JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 22

Square. $A^2=\begin{pmatrix}1&a(b-1)\\0&b^2\end{pmatrix}$.

$n=1$ needs $A^2=I$: $b^2=1$, so $b=1$, and then $a(b-1)=0$ for every $a$.

All $n$. $n(n+1)$ is even, so $A^2=I$ gives $A^{n(n+1)}=I$ for every $n$. The intersection is $\{b=1\}$, with $a$ free.

Count 100.

Answer: 100