Let $ \alpha $ be a root of the equation $ x^{2}+x+1=0 $ and the matrix $ A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \ 1 & \alpha & \alpha^{2} \ 1 & \alpha^{2} & \alpha^{4}\end{array}\right] $, then the matrix $ A^{31} $ is equal to

Question

Let $ \alpha $ be a root of the equation $ x^{2}+x+1=0 $ and the matrix $ A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \ 1 & \alpha & \alpha^{2} \ 1 & \alpha^{2} & \alpha^{4}\end{array}\right] $, then the matrix $ A^{31} $ is equal to, $ A^{3} $, $ I_{3} $, $ A^{2} $, $ A $

MARKS App · Accepted Answer

Here, α = ω , ω 2 where ω is complex cube root of unity. A 2 = 1 3 1 1 1 1 ω ω 2 1 ω 2 ω 1 1 1 1 ω ω 2 1 ω 2 ω = 1 0 0 0 0 1 0 1 0 ⇒ A 4 = 1 0 0 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 = 1 0 0 0 1 0 0 0 1 ⇒ A 31 = A 28 × A 3 = I × A 3 = A 3

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