Animated Solution for Mathematics - Matrices and Determinants: Let $S=\{n\in \mathbb{N}:{\left(\begin{array}{cc}0& i\\ 1& 0\end{array}\right)}^n\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\forall a,b,c,d\in \mathbb{R}\}$, where $i=\sqrt{-1}$. Then the number of 2-digit numbers in the set $S$ is

The problem presents a matrix $A=\left(\begin{array}{cc}0& i\\ 1& 0\end{array}\right)$ and requires that $A^{n}X=X$ for any arbitrary matrix $X$.

If this condition holds for all $X$, then $A^n$ must be the identity matrix $I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$. This is because the identity matrix is the unique matrix that acts as the multiplicative identity for all other matrices.

Our objective is to determine the smallest positive integer $n$ such that $A^n=I$.

Let us begin by calculating the powers of $A$. First, we compute $A^2$:

We can factor out the scalar $i$ to observe that:

Next, we calculate $A^4$ using the laws of exponents:

Since $A^4=-I$, we proceed to calculate $A^8$:

We have successfully determined that the matrix $A$ has a period of $8$. Consequently, $A^n=I$ if and only if $n$ is a multiple of $8$.

We are tasked with finding the number of $2$-digit integers $n$ such that $n$ is a multiple of $8$. The set of such numbers is $S=\{n\in \mathbb{N}:n=8k,10\le n\le 99\}$.

The sequence of multiples of $8$ within this range starts at $16$ ($8\times 2$) and ends at $96$ ($8\times 12$). This forms an arithmetic progression where the first term $a=16$, the common difference $d=8$, and the last term $l=96$.

Using the formula for the $k$-th term of an arithmetic progression, $l=a+(k-1)d$, we substitute our known values:

Subtracting $16$ from both sides yields:

Dividing by $8$, we find:

There are exactly 11 such numbers.