Communication and dominance matrices: VCE General Mathematics Unit 4 Matrices

Each consecutive pair in the sequence must be a 1 in the matrix, reading row = sender, column = receiver.

• A: $A \to C$ is 0 (row $A$, column $C$). Fails at the first step.
• B: $A \to D$ is 1 and $D \to B$ is 1, but $B \to E$ is 0.
• C: $A \to B$ is 1, $B \to D$ is 1 and $D \to E$ is 1. Works.
• D: $A \to F$ is 1 but $F \to D$ is 0.

The answer is C ($93\%$ correct). Note that $A - D - E$ also works and is shorter; it was not an option.

Source: VCAA 2025 General Mathematics Examination 1, Question 28, and the 2025 examination report.

Row totals are wins. The ranking needs $U > V > S > T$, so with four teams the wins must be $U\ 3$, $V\ 2$, $S\ 1$, $T\ 0$.

• A: row totals $S\ 3$, $T\ 2$, $U\ 1$, $V\ 0$. Wrong order.
• B: $S$ beats $U$ and $U$ beats $S$ (both elements are 1). Impossible.
• C: totals $1, 1, 2, 2$: two ties, not the given ranking.
• D: totals $S\ 1$, $T\ 0$, $U\ 3$, $V\ 2$, and every pair has exactly one winner. Correct.
• E: a 1 on the leading diagonal ($U$ beats $U$). Impossible.

The answer is D ($69\%$). $19\%$ chose E, missing the diagonal check.

Source: VCAA 2023 General Mathematics Examination 1, Question 28, and the 2023 examination report.

Step 1: complete $D$. Kyle beat $M$ and $N$, which is already two wins, so $K$ lost to $L$: $d_{KL} = 0$, $d_{LK} = 1$. That is Lian's one win, so $L$ lost to $N$: $d_{LN} = 0$, $d_{NL} = 1$. Kyle beat Neil, so $d_{NK} = 0$.

$D = \begin{bmatrix} 0&0&1&1&0 \\ 1&0&0&0&0 \\ 0&1&0&1&1 \\ 0&1&0&0&0 \\ 1&1&0&1&0 \end{bmatrix}$ (row totals $2, 1, 3, 1, 3$).

Step 2: square and add (CAS). $T = D + D^2 = \begin{bmatrix} 0&2&1&2&1 \\ 1&0&1&1&0 \\ 2&3&0&2&1 \\ 1&1&0&0&0 \\ 2&2&1&2&0 \end{bmatrix}$, which is option B ($63\%$ correct; $21\%$ chose D).

Row totals of $T$: $K\ 6$, $L\ 3$, $M\ 8$, $N\ 2$, $O\ 7$. Maggie wins the tournament, ahead of Ophelia on two-step dominances.

Source: VCAA 2025 General Mathematics Examination 1, Question 32, and the 2025 examination report.

(a) Customers can only send to sales staff ($C \to S$). Sales staff send to managers or customers ($S \to M$). Managers send to directors ($M \to D$). Shortest sequence: C - S - M - D. (1 mark; $74\%$.)

(b) i. $H = G^2 = \begin{bmatrix} 1&2&1&2&2 \\ 0&3&1&2&2 \\ 1&0&1&1&1 \\ 1&0&1&2&1 \\ 0&1&0&0&1 \end{bmatrix}$. (1 mark; $66\%$.)

(b) ii. $g_{21} = 1$: managers can communicate directly with directors. $h_{21} = 0$: managers cannot communicate with directors through one other person (a two-step link). (1 mark; only $25\%$. Deal with each element separately and say what it means for managers and directors.)

Source: VCAA 2023 General Mathematics Examination 2, Question 10, and the 2023 examination report.

(a) $C = \begin{bmatrix} 0&1&1&0 \\ 0&0&1&1 \\ 1&0&0&0 \\ 0&0&1&0 \end{bmatrix}$ (1 mark)

(b) No. For example $c_{12} = 1$ but $c_{21} = 0$: P can send to Q but Q cannot send to P. Some links are one-way. (1 mark) A matrix of two-way links, such as the headset matrix in 2025 Examination 1 Question 28, is symmetric.

(a) Row X total: $0 + 0 + 1 + 1 = 2$ wins. Column X total: $1 + 0 + 0 + 0 = 1$ loss (to W). (1 mark)

(b) Each game gives exactly one 1, so the number of games is the sum of all elements: $2 + 2 + 1 + 1 = 6$, which matches $\frac{4 \times 3}{2} = 6$. (1 mark)

(c) W and X have 2 wins each; Y and Z have 1 each. (1 mark)

(a) $D^2 = \begin{bmatrix} 0&0&1&2 \\ 1&0&0&1 \\ 1&0&0&0 \\ 0&1&1&0 \end{bmatrix}$, $D + D^2 = \begin{bmatrix} 0&1&2&2 \\ 1&0&1&2 \\ 1&0&0&1 \\ 1&1&1&0 \end{bmatrix}$. (1 mark)

(b) Row totals of $D + D^2$: W 5, X 4, Y 2, Z 3. Ranking: W, X, Z, Y. The two-step wins separate W from X and lift Z above Y. (1 mark)

(c) It is 2: W has two two-step dominances over Z, because W beat X, who beat Z, and W beat Y, who beat Z. (1 mark)

$C^2 = \begin{bmatrix} 1&0&1&1 \\ 1&0&1&0 \\ 0&1&1&0 \\ 1&0&0&0 \end{bmatrix}$ and $C + C^2 = \begin{bmatrix} 1&1&2&1 \\ 1&0&2&1 \\ 1&1&1&0 \\ 1&0&1&0 \end{bmatrix}$.

(a) Row P, column R of $C^2$ is 1: one two-step path (P to Q to R). (1 mark)

(b) Zeros off the diagonal of $C + C^2$: R cannot reach S and S cannot reach Q in one or two steps. (1 mark) Both can be done in three steps (for example R to P to Q to S).

(a) F's single win is over G, so F lost to J. J's single win is therefore over F, so J lost to H. G has 2 wins; G beat J, so G also beat H.

$D = \begin{bmatrix} 0&1&0&0 \\ 0&0&1&1 \\ 1&0&0&1 \\ 1&0&0&0 \end{bmatrix}$ (row totals $1, 2, 2, 1$; every pair has exactly one 1). (2 marks)

(b) $D + D^2 = \begin{bmatrix} 0&1&1&1 \\ 2&0&1&2 \\ 2&1&0&1 \\ 1&1&0&0 \end{bmatrix}$, row totals F 3, G 5, H 4, J 2. Ranking: G, H, F, J. (1 mark)

(a) Every row of $D$ totals 2 and every row of $D^2$ totals 4 (for example A beat B and C, who between them beat C, D, D and E), so every row of $D + D^2$ totals 6. (1 mark)

(b) Every player won exactly two games, and the results form a perfectly balanced cycle, so each player's two-step wins are also equal. The dominance method can only break a tie when the wins are spread unevenly; here a different rule (such as head-to-head results or points margins) would be needed. (1 mark)