JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 17)
Let $$S = \left\{ {\left( {\matrix{
{ - 1} & a \cr
0 & b \cr
} } \right);a,b \in \{ 1,2,3,....100\} } \right\}$$ and let $${T_n} = \{ A \in S:{A^{n(n + 1)}} = I\} $$. Then the number of elements in $$\bigcap\limits_{n = 1}^{100} {{T_n}} $$ is ___________.
Answer
100
Explanation
$$
\begin{aligned}
&\mathrm{A}=\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right] \\\\
&\mathrm{A}^2=\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right]\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right] \\\\
&=\left[\begin{array}{cc}
1 & -\mathrm{a}+\mathrm{ab} \\\\
0 & \mathrm{~b}^2
\end{array}\right] \\\\
&\therefore \mathrm{T}_{\mathrm{n}}=\left\{\mathrm{A} \in \mathrm{S} ; \mathrm{A}^{\mathrm{n}(\mathrm{n}+1)}=\mathrm{I}\right\}
\end{aligned}
$$
$\therefore$ 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 .
$\therefore$ 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 .
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