QJEEM22Ma9991: JEE Main 2022 Mathematics - Class 12 JEEM22Ma9991 Solution: Let S = { ( − 1 a 0 b ) ; a , b ∈ { 1 , 2 , 3 , . . . .100 } }…

$$\begin{array}{rl}& \mathrm{A}=\left[\begin{array}{cc}-1& \mathrm{a}\\ \\ 0& \mathrm{b}\end{array}\right]\\ \\ & {\mathrm{A}}^2=\left[\begin{array}{cc}-1& \mathrm{a}\\ \\ 0& \mathrm{b}\end{array}\right]\left[\begin{array}{cc}-1& \mathrm{a}\\ \\ 0& \mathrm{b}\end{array}\right]\\ \\ & =\left[\begin{array}{cc}1& -\mathrm{a}+\mathrm{ab}\\ \\ 0& {\mathrm{b}}^2\end{array}\right]\\ \\ & \therefore {\mathrm{T}}_{\mathrm{n}}=\{\mathrm{A}\in \mathrm{S};{\mathrm{A}}^{\mathrm{n}(\mathrm{n}+1)}=\mathrm{I}\}\end{array}$$

$\therefore$ b must be equal to 1

$\therefore$ In this case $\mathrm{A}^2$ will become identity matrix and a can take any value from 1 to 100

$\therefore$ Total number of common element will be 100 .