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If $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$, then $A+A^{\prime}=I$, if the value of $\alpha$ is
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3}$
(c) $\pi$
(d) $\frac{3 \pi}{2}$
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3}$
(c) $\pi$
(d) $\frac{3 \pi}{2}$
Solution:
1288 Upvotes
Verified Answer
Now
$$
\begin{aligned}
&\mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \\
&\quad=\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\mathrm{I}=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
&\therefore \quad 2 \cos \alpha=1 \Rightarrow \cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}
\end{aligned}
$$
Thus option (b) is correct.
$$
\begin{aligned}
&\mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \\
&\quad=\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\mathrm{I}=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
&\therefore \quad 2 \cos \alpha=1 \Rightarrow \cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}
\end{aligned}
$$
Thus option (b) is correct.
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