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If \( \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \), then \( A A^{\prime}= \)
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Verified Answer
The correct answer is:
\( 11 \)
Given that \( A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \)
So, \( A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
Then,
\[
\begin{array}{l}
A \cdot A=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
=\left[\begin{array}{cc}
\cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\
-\cos \alpha \sin \alpha+\cos \alpha \sin \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I
\end{array}
\]
So, \( A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
Then,
\[
\begin{array}{l}
A \cdot A=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
=\left[\begin{array}{cc}
\cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\
-\cos \alpha \sin \alpha+\cos \alpha \sin \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I
\end{array}
\]
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