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

Given the equation \( A^m + A^m = 3I - A^{-6} \), where:

\(A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}\), we need to find the integer set \( S \) and determine \( n(S) \).

Step 1: Analyze the Problem

\(A^m + A^m = 2A^m = 3I - A^{-6}\). Our goal is to find such \( m \) that makes the equation hold true.

Step 2: Compute \( A^2 \)

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

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

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

Given \( A^3 = I \), then \( A^{-3} = A^0 = I \), hence \( A^{-6} = (A^{-3})^2 = I \).

Step 4: Solve the Equation

Original equation becomes:

\( 2A^m = 3I - A^{-6} = 2I \)

From \( A^m = I \), solutions are \( m = 0 \) or \( m = 3 \), as indices repeat every 3.

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

\(S = \{0, 3\}\), hence \( n(S) = 2 \).

The result is consistent with the given range (2,2).

Conclusion

Thus, the number of elements \( n(S) \) is equal to 2, and it falls within the specified range.

Answer 2

Given matrix:

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

Computing successive powers:

\[ A^2 = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix}, \quad A^3 = \begin{bmatrix} 4 & -3 \\ 3 & -2 \end{bmatrix}, \quad A^4 = \begin{bmatrix} 5 & -4 \\ 4 & -3 \end{bmatrix} \]

Continuing further:

\[ A^6 = \begin{bmatrix} 7 & -6 \\ 6 & -5 \end{bmatrix} \]

For general exponentiation:

\[ A^m = \begin{bmatrix} m+1 & -m \\ m & -m+1 \end{bmatrix} \]

For squared exponentiation:

\[ A^{m^2} = \begin{bmatrix} m^2 +1 & -m^2 \\ m^2 & -(m^2-1) \end{bmatrix} \]

Adding the two matrices:

\[ A^{m^2} + A^m = 3I - A^{-6} \]

\[ \begin{bmatrix} m^2+1 & -m^2 \\ m^2 & -(m^2-1) \end{bmatrix} + \begin{bmatrix} m+1 & -m \\ m & -m+1 \end{bmatrix} \]

Substituting values:

\[ = 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -5 & 6 \\ -6 & 7 \end{bmatrix} \]

\[ = \begin{bmatrix} 8 & -6 \\ 6 & -4 \end{bmatrix} \]

Equating to given conditions:

\[ m^2 + 1 + m + 1 = 8 \]

\[ m^2 + m -6 = 0 \Rightarrow m = -3, 2 \]

Thus, the number of solutions:

\[ n(s) = 2 \]