Let $$S = \left\{ {\left( {\matrix{ { - 1} & a \cr 0 & b \cr } } \right);a,b \in \{ 1,2,3,....100\} } \right\}$$ and let ${T_n} = \{ A \in S:{A^{n(n + 1)}} = I\}$. Then the number of elements in $\bigcap\limits_{n = 1}^{100} {{T_n}}$ is ___________.

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