Given that $A$ is a non-zero skew symmetric even order matrix.Then which od following is true
$1).$$Ax=0$ has infinitely many solutions
$2).$$Ax=\lambda x$ has solution for every non-zero $\lambda \in \mathbb{R}$
$3).$$Ax=\lambda x$ has unique solution for all $\lambda \in \mathbb{R^{*}}$
$4).$ $Ax=\lambda x$ may not have unique solution for some $\lambda \in \mathbb{R}$
The solution I Tried-Given matrix is of even order so by property of skew-symmetric matrix it may or may not have determinent zero.so option 1 is discarded
Rest of options are making me confuse because in these options $\lambda $ is not eigenvalues and $x$ is not eigenfunction,How to think further Please help
Thank you