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If $E(\theta)=\begin{bmatrix}\cos^2\theta&\cos \theta\sin \theta\\\cos \theta\sin\theta&\sin^2\theta\end{bmatrix}$ and $\theta$ and $\phi$ differ by an odd multiple of $\large\frac{\pi}{2}$ then $E(\theta)E(\phi)$ is a

$(a)\;unit\;vector\qquad(b)\;null\;matrix\qquad(c)\;diagonal\;matrix\qquad(d)\;None\;of\;these$

1 Answer

Given :
$E(\theta)=\begin{bmatrix}\cos^2\theta&\cos \theta\sin \theta\\\cos \theta&\sin^2\theta\end{bmatrix}$
$E(\theta)E(\phi)=\begin{bmatrix}\cos^2\theta&\cos \theta\sin \theta\\\cos \theta\sin \theta&\sin ^2\theta\end{bmatrix}\begin{bmatrix}\cos^2\phi&\cos \phi\sin \phi\\\cos \phi\sin \phi&\sin ^2\phi\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\cos ^2\theta\cos^2\phi+\cos \theta\sin\theta\cos \phi\sin \phi&\cos^2\theta\cos \phi+\cos \phi\sin \theta\sin ^2\phi\\\cos \theta\sin \theta\cos^2\phi+\sin^2\theta\cos \phi\sin \phi&\cos \theta\sin \theta\cos \phi\sin \phi+\sin ^2\theta+\sin ^2\phi\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\cos \theta\cos \phi\cos(\theta-\phi)&\cos \theta\sin \phi\cos(\theta-\phi)\\\cos \phi\sin \theta\cos(\theta-\phi)&\sin \theta\sin\phi\cos(\theta-\phi)\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\cos \theta\cos\phi\cos(2n+1)\large\frac{\pi}{2}&\cos \theta\sin \phi\cos(2n+1)\large\frac{\pi}{2}\\\cos\phi\sin \theta\cos(2n+1)\large\frac{\pi}{2}&\sin \theta\sin \phi\cos(2n+1)\large\frac{\pi}{2}\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\therefore \cos(2n+1)\large\frac{\pi}{2}$$=0$
Hence (b) is the correct answer.
answered Nov 25, 2013 by sreemathi.v
 

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