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 ______.

Hint:

When solving problems with matrix powers and sets, first compute the necessary powers of matrices, then use the given conditions to find the valid elements in the set.

Answer 1

Correct Answer: 20

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:

\[ n(s) = 2 \]