Is it possible to construct a $3*3$ MDS matrix in $GF(2^4)$?
A matrix $M$ of order $n$ is an MDS (Maximum Distance Separable) matrix if and only if every sub-matrix of $M$ is non-singular. Therefore, if you have an $n\times n$ MDS matrix $M$, constructing the $(n-1)\times (n-1)$ MDS matrix $M'$ is so easy. $M'$ can be constructed by choosing an arbitrary $(n-1)\times (n-1)$ sub-matrix of $M$. Here you can find some information about constructing the MDS matrices over $GF(2^q)$.