No. Let $e_i$ denotes the vector with a $1$ at the $i$-th position and zeros elsewhere. If $A$ is not symmetric, then $a_{ij}\
e a_{ji}$ for some $i\
e j$, but then $e_i^TAe_j=a_{ij}\
e a_{ji}=e_j^TAe_i$.
No. Let $e_i$ denotes the vector with a $1$ at the $i$-th position and zeros elsewhere. If $A$ is not symmetric, then $a_{ij}\
e a_{ji}$ for some $i\
e j$, but then $e_i^TAe_j=a_{ij}\
e a_{ji}=e_j^TAe_i$.