JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(\mathrm{A}^{31}\) is equal to
- A \(A^3\)
- B \(A\)
- C \(A^2\)
- D \(I_3\)
Answer & Solution
Correct Answer
(A) \(A^3\)
Step-by-step Solution
Detailed explanation
\(x^{2}+x+1=0\) \(\alpha=\omega\) \(\alpha^{2}=\omega^{2}\) \(A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {\omega} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega}\end{array}\right]\)…
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