This paper says that, each quasigroup of order 4 can be represented in matrix form using the following equation, \begin{equation} x \ast y \equiv m^T +Ax^T +By^T +CA\cdot x^T \circ CB\cdot y^T \end{equation} where, $A = \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}, \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}$ are non-singluar Boolean matrices and $m = [m_1, m_2]$ is a Boolean vector. Note that we also need to consider the boolean representation of the elements $x$ and $y$ for the equation to make sense. Now, since we are dealing with boolean representations of the elements, I interpreted all the operations in the above equation as boolean operations '+', '.' and $'\circ'$ is a dot product which would mean operating the corresponding elements of the matrices using '$\cdot$'. Again, as per this, any arbitrary non-singular matrices A, B, a boolean vector m and some particular C, we obtain corresponding quasigroups of order 4. As an experiment, we take, $m=[0, 0]$, $A=\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}, B=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} $. $C= \begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$ (this matrix has been specified in the paper to obtain quadratic quasigroups of order 4). Now we wish to generate a Latin square for the quasigroup $Q$ having elements, $\{0, 1, 2, 3\}$. Using the equation and the boolean representations of the elements, $(0 \equiv 00, 1 \equiv 01, 2 \equiv 10, 3 \equiv 11)$ we perform the following operations: $0 \ast 0 = (0, 0) \ast (0, 0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0\\ 0 \end{bmatrix} + \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0\\ 0 \end{bmatrix} + \begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 0 \end{bmatrix} \circ \begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 0 \end{bmatrix} = [0 \; 0]^T$ $0 \ast 1 = (0, 0) \ast (0, 1) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0\\ 0 \end{bmatrix} + \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix} + \begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 0 \end{bmatrix} \circ \begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 1 \end{bmatrix} = [0\; 1]^T$
Similarly, $0 \ast 2 = [1 \;0]^T and, 0 \ast 3 = [1 \;1]^T.$ Next,
$1 \ast 0 = [1 \; 0]^T$;
$1 \ast 1 = [1 \; 1]^T$;
$1 \ast 2 = [1 \; 1]^T$
We know that no element can be repeated within a row or a column of a Latin square. But in the example above, the row corresponding to the element 1 will have repeated elements because, $1 \ast 1 = 1 \ast 2$. Could there thus be some ambiguity in my interpretation or the implementation of the matrix operations? Please help me identify my mistake here.