 1 = I . Then…

Given,

$A=\left[\begin{array}{cc}-1& a\\ 0& b\end{array}\right]$

$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]$

$\therefore T_n=\left\{A\in S;A^{n\left(a+1\right)}=I\right\}$

$\therefore b$ 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 $100$.