Question

Let A =$\left[\begin{array}{cc}co{s}^2\theta & sin\theta cos\theta \\ cos\theta sin\theta & si{n}^2\theta \end{array}\right]$ and $B=\left[\begin{array}{cc}co{s}^2\varphi & sin\varphi cos\varphi \\ cos\varphi sin\varphi & si{n}^2\varphi \end{array}\right]$,then $AB=0$,if

• A. $\theta =n\varphi ,n=0,1,2,....$
• B. $\theta +\varphi =n\pi ,n=0,1,2,....$
• C. $\theta =\varphi +(2n+1)\frac{\pi }{2},n=0,1,2,....$
• D. $\theta =\varphi +\frac{n\pi }{2},n=0,1,2,....$

Answer 1

Answer: (C)

AB = $\left[\begin{array}{cc}co{s}^2\theta & sin\theta cos\theta \\ cos\theta sin\theta & si{n}^2\theta \end{array}\right]\left[\begin{array}{cc}co{s}^2\varphi & sin\varphi cos\varphi \\ cos\varphi sin\varphi & si{n}^2\varphi \end{array}\right]$
.$=\left[\begin{array}{cc}co{s}^2\theta co{s}^2\varphi +sin\theta cos\varphi cos\theta sin\varphi & co{s}^2\theta sin\varphi cos\varphi +si{n}^2\varphi sin\theta cos\theta \\ co{s}^2\varphi cos\theta sin\theta +si{n}^2\theta sin\varphi cos\varphi & cos\theta sin\theta sin\varphi cos\varphi +si{n}^2\theta si{n}^2\varphi \end{array}\right]$
$=\left[\begin{array}{cc}cos\theta cos\varphi cos\left(\theta -\varphi \right)& sin\varphi cos\theta cos\left(\theta +\varphi \right)\\ sin\theta cos\varphi cos\left(\theta +\varphi \right)& sin\theta sin\varphi cos\left(\theta -\varphi \right)\end{array}\right]$
$\therefore AB=0$
$\Rightarrow cos(\theta -\varphi )=0$
$\Rightarrow cos(\theta -\varphi )=cos(2n+1)\frac{\pi }{2}$
$\Rightarrow \theta =(2n+1)\frac{\pi }{2}+\varphi$ ,where $n=0,1,2,......$