Key Concepts and Formulas

• Identity Matrix: A matrix $I$ such that for any matrix $A$ of compatible dimensions, $AI = IA = A$. For a 2x2 matrix, $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
• Matrix Powers: $A^n = A \cdot A \cdot ... \cdot A$ ($n$ times).
• Cyclic Nature of Powers: Some matrices, when raised to successive powers, exhibit a cyclic pattern, eventually returning to the identity matrix.

Step-by-Step Solution

Step 1: Understanding the Problem and Setting up the Condition

We are given the set S = \left\{ {n \in N\left| {{{\left( {\matrix{ 0 & i \cr 1 & 0 \cr } } \right)}^n}\left( {\matrix{ a & b \cr c & d \cr } } \right) = \left( {\matrix{ a & b \cr c & d \cr } } \right)\forall a,b,c,d \in R} \right.} \right\}. We want to find the number of 2-digit numbers in this set. Let $M = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix}$. The condition given translates to: $M^n X = X$ for all 2x2 real matrices $X$. This implies $M^n$ must be the identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

WHY: If $M^n X = X$ for all $X$, then $M^n$ must be the identity matrix. If we choose $X=I$, then $M^n I = I$, which implies $M^n = I$.

Step 2: Calculating Powers of Matrix M

We now compute powers of $M$ to find the smallest $n$ such that $M^n = I$.

• n = 1: $M^1 = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix}$
• n = 2: $M^2 = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix} = iI$
• n = 3: $M^3 = M^2 \cdot M = iI \cdot M = iM = \begin{pmatrix} 0 & -1 \\ i & 0 \end{pmatrix}$
• n = 4: $M^4 = M^2 \cdot M^2 = (iI)(iI) = i^2 I = -I = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$
• n = 5: $M^5 = M^4 \cdot M = -I \cdot M = -M = \begin{pmatrix} 0 & -i \\ -1 & 0 \end{pmatrix}$
• n = 6: $M^6 = M^4 \cdot M^2 = (-I)(iI) = -iI = \begin{pmatrix} -i & 0 \\ 0 & -i \end{pmatrix}$
• n = 7: $M^7 = M^4 \cdot M^3 = (-I)(iM) = -iM = \begin{pmatrix} 0 & 1 \\ -i & 0 \end{pmatrix}$
• n = 8: $M^8 = M^4 \cdot M^4 = (-I)(-I) = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Therefore, the smallest $n$ for which $M^n = I$ is $n = 8$.

WHY: We are explicitly computing powers of $M$ until we arrive at the identity matrix. This shows the cyclic nature.

Step 3: Determining the General Form of n

Since $M^8 = I$, it follows that $M^{8k} = (M^8)^k = I^k = I$ for any positive integer $k$. Therefore, $n$ must be a multiple of 8. So $n = 8k$ for $k \in N$.

WHY: If $M^8 = I$, then raising both sides to any integer power $k$ preserves the equality. This means any multiple of 8 will also result in the identity matrix.

Step 4: Finding the Number of 2-Digit Multiples of 8

We need to find the number of 2-digit numbers in the set $S$, which are multiples of 8. 2-digit numbers range from 10 to 99 inclusive. We want to find how many multiples of 8 fall within this range.

The smallest 2-digit multiple of 8 is $8 \times 2 = 16$. The largest 2-digit multiple of 8 is $8 \times 12 = 96$.

Therefore, the multiples of 8 that are 2-digit numbers are $8 \times 2, 8 \times 3, ..., 8 \times 12$. The number of such multiples is $12 - 2 + 1 = 11$.

WHY: We find the lower and upper bounds by finding the smallest and largest integer $k$ such that $8k$ is within the range of 2-digit numbers.

Common Mistakes & Tips

• Be careful with matrix multiplication, especially with complex numbers.
• Remember that $M^n X = X$ for all $X$ implies $M^n = I$.
• Ensure you are looking for 2-digit numbers (10-99 inclusive).

Summary

We are looking for the number of 2-digit integers $n$ such that $M^n = I$, where $M = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix}$. We calculated powers of $M$ and found that $M^8 = I$. Thus, $n$ must be a multiple of 8. The 2-digit multiples of 8 are $16, 24, \dots, 96$, and there are 11 such numbers.

Final Answer

The final answer is \boxed{11}.