Let $$S = \{n \in N, \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix} a & b \\ c & d \end{pmatrix}= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ $$\forall a, b, c, d \in R$$, where $$i = \sqrt{-1}\}$$. Then the number of 2-digit numbers in the set $$S$$ is ___.

Let $$M=\begin{pmatrix} 0&i\\ 1&0 \end{pmatrix}$$

We require

$$M^nA=A \qquad\forall A$$

This is possible only when $$M^n=I$$

Now,

$$M^2=\begin{pmatrix}0&i\\1&0\end{pmatrix}\begin{pmatrix}0&i\\1&0\end{pmatrix}=\begin{pmatrix}i&0\\0&i\end{pmatrix}=iI$$

Hence,

$$M^4=(iI)^2=-I$$ and $$M^8=(-I)^2=I$$

Therefore,

$$M^n=I \iff 8\mid n$$

Now count two-digit multiples of $$8.$$

Smallest two-digit multiple: $$16$$

Largest two-digit multiple: $$96$$

Thus number of terms is $$\frac{96-16}{8}+1$$

$$=10+1$$

$$=11$$

Hence, the required answer is $$\boxed{11}$$