Build a binary communication or dominance matrix, square it to count two-step links, add the matrix and its square to combine one-step and two-step connections, and rank competitors by their total

What this dot point is asking

VCAA wants you to use a matrix of zeros and ones to model direct links between people, teams or objects. A communication matrix records who can pass a message to whom; a dominance matrix records which competitor beats which. Squaring the matrix counts the two-step (indirect) links, and adding the matrix to its square combines one-step and two-step connections so you can rank the entries. This is the matrix module's link to networks.

Building the binary matrix

Suppose four teams play each other and the results are: A beats B, B beats C, C beats A, and A beats C. Reading "row beats column", the dominance matrix is

for teams A, B, C. A $1$ in row A, column B means A beats B. The diagonal is zero. A communication matrix is built the same way, with $1$ meaning a direct channel from the row to the column.

Two-step links from the square

Direct wins alone often leave teams tied. Squaring the matrix, $M^2$, counts the two-step connections: in dominance terms, how many teams you beat that in turn beat a third team. These indirect wins break ties.

Communication interpretation

For a communication matrix the same arithmetic answers a different question: an entry of $M^2$ gives the number of two-step paths (relayed messages) from one person to another. Adding $M + M^2$ gives the total number of ways a message can reach someone in at most two steps.

Why squaring counts two-step links

The reason $M^2$ counts two-step connections comes straight from matrix multiplication. The entry in row $i$, column $j$ of $M^2$ is the sum $\sum_k M_{ik}M_{kj}$. Each term in that sum is $1$ only when both $M_{ik} = 1$ and $M_{kj} = 1$, that is when $i$ links to some intermediate $k$ and $k$ links to $j$. So the entry counts exactly the number of two-step paths from $i$ to $j$ through any single intermediary. The same logic extends to $M^3$ for three-step links, although VCE questions stop at two steps. Understanding this is what lets you interpret a particular entry of $M^2$ in words, which is a common short-answer task.

Choosing how many steps to combine

The standard VCE ranking uses $M + M^2$, which credits a competitor for both whom they beat directly and whom those teams beat in turn. Direct wins are worth one and two-step wins are also counted once each, so a team that beats strong opponents is rewarded over one that beats weak opponents with the same number of direct wins. Occasionally a question gives weights or asks only for one-step or only for two-step totals; read carefully which matrix the ranking is built from. The reasoning mark is for stating that $M + M^2$ combines direct and indirect dominance, so always name what the totals represent rather than just listing numbers.

Reading the diagonal and the totals

The leading diagonal of $M$ is always zero, since no competitor beats itself, but the diagonal of $M^2$ need not be zero: a non-zero diagonal entry signals a cycle of dominance (for instance $i$ beats $j$ and $j$ beats $i$ across two steps), which is a sign that the results are not perfectly ordered. When ranking, you total each full row of $M + M^2$, diagonal included, and the largest row total is the strongest competitor. Stating the ranking from strongest to weakest with the supporting totals, and noting any ties broken by the two-step contribution, is exactly the structure markers expect.

Why this matters for the exams

Communication and dominance questions test careful matrix reading and the meaning of $M^2$ as a two-step count. Markers reward the correct binary matrix, an accurate square, and a ranking based on $M + M^2$ with a one-line justification. The technique reuses matrix multiplication and connects the Matrices module to the path-counting ideas in Networks.