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Let $S=\left\{\left(\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right) ; a, b \in\{1,2,3, \ldots 100\}\right\}$ and let $T_{n}=\left\{A \in S: A^{n(n+1)}=I\right\}$. Then the number of elements in $\bigcap \limits_{n=1}^{100} T_{n}$ is
$50$
$85$
$100$
$137$
Solution
$A=\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]$
$A^{2}=\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]$
$=\left[\begin{array}{cc}1 & -a+a b \\ 0 & b^{2}\end{array}\right]$
$\therefore T _{ n }=\left\{ A \in S ; A ^{ n ( n +1)}= I \right\}$
$\therefore$ $b$ 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 $100$ .
