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 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.