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

$$\begin{array}{1 1} (A) \; \frac{\pi}{6} & (B) \; \frac{\pi}{3} \\ (C) \; \pi & (D) \; \frac{3\pi}{2} \end{array}$$

Toolbox:
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
Given:
$A = \begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$
Interchange the rows into column.
$A' = \begin{bmatrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{bmatrix}$
$A+A'= \begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix} +\begin{bmatrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 2cos\alpha &- sin\alpha+\sin\alpha \\ sin\alpha-sin\alpha & cos\alpha+cos\alpha \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 2cos\alpha& 0 \\ 0 & 2cos\alpha \end{bmatrix}=I$
$\begin{bmatrix} 2cos\alpha& 0 \\ 0 & 2cos\alpha \end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\Rightarrow 2cos\alpha=1,cos\alpha=\frac{1}{2}.$
$\alpha=\frac{\pi}{3}$
so the correct option is B
answered Mar 15, 2013