If A= [ cc & - & ], then A+A^ =I, if the value of is - Vidyadip

MCQ

If $\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right],$ then $\mathrm{A}+\mathrm{A}^{\prime}=\mathrm{I},$ if the value of $\alpha$ is

A
$\frac{\pi}{6}$
B
$\frac{3\pi}{2}$
C
${\pi}$
✓
$\frac{\pi}{3}$

✓

Answer

Correct option: D.

$\frac{\pi}{3}$

d
$A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$\Rightarrow A^{\prime}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

Now $A+A^{\prime}=1$

$\therefore $ $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]+\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow $ $\left[\begin{array}{cc}2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Comparing the corresponding elements of the two matrices, we have :

$\cos \alpha=\frac{1}{2}$

$\alpha=\cos ^{-1}\left(\frac{1}{2}\right)$

$\therefore $ $\alpha=\frac{\pi}{3}$

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