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

Answer

Given $A=\left[\begin{array}{cc} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]$ 
Therefore, $A^{\prime}=\left[\begin{array}{cc} {\cos \alpha} & {\sin \alpha} \\ {-\sin \alpha} & {\cos \alpha} \end{array}\right]$ 
Also given that $A + A' = I    ...(1)$
$($Putting the values in equation $(1))$
$\left[\begin{array}{cc} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]+\left[\begin{array}{cc} {\cos \alpha} & {\sin \alpha} \\ {-\sin \alpha} & {\cos \alpha} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$ 
$\Rightarrow$ $\left[\begin{array}{cc} {\cos \alpha+\cos \alpha} & {-\sin \alpha+\sin \alpha} \\ {\sin \alpha-\sin \alpha} & {\cos \alpha+\cos \alpha} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$ 
$\Rightarrow$ $\left[\begin{array}{cc} {2 \cos \alpha} & {0} \\ {0} & {2 \cos \alpha} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$ 
We know the two matrices are equal only when all their corresponding elements or entries are equal i.e. if A$ = B,$ then $a_{ij}$ and $b_{ij}$ for all $i$ and $j.$
This implies,
$2 \cos \alpha=1$ 
$\Rightarrow$ $\cos \alpha=\frac{1}{2}$ 
$\Rightarrow$ $\cos \alpha=\cos \frac{\pi}{3}~~~ ...\left(\because \cos \frac{\pi}{3}=\frac{1}{2}\right)$ 
$\Rightarrow$ $\alpha=\frac{\pi}{3}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free