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Mathematics Question on Matrices

An orthogonal matrix is

A

[cos⁡ α2 sin⁡ α\[0.3em]−2 sin⁡ αcos⁡ α]\begin{bmatrix} \cos\,\alpha & 2\,\sin\,\alpha \\\[0.3em] -2\,\sin\,\alpha & \cos\,\alpha \end{bmatrix}

B

[cos⁡ αsin⁡ α\[0.3em]−sin⁡ αcos⁡ α]\begin{bmatrix} \cos\,\alpha & \sin\,\alpha \\\[0.3em] -\sin\,\alpha & \cos\,\alpha \end{bmatrix}

C

[cos⁡ αsin⁡ α\[0.3em]sin⁡ αcos⁡ α]\begin{bmatrix} \cos\,\alpha & \sin\,\alpha \\\[0.3em] \sin\,\alpha & \cos\,\alpha \end{bmatrix}

D

[11\[0.3em]11]\begin{bmatrix} 1& 1\\\[0.3em] 1& 1\end{bmatrix}

Answer

[cos⁡ αsin⁡ α\[0.3em]−sin⁡ αcos⁡ α]\begin{bmatrix} \cos\,\alpha & \sin\,\alpha \\\[0.3em] -\sin\,\alpha & \cos\,\alpha \end{bmatrix}

Explanation

Solution

Let A = [cos⁡ α2 sin⁡ α\[0.3em]−2 sin⁡ αcos⁡ α]\begin{bmatrix} \cos\,\alpha & 2\,\sin\,\alpha \\\[0.3em] -2\,\sin\,\alpha & \cos\,\alpha \end{bmatrix} ,then A' = [cos⁡ α−sin⁡ α\[0.3em]sin⁡ αcos⁡ α]\begin{bmatrix} \cos\,\alpha & -\sin\,\alpha \\\[0.3em] \sin\,\alpha & \cos\,\alpha \end{bmatrix} ∴\therefore AA' = [cos⁡ α−sin⁡ α\[0.3em]−sin⁡ αcos⁡ α][cos⁡ α−sin⁡ α\[0.3em]sin⁡ αcos⁡ α]\begin{bmatrix} \cos\,\alpha & -\sin\,\alpha \\\[0.3em] -\sin\,\alpha & \cos\,\alpha \end{bmatrix} \begin{bmatrix} \cos\,\alpha & -\sin\,\alpha \\\[0.3em] \sin\,\alpha & \cos\,\alpha \end{bmatrix} = [cos2 α+sin2 α−cos α sin α +sin α cos α\[0.3em]−sin α cos α+cos α sin αsin2 α+cos2 α]\begin{bmatrix} cos^2\,\alpha + sin^2 \, \alpha & -cos \, \alpha \,sin\,\alpha\,+sin\,\alpha\,cos\,\alpha \\\[0.3em] -sin\,\alpha \, cos \, \alpha + cos \, \alpha \, sin \, \alpha &sin^2 \, \alpha + cos^2 \, \alpha \end{bmatrix} = [10 01]\begin{bmatrix}1&0\\\ 0&1\end{bmatrix} = I