JEE Main 2022 (Online) 24th June Evening ShiftQuestion 1
Let $S=\left\{\begin{pmatrix}-1&a\\0&b\end{pmatrix}:a,b\in\{1,2,3,\ldots,100\}\right\}$ and let $T_n=\{A\in S:A^{n(n+1)}=I\}$. Then the number of elements in $\bigcap_{n=1}^{100}T_n$ is $\_\_\_\_\_\_$ .
Matrices
Answer$100$
Solution
$A=\begin{pmatrix}-1&a\\0&b\end{pmatrix}$ is upper triangular so $A^k$ has diagonal $(-1)^k,b^k$. $n(n+1)$ is even, so $(-1)^k=1$; need $b^k=1$ with $b\in\{1,\dots,100\}$ positive, so $b=1$. For $b=1$: $A^2=\begin{pmatrix}1&-a+a\\0&1\end{pmatrix}=I$, so $A^{k}=I$ for all even $k$. So every $A$ with $b=1$ lies in every $T_n$, $a$ arbitrary: $100$ elements.