Let , where . Then n(S) is equal to __________. [2025]

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Solution

Q.
Let $S = {m \in Z : A^{m^{2}} + A^{m} = 3 I - A^{- 6}}$ , where $A = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}]$ . Then n(S) is equal to __________. [2025]

Ans.
(2)

$A = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}], A^{2} = [\begin{matrix} 3 & - 2 \\ 2 & - 1 \end{matrix}], A^{3} = [\begin{matrix} 4 & - 3 \\ 3 & - 2 \end{matrix}]$ ,

$A^{4} = [\begin{matrix} 5 & - 4 \\ 4 & - 3 \end{matrix}]$ and so on $A^{6} = [\begin{matrix} 7 & - 6 \\ 6 & - 5 \end{matrix}]$

$\Rightarrow A^{m} = [\begin{matrix} m + 1 & - m \\ m & - m + 1 \end{matrix}] and A^{m^{2}} = [\begin{matrix} m^{2} + 1 & - m^{2} \\ m^{2} & - (m^{2} - 1) \end{matrix}]$

Now, $A^{m^{2}} + A^{m} = 3 I - A^{- 6}$

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

$= 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} - 5 & 6 \\ - 6 & 7 \end{matrix}] = [\begin{matrix} 8 & - 6 \\ 6 & - 4 \end{matrix}]$

$\Rightarrow m^{2} + 1 + m + 1 \Rightarrow 8 = m^{2} + m - 6 = 0 \Rightarrow m = - 3, 2$

$∴ n (S) = 2$ .

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