Let S = \ n ∈ N, ( cases 0 & i 1 & 0 cases ) ^ n ( cases a & b c &...

Let X = \left(\begin{cases}a & b \\ c & d\end{cases}\right) \text{ and } A = {\left(\begin{cases}0 & i \\ 1 & 0\end{cases}\right)}^{n}

$⇒ AX = IX$

$⇒ A = I$

⇒ {\left(\begin{cases}0 & i \\ 1 & 0\end{cases}\right)}^{n} = I

⇒ {A}^{8} = \left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right]

$⇒ n$ is multiple of $8 .$

So number of $2$ digit numbers in the set

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

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

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

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

Consider, $l = a + \left(n - 1\right)d = 96$

$⇒ 16 + \left(n - 1\right)8 = 96$

$⇒ \left(n - 1\right)8 = 80$

$⇒ \left(n - 1\right) = 10$

$⇒ n = 11$

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