Dominance matrices
Study dominance matrices for Year 12 Specialist Mathematics in Queensland (QCAA). A dominance matrix turns a round-robin tournament, food web or preference network into a simple grid of ones and zeros that records exactly who beats whom.
You will learn to build the matrix from results or a directed graph, read one-step dominance scores from the row sums, find two-step dominances from \(M^2\), and rank the competitors with the combined matrix \(M+M^2\) — a key application of matrices in Unit 3.
Theory
A dominance matrix records "who beats whom" in a round-robin or network for Year 12 Specialist Mathematics (QCAA, Queensland). Its row sums give the one-step dominance scores, the square \(M^2\) gives the two-step dominances, and the combined matrix \(M+M^2\) ranks the competitors — separating any left tied on one-step wins alone.
A dominance matrix \(M\) models a set of competitors (teams, players, species) in which every pair has a clear winner — a round-robin tournament, a food web, or a preference network. If the rows and columns are listed in the same order, the entry \(m_{ij}=1\) when competitor \(i\) dominates (beats/preys on) competitor \(j\), and \(m_{ij}=0\) otherwise. The diagonal is always zero because no competitor dominates itself.
The one-step dominance score of a competitor is the sum of its row in \(M\): it counts the competitors it beats directly. Ranking on one-step scores alone often leaves ties, because it ignores who each competitor beat.
A two-step dominance is an indirect win: \(i\) beats some \(k\) who beats \(j\). The number of such two-step paths from \(i\) to \(j\) is the entry \((M^2)_{ij}\) of the matrix square. Row sums of \(M^2\) are the two-step dominance scores.
The combined dominance matrix \(M+M^2\) adds direct and indirect influence. Its row sums give each competitor a combined score, and ordering them from largest to smallest produces a ranking that usually breaks the ties left by \(M\) on its own. The syllabus keeps to \(M\) and \(M^2\).
For competitors listed in a fixed order, the dominance matrix has entries
The one-step dominance score of competitor \(i\) is its row sum, and the two-step score uses the matrix square \(M^2\):
The combined dominance matrix and the score used to rank the competitors are
Ranking with a dominance matrix
- Build \(M\): fix the order of competitors, then put \(m_{ij}=1\) where \(i\) beats \(j\), \(0\) elsewhere, and \(0\) on the diagonal (read it straight from a results table or a digraph).
- One-step scores: add each row of \(M\) to count direct wins.
- Two-step matrix: compute \(M^2\) (by hand or with technology); its row sums are the two-step scores, and \((M^2)_{ij}\) counts the indirect wins of \(i\) over \(j\).
- Combine and rank: form \(M+M^2\), add each row, and order the totals from largest to smallest to rank the competitors.
Put a 1 in row \(i\), column \(j\) for each win of \(i\) over \(j\); all other off-diagonal entries are 0:
| \(m_{NO}\) | \(=\) | \(1 \quad(\text{Nina beat Omar})\) |
| \(m_{NP}\) | \(=\) | \(1 \quad(\text{Nina beat Pia})\) |
| \(m_{PO}\) | \(=\) | \(1 \quad(\text{Pia beat Omar})\) |
| \(M\) | \(=\) | \(\begin{pmatrix}0&1&1\\0&0&0\\0&1&0\end{pmatrix}\) |
\(M=\begin{pmatrix}0&1&1\\0&0&0\\0&1&0\end{pmatrix}\) (row sums \(2,0,1\)).
The one-step score is the row sum of \(M\) (the direct wins):
| \(\text{score}(A)\) | \(=\) | \(0+1+0+1=2\) |
| \(\text{score}(B)\) | \(=\) | \(0+0+1+1=2\) |
| \(\text{score}(C)\) | \(=\) | \(1+0+0+0=1\) |
| \(\text{score}(D)\) | \(=\) | \(0+0+1+0=1\) |
Scores \(A{:}2,\ B{:}2,\ C{:}1,\ D{:}1\); \(A\) and \(B\) are tied on one-step wins.
Each entry of \(M^2\) is a row of \(M\) times a column of \(M\):
| \((M^2)_{XZ}\) | \(=\) | \((0,1,1)\cdot(1,1,0)=1\) |
| \((M^2)_{XY}\) | \(=\) | \((0,1,1)\cdot(1,0,0)=0\) |
| \(M^2\) | \(=\) | \(\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\) |
The two-step score of \(X\) is its row sum in \(M^2\):
| \(s^{(2)}_X\) | \(=\) | \(0+0+1=1\) |
\(M^2=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\); \(X\) beats \(Z\) in two steps (\(X\to Y\to Z\)).
First square \(M\); its row sums are the two-step scores:
| \(M^2\) | \(=\) | \(\begin{pmatrix}0&0&2&1\\1&0&1&0\\0&1&0&1\\1&0&0&0\end{pmatrix}\) |
| \(\text{two-step}\) | \(=\) | \((3,\ 2,\ 2,\ 1)\) |
Add each team's one-step and two-step scores (the row sums of \(M+M^2\)):
| \(\text{combined}(A)\) | \(=\) | \(2+3=5\) |
| \(\text{combined}(B)\) | \(=\) | \(2+2=4\) |
| \(\text{combined}(C)\) | \(=\) | \(1+2=3\) |
| \(\text{combined}(D)\) | \(=\) | \(1+1=2\) |
Order the combined scores from largest to smallest:
| \(5>4>3>2\) | \(\Rightarrow\) | \(A,\ B,\ C,\ D\) |
Ranking \(A, B, C, D\); the one-step tie between \(A\) and \(B\) is broken in \(A\)'s favour.
Common pitfalls
Frequently asked questions
What is a dominance matrix?
A square matrix that records "who beats whom": with competitors in a fixed order, \(m_{ij}=1\) if \(i\) beats \(j\) and \(0\) otherwise, with zeros on the diagonal.
What is a one-step dominance score?
The row sum of \(M\). It counts the competitors a team beats directly — its number of direct wins.
What does M squared tell you?
The entry \((M^2)_{ij}\) counts the two-step dominances of \(i\) over \(j\): the number of competitors \(k\) with \(i\) beating \(k\) and \(k\) beating \(j\). Its row sums are the two-step scores.
Why use M plus M squared to rank competitors?
Adding \(M^2\) includes indirect (two-step) wins as well as direct ones, so beating strong competitors counts. The combined row sums usually separate teams tied on \(M\) alone.
How do you rank the competitors?
Form \(M+M^2\), add each row to get a combined score, then order the scores from largest to smallest.
Do you go past M squared to higher powers?
Not in this course. The Specialist syllabus ranks competitors using the one-step and two-step dominances, \(M\) and \(M^2\), only.