If A = &- & ,then A + A'=I, if the value of a is: 1. 6 2. 3 3. n … - Vidyadip

Question

If $\text{A} = \begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix},\text{then}\ \text{A + A}'=\text{I}$, if the value of a is:

$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\text{n}$
$\frac{3\pi}{2}$

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Answer

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

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