Question

Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where

\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]

Then \( n(S) \) is equal to ______.

Answer 1

Correct Answer: 2

For the equation \( A^m + A^m = 3I - A^{-6} \), with matrix \(A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}\), we aim to find the integer set \( S \) and its cardinality \( n(S) \).

Step 1: Problem Analysis

The equation simplifies to \( 2A^m = 3I - A^{-6} \). We seek values of \( m \) that satisfy this equation.

Step 2: Compute \( A^2 \) and \( A^3 \)

\(A^2 = A \cdot A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix}\)

\(A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I\)

Step 3: Calculate \( A^{-6} \)

Since \( A^3 = I \), it follows that \( A^{-3} = I \). Therefore, \( A^{-6} = (A^{-3})^2 = I^2 = I \).

Step 4: Solve the Equation for \( m \)

Substituting \( A^{-6} = I \) into the simplified equation \( 2A^m = 3I - A^{-6} \) yields \( 2A^m = 3I - I = 2I \). This further simplifies to \( A^m = I \).

Given that \( A^3 = I \) and the powers of \( A \) are periodic with period 3, the integer solutions for \( m \) are \( m = 0 \) and \( m = 3 \).

Step 5: Determine \( n(S) \)

The set of solutions is \( S = \{0, 3\} \). The cardinality of this set is \( n(S) = 2 \).

The calculated value \( n(S) = 2 \) is consistent with the provided range (2,2).

Conclusion

The number of elements in the solution set \( S \), denoted as \( n(S) \), is 2, which falls within the specified range.