Question

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

Hint:

If a square matrix $A$ satisfies $A^k = cI$, then $A^{2k} = c^2I$. Power-of-two iterations are often faster than computing $A^2, A^3, A^4$ linearly.

Answer 1

Correct Answer: 11

Step 1: Understanding the Concept:
If \(A^n X = X\) for any matrix \(X\), then \(A^n\) must be the identity matrix \(I\).
We compute the powers of the given matrix \(A\) to find the cycle of its powers.
Step 3: Detailed Explanation:
Let \(A = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix}\).
Compute \(A^2\):
\[ A^2 = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix} = i I \]
Compute higher powers:
\(A^4 = (A^2)^2 = (i I)^2 = -I\).
\(A^8 = (A^4)^2 = (-I)^2 = I\).
Thus, the smallest positive integer \(n\) for which \(A^n = I\) is 8.
The matrix \(A^n\) will equal \(I\) for any \(n\) that is a multiple of 8.
We need the number of 2-digit natural numbers in set \(S\).
The 2-digit multiples of 8 are:
\(16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\).
Counting these values:
The sequence is \(8 \times 2, 8 \times 3, ......., 8 \times 12\).
Number of terms \(= 12 - 2 + 1 = 11\).
Step 4: Final Answer:
The number of such 2-digit numbers is 11.