JEE Main 2021 — Mathematics Algebra

Let X=(\begin{matrix}a & b \\ c & d\end{matrix})&A={(\begin{matrix}0 & i \\ 1 & 0\end{matrix})}^{n}

$$\Rightarrow AX=IX$$⇒AX=IX

$$\Rightarrow A=I$$⇒A=I

$$\Rightarrow {(\begin{matrix}0 & i \\ 1 & 0\end{matrix})}^{n}=I$$⇒(01​i0​)n=I

$$\Rightarrow {A}^{8}=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]$$⇒A8=[10​01​]

$\Rightarrow n$⇒n is multiple of $8.$8.

So number of $2$2 digit numbers in the set

$$S={16,24,32,\ldots ,96}$$S=16,24,32,…,96

Clearly, $16,24,32,\ldots ,96$16,24,32,…,96 are in arithmetic progression.

Here, $a=16,d=24-16=8$a=16,d=24−16=8 and $l=a+(n-1)d=96$l=a+(n−1)d=96

Where, $a$a is the first term, $d$d is the common difference, $l$l is the last term and $n$n is the number of terms.

Consider, $l=a+(n-1)d=96$l=a+(n−1)d=96

$$\Rightarrow 16+(n-1)8=96$$⇒16+(n−1)8=96

$$\Rightarrow (n-1)8=80$$⇒(n−1)8=80

$$\Rightarrow (n-1)=10$$⇒(n−1)=10

$$\Rightarrow n=11$$⇒n=11

Hence, number of two-digit numbers in set $S$S is $11$11.