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Given $f(\theta)=\left[\begin{matrix}cos^2\theta&cos\theta sin\theta&-sin\theta\\cos\theta sin\theta&sin^2\theta&cos\theta\\sin\theta&-cos\theta&0\end{matrix}\right]$ Then $f(\dfrac{\pi}{7})=?$

Options given :

  1. $Symmetric$

  2. $skew-symmetric$

  3. $Singular$

  4. $Non-singular$

Now I have already tested

$\begin{vmatrix} cos^2\theta & cos\theta sin\theta & -sin\theta \\ cos\theta sin\theta & sin^2\theta & cos\theta \\ sin\theta & -cos\theta & 0 \\ \notag \end{vmatrix}=1$

Hence option c $\implies$ singular ruled out.

Now in order to be skew-symmetric we must have $A_{ij}=-A_{ji}$

In the above example that $f(\dfrac{\pi}{7})$ is most probably a useless diversion and more importantly that $\dfrac{\pi}{7}$ lies in the first quadrant. Hence both $cos\theta \space and \space sin\theta$ are positive there on that value.

Thus we are testing the original matrix $f(\theta)$

Here we see that all matrix elements obey the skew symmetric principle except(which is my slight nonsensical childish idiotic(I am an Idiot anyway) doubt anyway) and that element is $A_{12}=A_{21}=cos\theta sin\theta$. Now I know that $cos-\theta=cos\theta$. So when I apply the skew symmetric property for testing I should check for the element $A_{ij}$ right? In this case is it applied on that $\theta$. I am a bit confused. It is most likely that answer to this question is that it is option 4-> non-singular.

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You have , $f(\theta)=\left[\begin{matrix}cos^2\theta&cos\theta sin\theta&-sin\theta\\cos\theta sin\theta&sin^2\theta&cos\theta\\sin\theta&-cos\theta&0\end{matrix}\right]$

the transpose of $f(\theta)$ is ;

$f^T(\theta) =\begin{bmatrix}\cos^2(\theta)&&\cos(\theta)\sin(\theta)&&\sin(\theta)\\\cos(\theta)\sin(\theta)&&\sin^2(\theta)&&-\cos(\theta)\\-\sin(\theta)&&\cos(\theta)&&0\end{bmatrix} \ne -f(\theta)$

The transpose in also not equal to $f(\theta)$ hence it is neither symmetric nor skew symmetric . So option (4)is the correct answer as the matrix is non singular with $det(f(\theta)) = 1$

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  • $\begingroup$ So @pranavB23 I guess I was right from the beginning - only that my rights were not confirmed about me being right. ! $\endgroup$ – Saradamani Apr 15 '18 at 15:00
  • $\begingroup$ Yeah , you need to start believing in yourself more i think.:) $\endgroup$ – The Integrator Apr 15 '18 at 15:02

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