If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, then $A + A' =…
If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, then $A + A' = I$, if the value of $a$ is:
$\frac{3\pi}{2}$
$n$
$\frac{\pi}{6}$
$\frac{\pi}{3}$
Solution
The correct answer is B.
$A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix}$
$\Rightarrow A'=\begin{bmatrix}\cos\alpha&\sin\alpha\\ -\sin\alpha&\cos\alpha \end{bmatrix}$
Now, $A + A'=I$
$\therefore \begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}+\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Comparing the corresponding elements of the two matrices, we have:
$2\cos\alpha=1$
$\Rightarrow \cos\alpha=\frac{1}{2}=\cos\frac{\pi}{3}$
$\therefore \alpha=\frac{\pi}{3}$