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JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant

यदि समीकरण \(x^{2}+x+1=0\) का एक मूल \(\alpha\) है तथा आव्यूह \(A =\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \alpha^{2} & \alpha^{4}\end{array}\right]\) है, तो आव्यूह \(A ^{31}\) बराबर है

  1. A \(A^3\)
  2. B \(A\)
  3. C \(A^2\)
  4. D \(I_3\)
Verified Solution

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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