Question

Let $S=\left\{\begin{pmatrix}-1 & a \\ 0 & b \end{pmatrix} ; a , b \in\{1,2,3, \ldots 100\}\right\}$ and let $T_{n}=\left\{A \in S: A^{n(n+1)}=I\right\}$. Then the number of elements in $\displaystyle\bigcap_{ n =1}^{100} T _{ n }$ is _____

Answer 1

$A =\begin{bmatrix}-1 & a \\0 & b\end{bmatrix}$
$ A ^{2}=\begin{bmatrix}-1 & a \\0 & b\end{bmatrix} \begin{bmatrix}-1 & a \\0 & b\end{bmatrix} $
$=\begin{bmatrix}1 & - a + ab \\0 & b ^{2}\end{bmatrix}$
$\therefore T _{ n }=\left\{ A \in S ; A ^{ n ( n +1)}= I \right\}$
$\therefore$ 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 $100$ .