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 ______.
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 ______.
Correct Answer: 2
Approach Solution - 1
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.
Approach Solution -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:
\[ n(s) = 2 \]
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