\begin{array}{c|c}  \begin{array}{ccc} 0&.&. \\ |&.&. \\
|&0&- \end{array}  &  \begin{array}{ccc} .&|&. \\ .&0&.
\\ \!-\!&-&0 \end{array} \\ \hline  \begin{array}{ccc} |&.&.
\\ 0&-&- \\ |&.&. \end{array}  &  \begin{array}{ccc}
.&|&. \\ 0&|&- \\ .&0&. \end{array}  \end{array} \right)$$

Here, the outer matrix corresponds to the measurement choice ($x$ vertically, $y$ horizontally), and the inner submatrices correspond to the outcomes ($a$ vertically, $b$ horizontally). By comparing pairs of product states as above, deduce that each line of zeroes can be extended perpendicularly, yielding a row and column of zeroes in each submatrix: $$\left( \begin{array}{c|c}  \begin{array}{ccc} 0&0&0 \\ .&0&. \\
.&0&. \end{array} &  \begin{array}{ccc} .&.&0 \\ 0&0&0
\\ .&.&0 \end{array} \\ \hline \begin{array}{ccc} 0&.&. \\
0&0&0 \\ 0&.&. \end{array}  &  \begin{array}{ccc} 0&.&.
\\ 0&.&. \\ 0&0&0 \end{array}  \end{array} \right)$$

It then follows that all of the unknown elements must also be zero. For example, in order to show that $M(1 0| 0 0)=0$ (the element below the top left element), consider the states ${\mathbf{P}}$ and ${\mathbf{P}}'$ as follows (non-zero elements of ${\mathbf{P}}$ are shown boxed and non-zero elements of ${\mathbf{P}}'$ are shown in bold): $$\left( \begin{array}{c|c}  \begin{array}{ccc} \textbf{1}&0&0 \\
\fbox{1}&0&0 \\ 0&0&0 \end{array}  &
\begin{array}{ccc} 0&0&\textbf{1} \\ 0&0&\fbox{1}  \\ 0&0&0
\end{array} \\ \hline  \begin{array}{ccc}
\fbox{\textbf{1}}&0&0 