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$A_{(\alpha)}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right], A_{(\beta)}=\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
Which one of the following is correct?
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Which one of the following is correct?
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Verified Answer
The correct answer is:
$A_{(\alpha)}+A_{(\beta)}=A_{(\alpha+\beta)}$
As given $\mathrm{A}(\alpha)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
and $\quad A(\beta)=\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
Hence,
$\mathrm{A}(\alpha) \cdot \mathrm{A}(\beta)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
$=\left[\begin{array}{cc}\cos \alpha \cos \beta-\sin \alpha \sin \beta & -\cos \alpha \sin \beta-\sin \alpha \cos \beta \\ \sin \alpha \cos \beta+\cos \alpha \sin \beta & \sin \alpha \sin \beta+\cos \alpha \cos \beta\end{array}\right]$
$=\left|\begin{array}{cc}\cos (\alpha+\beta) & -\sin (\alpha+\beta) \\ \sin (\alpha+\beta) & \cos (\alpha+\beta)\end{array}\right|=\mathbf{A}_{(\alpha+\beta)}$
and $\quad A(\beta)=\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
Hence,
$\mathrm{A}(\alpha) \cdot \mathrm{A}(\beta)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
$=\left[\begin{array}{cc}\cos \alpha \cos \beta-\sin \alpha \sin \beta & -\cos \alpha \sin \beta-\sin \alpha \cos \beta \\ \sin \alpha \cos \beta+\cos \alpha \sin \beta & \sin \alpha \sin \beta+\cos \alpha \cos \beta\end{array}\right]$
$=\left|\begin{array}{cc}\cos (\alpha+\beta) & -\sin (\alpha+\beta) \\ \sin (\alpha+\beta) & \cos (\alpha+\beta)\end{array}\right|=\mathbf{A}_{(\alpha+\beta)}$
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