# Confirming whether a matrix is skew symmetric or not

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. $Non-singular$

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

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.

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 $det(f(\theta)) = 1$