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Mathematics Question on Matrices

If A=[cosα−sinα sinαcosα]A=\begin{bmatrix}cosα& -sinα\\\ sinα& cosα\end{bmatrix},thenA+A=I,if the value of αα is

A

π6\frac{π}{6}

B

π3\frac{π}{3}

C

π

D

3π2\frac{3π}{2}

Answer

π3\frac{π}{3}

Explanation

Solution

The correct option is(B): π3\frac{π}{3}

A=[cosα−sinα sinαcosα]A=\begin{bmatrix}cosα& -sinα\\\ sinα& cosα\end{bmatrix}
A′=[cosαsinα −sinαcosα]A'=\begin{bmatrix}cosα& sinα\\\ -sinα& cosα\end{bmatrix}

Now A+A=I
[cosα−sinα sinαcosα]+[cosαsinα −sinαcosα]=[10\01]\begin{bmatrix}cosα& -sinα\\\ sinα& cosα\end{bmatrix}+\begin{bmatrix}cosα& sinα\\\ -sinα& cosα\end{bmatrix}=\begin{bmatrix}1&0\\\0&1\end{bmatrix}
=>[2cosα0 02cosα]=[10\01]=>\begin{bmatrix}2cosα& 0\\\ 0& 2cosα\end{bmatrix}=\begin{bmatrix}1&0\\\0&1\end{bmatrix}

Comparing the corresponding elements of the two matrices, we have:
2cosα=1
=>cosa=12=cosπ3cos a=\frac{1}{2}=cos\frac{\pi}{3}
Therefore α=π3α=\frac{π}{3}