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AP EAMCET · Maths · Matrices

Let \(\omega\) be a complex cube root of unity with \(\omega \neq 1\) and \(P=\left[p_{i j}\right]\) be a \(2 \times 2\) matrix with \(p_{i j}=\omega^{i+j}\). For \(P^2 \neq 0\) if \(P^k=P\), then \(k\) is equal to

  1. A 57
  2. B 54
  3. C 58
  4. D 56
Verified Solution

Answer & Solution

Correct Answer

(A) 57

Step-by-step Solution

Detailed explanation

\(P=\begin{pmatrix} \omega^2 & \omega^3 \\ \omega^3 & \omega^4 \end{pmatrix}=\begin{pmatrix} \omega^2 & 1 \\ 1 & \omega \end{pmatrix}\)…
From AP EAMCET
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