(b)
Let $A = \lbrack a_{ij}\rbrack$. Since A is skew-symmetric, therefore
$a_{ij} = 0$ and $a_{ij} = - a_{ij}(i \neq j)$
$A$ is symmetric as well, so $a_{ij} = a_{ji}$ for all $i$ and $j$
$\therefore a_{ij} = 0\$for all$\ i \neq j$
Hence, $a_{ij} = 0$ for all $i$ and $j$, i.e., $A$ is null matrix.