If A = 1 2 [ 1 3 - 3 1 ] , then [2023]

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Solution

Q.
If $A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}],$ then [2023]

1 $A^{30} - A^{25} = 2 I$

2 $A^{30} = A^{25}$

3 $A^{30} + A^{25} - A = I$

4 $A^{30} + A^{25} + A = I$

Ans.
(3)

$A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}] \Rightarrow A = [\begin{matrix} \cos 60 ° & \sin 60 ° \\ - \sin 60 ° & \cos 60 ° \end{matrix}] \begin{matrix} \end{matrix}$

Let $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}], where α = \frac{π}{3}$

$A^{2} = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] \begin{matrix} \end{matrix}$

$= [\begin{matrix} \cos 2 α & \sin 2 α \\ - \sin 2 α & \cos 2 α \end{matrix}] \begin{matrix} \end{matrix}$

$A^{30} = [\begin{matrix} \cos 30 α & \sin 30 α \\ - \sin 30 α & \cos 30 α \end{matrix}]; A^{30} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$

$A^{25} = [\begin{matrix} \cos 25 α & \sin 25 α \\ - \sin 25 α & \cos 25 α \end{matrix}] = [\begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{- \sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$

$\Rightarrow A^{25} = A \Rightarrow A^{25} - A = 0$

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