JEE Main 2022 — Mathematics Algebra

We have $A=[\begin{matrix}1+i & 1 \\ -i & 0\end{matrix}]$A=[1+i−i​10​]

$${A}^{2}=A\cdot A=[\begin{matrix}1+i & 1 \\ -i & 0\end{matrix}][\begin{matrix}1+i & 1 \\ -i & 0\end{matrix}]=[\begin{matrix}i & 1+i \\ -i+1 & -i\end{matrix}]$$A2=A⋅A=[1+i−i​10​][1+i−i​10​]=[i−i+1​1+i−i​]

$${A}^{3}={A}^{2}\cdot A=[\begin{matrix}i & 1+i \\ 1-i & -i\end{matrix}][\begin{matrix}1+i & 1 \\ -i & 0\end{matrix}]=[\begin{matrix}0 & i \\ 1 & 1-i\end{matrix}]$$A3=A2⋅A=[i1−i​1+i−i​][1+i−i​10​]=[01​i1−i​]

$${A}^{4}={A}^{3}\cdot A=[\begin{matrix}0 & i \\ 1 & 1-i\end{matrix}][\begin{matrix}1+i & 1 \\ -i & 0\end{matrix}][\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]=I$$A4=A3⋅A=[01​i1−i​][1+i−i​10​][10​01​]=I

$$\therefore {A}^{4}=I$$∴A4=I

So, ${A}^{5}={A}^{4}\cdot A=I\cdot A=A$A5=A4⋅A=I⋅A=A

${A}^{6}={A}^{4}\cdot {A}^{2}=I\cdot {A}^{2}={A}^{2}$A6=A4⋅A2=I⋅A2=A2 and so on

$$\because {A}^{1}={A}^{5}={A}^{9}=.....={A}^{97}=A$$∵A1=A5=A9=.....=A97=A

Hence, possible values of $n$n, such that ${A}^{n}=A$An=A

$$={1,5,9,\ldots ,97}$$=1,5,9,…,97

Clearly, above sequence is in A.P. where

a=1,d=4&{t}_{n}=97 $\Rightarrow a+(n-1)d=97$⇒a+(n−1)d=97

$\Rightarrow 1+(n-1)4=97$⇒1+(n−1)4=97 $\Rightarrow n=25$⇒n=25

$\therefore$∴ The number of elements in the given set $=25$=25.