Q.If $A = \begin{pmatrix} 1 & 0 & 0 \ 0 & \cos \theta & \sin \theta \ 0 & -\sin \theta & \cos \theta \end{pmatrix}$ , then $|A^{50}|$ equals

\cos 50\theta

\sin 50\theta

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.If A=(100 0cos⁡θsin⁡θ 0−sin⁡θcos⁡θ)A = \begin{pmatrix} 1 & 0 & 0 \ 0 & \cos \theta & \sin \theta \ 0 & -\sin \theta & \cos \theta \end{pmatrix}A=(1​0​0 0​cosθ​sinθ 0​−sinθ​cosθ​), then ∣A50∣|A^{50}|∣A50∣ equals