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NDA 2024 Maths · Mathematics · Matrices · medium

If A=beginpmatrixcos θ&sin θ\\-sin θ&cos θendpmatrix with θ=(π)/(6), then A^6 equals

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If A=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)A=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} with θ=π6\theta=\dfrac\pi6, then A6A^6 equals
  1. A. (1001)\begin{pmatrix}1&0\\0&1\end{pmatrix}
  2. B. (01−10)\begin{pmatrix}0&1\\-1&0\end{pmatrix}
  3. C. (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}
  4. D. (−100−1)\begin{pmatrix}-1&0\\0&-1\end{pmatrix}
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Answer: (D) (−100−1)\begin{pmatrix}-1&0\\0&-1\end{pmatrix}

An=(cos⁡nθsin⁡nθ−sin⁡nθcos⁡nθ)A^n=\begin{pmatrix}\cos n\theta&\sin n\theta\\-\sin n\theta&\cos n\theta\end{pmatrix}. For n=6n=6, nθ=πn\theta=\pi, so cos⁡π=−1\cos\pi=-1 and sin⁡π=0\sin\pi=0 give −I-I. Trap: 2π2\pi would give II.
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