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
- A 57
- B 54
- C 58
- D 56
Answer & Solution
Correct Answer
(A) 57
Step-by-step Solution
Detailed explanation
\(P=\begin{pmatrix} \omega^2 & \omega^3 \\ \omega^3 & \omega^4 \end{pmatrix}=\begin{pmatrix} \omega^2 & 1 \\ 1 & \omega \end{pmatrix}\)…
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