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 $\underset{n=1}{\overset{100}{\cap }}{T}_n$ is

$A=\left[\begin{array}{cc}-1& a\\ 0& b\end{array}\right]\Rightarrow {\mathrm{A}}^2=\left[\begin{array}{cc}-1& \mathrm{a}\\ 0& \mathrm{b}\end{array}\right]\left[\begin{array}{cc}-1& \mathrm{a}\\ 0& \mathrm{b}\end{array}\right]==\left[\begin{array}{cc}1& -a+ab\\ 0& b^2\end{array}\right]$

$$\begin{gathered} \therefore {\mathrm{T}}_{\mathrm{n}}=\left\{\mathrm{A}\in \mathrm{S};{\mathrm{A}}^{\mathrm{n}(\mathrm{n}+1)}=\mathrm{I}\right\} \\ \therefore \text{ b must be equal to }1 \\ \therefore \text{ In this case }{\mathrm{A}}^2\text{ will become identity matrix and a } \\ \text{ can take any value from }1\text{ to }100 \\ \therefore \text{ Total number of common element will be }100\text{. } \end{gathered}$$

The number of elements in $\underset{n=1}{\overset{100}{\cap }}{T}_n$ is 100.