$\beta \in {{\mathbb{F}}}_q$ such that the requirements for the extension code are satisfied.

For this proof we form a $(q+1) \times (q+1)$ matrix ${{\cal M}}$ whose first row consists of the columns of the matrix $M_1$ in their given order. The other rows are indexed by the elements of ${{\mathbb{F}}}_q$. The row which are indexed by $\beta \in {{\mathbb{F}}}_q$ has ${\tiny \left( \hspace{-0.1cm} \begin{array}{c} 0 \\ 1 \end{array} \hspace{-0.1cm} \right)}$ in the first entry and ${\tiny \left( \hspace{-0.1cm} \begin{array}{c} 1 \\ \beta \end{array} \hspace{-0.1cm} \right)}$ in the second entry. The $i$-th entry, $3 \leq i \leq q+1$, will be ${\tiny \left( \hspace{-0.1cm} \begin{array}{c} 1 \\ \alpha^{i-3} + \beta \end{array} \hspace{-0.1cm} \right)}$. It is easy to verify that the $q \times q$ sub-matrix ${{\cal M}}'$ of ${{\cal M}}$ defined by removing the first row and first column of ${{\cal M}}$ is a Latin square (each row and each column is a permutation of the $q$ column vectors ${\tiny \left( \hspace{-0.1cm} \begin{array}{c} 1 \\ \beta \end{array} \hspace{-0.1cm} \right)}$, $\beta \in {{\mathbb{F}}}_q$). For each $i$, $3 \leq i \leq q+1$, the element in the $i$-th entry of the first row of ${{\cal M}}$ appears in the linear span of the $i$-th entry of 