This guide covers two ways matrices model real situations. A dominance matrix records who beat whom in a competition, and its powers rank the players. A Leslie matrix moves a female population forward in time, one age group at a time. Both turn up nearly every year, usually as simple familiar questions on the Technology-active paper. Work through it in order, and tick off each skill once you can do it without help.
Skills in this topic
Only tick a skill when you could do it in an exam.
Build a dominance matrix from a list of results, including draws and results you have to work out. Part 1
Rank players with one-step and two-step dominances, using a model such as $\mathbf{N}+\mathbf{N}^2$.Part 1
Spot a tie, explain the limitation of the model and suggest a refinement. Part 1
Build a Leslie matrix from a table or a description. Part 2
Project a population forward, counting the time steps correctly, and scale up for males. Part 2
Find unknown birth or survival rates from two population counts. Part 2
Part 1
Dominance matrices
A dominance matrix records the results of a competition where every player meets every other player once. Each row is a winner and each column is a loser. A 1 in row A, column B means A beat B, and a 0 means A did not. The leading diagonal is all zeros, because nobody plays themselves.
$\mathbf{M}=$
ABCD
ABCD
0101001010010100
Add along each row to get the one-step dominance vector $V_1$, which is the number of wins for each player.
A and C both have two wins, so one-step dominances cannot separate them. A two-step dominance is a win over a player who won a game. The entries of $\mathbf{M}^2$ count them. For example, row C, column B of $\mathbf{M}^2$ is 2, because C beat A, who beat B, and C beat D, who also beat B.
Now every player has a different total. C is first, A second, B third and D fourth.
The idea
Two-step wins break the tie
Ranking models
Exam questions name the model to use, so read it carefully. Find the whole matrix for the model, then rank by its row sums.
ModelWhat it counts
$\mathbf{N}+\mathbf{N}^2$One-step and two-step wins, counted equally
$\mathbf{N}+0.5\mathbf{N}^2$Two-step wins count for half, so direct wins matter more
$\mathbf{N}+\mathbf{N}^2+\mathbf{N}^3$Adds three-step wins, which separates more teams
If draws are possible, the question gives a scoring, such as 2 for a win, 1 for a draw and 0 for a loss in 2022 Paper 2, Question 2. Use exactly the numbers it gives. If two teams still have the same total, the model cannot separate them, and that is its limitation. To refine it, weight the two-step dominances (as in $\mathbf{N}+0.5\mathbf{N}^2$) or add a higher power such as $\mathbf{N}^3$.
Game plan: dominance matrices
1Write the matrix with winners as rows and losers as columns, in the same order both ways. Put 1 for a win and 0 otherwise, unless the question gives a different scoring.
2If some results are missing, use the totals you are given to work them out first.
3Calculate the matrix for the model the question names, using technology on a Technology-active paper.
4Add along each row and rank from the largest total.
5If teams are still tied, say so. That is the limitation, and a weighting or a higher power is the refinement.
Try it yourself
Rank a round robin
Pick a competition, then tap a match to change who won it. Switch between the ranking models and watch which teams stay tied.
Matches (tap to change the winner)
TeamOne-stepTwo-stepModelPlace
Practise
Exam questions: dominance matrices
2024 Paper 1, Question 6
Multiple choice · Technology-free · [1 mark]
Players P, Q, R and S played each other once in a competition where there were no draws. Only the following results are known.
Player P defeated players Q and R.
Player Q defeated two players.
Players R and S each defeated one player.
Based on these results, a dominance matrix $\mathbf{N}$ was partially constructed as shown.
$\mathbf{N}=$
PQRS
PQRS
011000011000
The completed matrix $\mathbf{N}$ is
Row Q needs two 1s. Q lost to P, so the P column is 0, and the Q column is 0 because Q cannot play itself. So Q beat R and S, and row Q is $0\ \ 0\ \ 1\ \ 1$.
Check it against the other rows. Row R has R losing to Q, which matches. (A) has Q beating P, which contradicts row P. (C) gives Q only one win, and (D) has a 1 on the leading diagonal.
Answer
(B)
2023 Paper 2, Question 7
Multiple choice · Technology-active · [1 mark]
Matrix $\mathbf{N}$ represents the results for a competition involving four teams.
$\mathbf{N}=$
PQRS
PQRS
0011100001000110
Rows are winning teams and columns are losing teams.
Key: Team P lost to team Q but won against teams R and S.
Using the ranking model $\mathbf{N}+0.5\mathbf{N}^2$, the teams that placed first, second and third respectively are
The row sums are A 4, B 7, C 4, D 9 and E 2. So D is first, B second, A and C are tied for third, and E is fifth.
1 markRanks the teams to show that teams A and C are tied.
Part c)
The model does not give individual places from first to fifth, because A and C are tied.
1 markIdentifies a limitation of the organisers’ model based on $\mathbf{N}+\mathbf{N}^2$.
Part d)
Include a weighting, for example $\mathbf{N}+\frac12\mathbf{N}^2$. Its row sums are A 2.5, B 5, C 3, D 6 and E 1.5, which separate A and C.
1 markCorrectly describes a suitable mathematical refinement.
The guide also accepts adding third-order dominances, as in $\mathbf{N}+\mathbf{N}^2+\mathbf{N}^3$. This is the third preset in the interactive above.
Answer
b) D, B, then A and C tied, Ec) A and C cannot be separatedd) a weighting such as $\mathbf{N}+\frac12\mathbf{N}^2$
Part 2
Leslie matrices
A Leslie matrix models the females of a population, split into age groups of equal length. The top row holds the fecundity rates, which are the average number of female offspring per female in each age group. The sub-diagonal holds the survival rates, the proportion of each age group that lives on into the next. Every other entry is 0.
The survival rate of the oldest group never appears, because there is no older group for it to move into. Tables often show it as 0, as in 2023 Paper 1, Question 4.
Projecting the population
Write the number of females in each age group as a column matrix. Multiplying by $\mathbf{L}$ moves the population on by one time step: the top row adds up the births, and the sub-diagonal moves the survivors along.
Starting from $\mathbf{N}_0$
$$\mathbf{N}_n=\mathbf{L}^n\mathbf{N}_0$$
Used in 2020 Paper 2, Question 2 and 2021 Paper 2, Question 14. $n$ is the number of time steps.
Starting from Year 1
$$\mathbf{N}_r=\mathbf{L}^{r-1}\mathbf{N}_1$$
Used in 2022 Paper 2, Question 12. Year 20 is 19 steps after Year 1.
Either way, count the steps between the two dates. From the start of 2021 to the start of 2025 is four steps, so use $\mathbf{L}^4$. The model only counts females. If the question gives a male to female ratio and asks for the whole population, scale up at the end. With males to females $1:2$, the total is $\frac32$ of the females.
Watch it on an exam question
Four years of thornbills
The idea
The same model, fifteen years on
When the rates are unknown
Sometimes you are given two population counts and have to find the rates, as in 2024 Paper 2, Question 19. Write $\mathbf{L}$ with letters for the unknown rates, form the matrix equation linking the two counts, and equate entries to get simultaneous equations.
Game plan: Leslie matrices
1Find the fecundity rates and the survival rates in the table or the description.
2Build $\mathbf{L}$ with fecundity rates in the top row, survival rates on the sub-diagonal and zeros everywhere else.
3Write the starting females in each age group as a column matrix.
4Count the time steps between the two dates and calculate $\mathbf{L}^n$ times the starting matrix with technology.
5Add the age groups for the female total, then scale for males if the question asks for the whole population.
Practise
Exam questions: Leslie matrices
2023 Paper 1, Question 4
Multiple choice · Technology-free · [1 mark]
The age-specific population distribution of a particular species of animal is shown.
Age (years)
0–1
1–2
2–3
3–4
Female population
94
82
37
6
Breeding rate
0
1.3
0.9
0.2
Survival rate
0.6
0.8
0.4
0
The Leslie matrix based on this data is
The breeding rates go in the top row: $0,\ 1.3,\ 0.9,\ 0.2$. The survival rates of the first three age groups go on the sub-diagonal: $0.6,\ 0.8,\ 0.4$. The population numbers do not belong in $\mathbf{L}$ at all.
(A) puts the populations in the top row, and (B) and (C) swap the roles of the two rates.
Answer
(D)
2020 Paper 2, Question 2
Multiple choice · Technology-active · [1 mark]
The Leslie matrix for a certain endangered species is given.
The total is about 4221, and the closest option is 4000.
$\mathbf{L}^6\mathbf{N}_0$ gives about 2851 and $\mathbf{L}^8\mathbf{N}_0$ gives about 6241. Those are options (A) and (D), so miscounting the years leads straight to a wrong answer.
Answer
(B) 4000
2022 Paper 2, Question 12
Short response · Technology-active · Simple familiar · [5 marks]
A scientist collects data for a species of tree frog in a protected area. Details for the female tree frog population are shown in the table.
Age (years)
0–1
1–2
2–3
3–4
Population in Year 1
150
101
84
62
Birth (breeding) rate
0.4
0.7
0.5
0.1
Survival rate
0.6
0.3
0.2
0
The scientist uses a Leslie matrix model to make predictions about the female tree frog population.
a)State the initial population matrix.[1 mark]
b)Determine the Leslie matrix.[1 mark]
A species is considered to be endangered if the female population in a restricted area is predicted to fall to less than 125 in the next 20 years.
c)Determine whether this species of tree frog is considered to be endangered.[3 marks]
1 markCalculates a matrix representing the female population within the 20-year period.
The female population in Year 20 is about $62.7+39.7+12.6+2.7\approx118$.
1 markCalculates the female population for a year within the 20-year period.
This is less than 125 within the 20 years, so the species is considered to be endangered.
1 markMakes a suitable decision about whether the species is considered endangered.
The guide accepts any year from Year 18 to Year 21. The total first drops below 125 in Year 19, at about 124.2.
Answer
Endangered: the female population falls to about 118 by Year 20.
2021 Paper 2, Question 14
Short response · Technology-active · Simple familiar · [5 marks]
The Tasmanian thornbill is a species of bird that has an average life span of three years. Female thornbills do not reproduce in their first year, but produce an average of four female offspring in each of their second and third years. The survival rate of each age group is estimated as 25% in their first year and 30% in their second year.
A Leslie matrix, $\mathbf{L}$, modelling the population distribution of the Tasmanian thornbill, has been partially completed.
1 markDetermines the female population in their first, second and third years at the start of 2025.
$$2166+938.25+81.45\approx3186$$
1 markDetermines the total female population at the start of 2025.
Males to females is $1:2$, so the whole population is $\frac32$ of the females.
$$\tfrac32\times3185.7\approx4779$$
1 markDetermines the total population at the start of 2025, rounded to a whole number.
The marking guide gives 4778 and accepts nearby values such as 4777 and 4781, so small rounding differences are fine.
Answer
a) $x=0.25,\ y=0.3$b) about 4779 thornbills
2024 Paper 2, Question 19
Short response · Technology-active · Complex unfamiliar · [6 marks]
An experiment researching the population changes of a certain species of insect was conducted over a four-week period. The insect has two distinct stages in its two-week lifespan. Each stage is approximately one week in length.
A constant proportion of females survive from stage 1 into stage 2.
The ratio of the reproduction rate for females in stage 2 to females in stage 1 is $2:1$. All offspring are born into stage 1.
The number of females in each stage was measured initially and then again after two weeks as shown.
Female population
Stage 1
Stage 2
Initially
48
32
After two weeks
25
21
Use a matrix approach to estimate the total number of females after four weeks.
Let $x$ be the birth rate for stage 1 females, so the stage 2 rate is $2x$, and let $y$ be the survival rate from stage 1 to stage 2.
The total number of females after four weeks is about 26.
1 markDetermines the total number of females at the end of the experiment.
1 markShows logical organisation of a fully attempted solution, communicating key steps.
The organisation mark needs your variables defined and matrix notation used throughout. The guide accepts 25 to 27 depending on rounding. This question is worked step by step in the complex unfamiliar collection.
1Dominance: winners as rows, losers as columns, zeros on the leading diagonal.
2Use the model the question names, and rank by the row sums of the whole model.
3If teams are still tied, say so, then suggest a weighting or a higher power.
4Leslie: fecundity rates in the top row, survival rates on the sub-diagonal, zeros everywhere else.
5Count the time steps between the two dates before you raise $\mathbf{L}$ to a power.
6Add the age groups, then scale for males if the question asks for the whole population.
Mistakes I see a lot
putting the losers in the rows
ranking from $\mathbf{N}^2$ alone instead of the whole model
putting the survival rates in the top row
including the survival rate of the oldest age group
using $\mathbf{L}^5$ for a four-year gap
giving the female population when the question asks for the total
Practice
Practice questions
Ten questions of my own, worth 52 marks in total. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Check whether each one is Technology-free or Technology-active before you start.
Question 1 of 10
Dominance matrix; ranking with $\mathbf{N}+\mathbf{N}^2$; a tie and a refinement · Technology-active · Simple familiar · [5 marks]
Teams A, B, C, D and E each played each other once in a competition with no draws. The results were:
A defeated E.
B defeated A and E.
C defeated A, B and E.
D defeated A, B and C.
E defeated D.
a)Allocating 1 to represent defeated and 0 otherwise, construct the dominance matrix $\boldsymbol{N}$, with winning teams as rows and losing teams as columns.[1 mark]
The organisers rank the teams using the model $\boldsymbol{N}+\boldsymbol{N}^2$.
b)Use the model $\boldsymbol{N}+\boldsymbol{N}^2$ to rank the teams.[2 marks]
c)Use your result from Question 1b) to identify a limitation of the organisers' ranking model.[1 mark]
d)State a mathematical refinement the organisers could consider to overcome the limitation identified in Question 1c).[1 mark]
Ranking: D first, C second, B and E tied third, A fifth
1 markRanks the teams to show that teams B and E are tied.
Part c)
The model does not provide individual positions from first to fifth, as B and E are tied for third.
1 markIdentifies a limitation of the model based on $\boldsymbol{N}+\boldsymbol{N}^2$.
Part d)
Include weightings in the calculation, e.g. $\boldsymbol{N}+0.5\boldsymbol{N}^2$ (row sums 1.5, 3, 5, 6, 2.5, giving D, C, B, E, A).
1 markCorrectly describes a suitable mathematical refinement.
Using the head-to-head result (B beat E) to split the tie also works, and so does extending the model to $\mathbf{N}+\mathbf{N}^2+\mathbf{N}^3$ (row sums 5, 8, 13, 16, 10). The two refinements put B and E in different orders, and either is accepted.
Answer
b) D, C, then B and E tied third, Ac) no individual third and fourth placesd) weightings, e.g. $\mathbf{N}+0.5\mathbf{N}^2$
Question 2 of 10
Leslie matrix from a description; projecting; including males · Technology-active · Simple familiar · [5 marks]
A species of possum has an average life span of three years. Female possums do not reproduce in their first year, but produce an average of 1.2 female offspring in their second year and 0.8 female offspring in their third year. The survival rate is estimated as 60% in the first year and 45% in the second year.
A Leslie matrix, $\boldsymbol{L}$, modelling the female population has been partially completed.
1 markCorrectly identifies the initial population matrix.
Start of 2031 is five years later.
Using technology: $\boldsymbol{N}_5=\boldsymbol{L}^5\boldsymbol{N}_0\approx\begin{bmatrix}305.34\\176.26\\90.59\end{bmatrix}$
1 markDetermines the female population in each age group at the start of 2031.
Total female population $\approx305.34+176.26+90.59\approx572$
1 markDetermines the total female population at the start of 2031.
Males : females $=2:3$, so total $=\tfrac53\times572.18\approx954$
The total possum population is approximately 954.
1 markDetermines the total population at the start of 2031, rounded to a whole number.
Using $\mathbf{L}^4\mathbf{N}_0$ or $\mathbf{L}^6\mathbf{N}_0$ counts the years wrongly. Multiplying by $\frac23$ or $\frac52$ instead of $\frac53$ loses the last mark.
Answer
a) $x=0.6,\ y=0.45$b) about 954 possums
Question 3 of 10
Leslie matrix from a table; one year by hand · Technology-free · Simple familiar · [5 marks]
The age-specific population distribution of the females of a species of fish is shown.
Age (years)
0–1
1–2
2–3
Female population
240
100
50
Breeding rate
0
1.5
1
Survival rate
0.5
0.4
0
a)Determine the Leslie matrix, $\boldsymbol{L}$, for this population.[1 mark]
b)Determine the number of females in each age group after one year.[2 marks]
c)Use your result from Question 3b) to determine the number of females aged 2–3 years after two years.[1 mark]
d)Explain why the survival rate of the 2–3 age group does not appear as an element of $\boldsymbol{L}$.[1 mark]
1 markDetermines the number of females in each age group.
Part c)
Females aged 2–3 after two years $=0.4\times120=48$
1 markDetermines the number of females aged 2–3.
Part d)
No female survives beyond three years, so there is no older age group for the 2–3 age group to survive into. Survival rates appear only on the sub-diagonal, as the proportion moving into the next age group.
1 markExplains why the survival rate does not appear.
Calculating the whole product $\mathbf{L}^2\mathbf{N}_0$ in part c) is fine too.
Answer
b) 200, 120 and 40c) 48d) no female survives beyond three years
Question 4 of 10
Dominance with missing results; one-step and two-step dominances · Technology-free · Simple familiar · [5 marks]
Players P, Q, R and S each played each other once in a competition where there were no draws. Only the following results are known.
Player P defeated player Q.
Player Q defeated players R and S.
Player R defeated player P.
Players R and S each defeated exactly one player.
a)Construct the dominance matrix, $\boldsymbol{M}$, for the competition, with winning players as rows and losing players as columns.[2 marks]
b)Explain why the players cannot be ranked using one-step dominances alone.[1 mark]
c)Use the sum of the one-step and two-step dominances to rank the players.[2 marks]
Part a)
R defeated exactly one player (P), so R lost to Q and S.
S defeated exactly one player (R), so S lost to P.
1 markCorrectly determines $\boldsymbol{M}^2$ and the two-step dominances.
Sum $=\begin{bmatrix}5\\4\\3\\2\end{bmatrix}$
Ranking: P first, Q second, R third, S fourth
1 markRanks the players.
Answer
b) P and Q tied, R and S tiedc) P, Q, R, S (totals 5, 4, 3, 2)
Question 5 of 10
Dominance with draws; a weighted model; meaning of an element of $\mathbf{M}^2$ · Technology-active · Simple familiar · [5 marks]
Teams W, X, Y and Z each played each other once in a netball competition. A win was allocated 2, a draw 1 and a loss 0. The results were:
W defeated X and Y, and lost to Z.
X defeated Y and drew with Z.
Y drew with Z.
a)Construct the matrix $\boldsymbol{M}$ that represents these results, with rows representing the points each team earned against the team in each column.[1 mark]
b)Explain why the total points earned cannot be used to rank all four teams.[1 mark]
c)Use the model $\boldsymbol{M}+0.5\boldsymbol{M}^2$ to rank the teams.[2 marks]
d)Explain what the element in row Z and column X of $\boldsymbol{M}^2$ represents in this context.[1 mark]
1 markDetermines $\boldsymbol{L}$ and the correct matrix product for Year 4.
Using technology: $\boldsymbol{N}_4\approx\begin{bmatrix}172.49\\79.30\\61.25\\16.80\end{bmatrix}$
Approximately 172, 79, 61 and 17 females
1 markDetermines the number of females in each age group.
Part b)
Using technology: total in Year 12 ($\boldsymbol{L}^{11}\boldsymbol{N}_1$) $\approx481$
total in Year 13 ($\boldsymbol{L}^{12}\boldsymbol{N}_1$) $\approx504$
1 markDetermines the total female population for consecutive years either side of 500.
The total first exceeds 500 in Year 13.
1 markStates the year.
Part c)
$\boldsymbol{N}_{10}=\boldsymbol{L}^9\boldsymbol{N}_1\approx\begin{bmatrix}228.51\\108.97\\72.80\\27.82\end{bmatrix}$, total $\approx438.11$
$\dfrac{228.51}{438.11}\times100\approx52.2\%$
1 markDetermines the percentage.
With Year 1 as $\mathbf{N}_1$, Year 4 is $\mathbf{L}^3\mathbf{N}_1$. Using $\mathbf{L}^4\mathbf{N}_1$ counts one year too many, but if you carry the same slip into part b) you can still get those marks.
Answer
a) about 172, 79, 61 and 17b) Year 13c) about 52.2%
Question 7 of 10
Leslie model in reverse; reasonableness of earlier populations · Technology-active · Simple familiar · [5 marks]
The female population of a species of bird is modelled by the Leslie matrix
1 markDetermines the population two years before the census.
Part d)
The model predicts $-100$ females aged 2–3, which is impossible. Using the model to estimate earlier populations is not reasonable beyond one year.
1 markComments on the reasonableness using the negative value.
Answer
a) $\det(\mathbf{L})=0.3$b) 200, 90 and 120c) 180, 200 and $-100$d) a negative number, so not reasonable
Question 8 of 10
Two-group Leslie model with unknown rates · Technology-free · Simple familiar · [5 marks]
The females of a species of insect are modelled in two age groups: juveniles and adults. Each year, every adult female produces an average of $b$ female offspring, and a proportion $s$ of juvenile females survives to become adults. Adult females do not survive to the following year. The model is
An initial census recorded 80 juvenile females and 60 adult females. One year later, a second census recorded 150 juvenile females and 48 adult females.
a)Determine the values of $b$ and $s$.[2 marks]
b)Show that $\boldsymbol{L}^2=1.5\boldsymbol{I}$, where $\boldsymbol{I}$ is the $2\times2$ identity matrix.[1 mark]
c)Use your result from Question 8b) to determine the total female population four years after the initial census.[1 mark]
d)Describe the long-term behaviour of the female population predicted by the model.[1 mark]
$\boldsymbol{N}_4=2.25\begin{bmatrix}80\\60\end{bmatrix}=\begin{bmatrix}180\\135\end{bmatrix}$, total $=315$
1 markUses $\boldsymbol{L}^2=1.5\boldsymbol{I}$ to determine the total female population.
Part d)
Every two years each age group is multiplied by 1.5, so the population grows without bound (by 50% every two years).
1 markDescribes the long-term behaviour.
Working out each year in turn in part c) only gets the mark if you use the result from part b).
Answer
a) $b=2.5,\ s=0.6$c) 315d) grows by a factor of 1.5 every two years
Question 9 of 10
Leslie model with males; evaluating a claim · Technology-active · Complex familiar · [6 marks]
A species of frog lives for up to four years. Females do not breed in their first year, and produce an average of 0.8, 1.4 and 0.6 female offspring in their second, third and fourth years respectively. Of the females, 40% survive their first year, 70% survive their second year and 50% survive their third year. No frog survives beyond four years.
At the start of 2026, a survey estimated that there were 400, 250, 180 and 90 females in their first, second, third and fourth years respectively, and that the ratio of males to females was approximately $2:3$.
A ranger claims that the total frog population will fall below 1000 within five years of the survey.
Evaluate the reasonableness of the ranger's claim. Justify your decision using mathematical reasoning.
$\boldsymbol{N}_0=\begin{bmatrix}400\\250\\180\\90\end{bmatrix}$ (start of 2026)
1 markCorrectly identifies the initial population matrix.
Five years after the survey (start of 2031):
Using technology: $\boldsymbol{N}_5=\boldsymbol{L}^5\boldsymbol{N}_0\approx\begin{bmatrix}328.68\\147.44\\103.94\\59.78\end{bmatrix}$, female total $\approx639.84$
1 markDetermines the female population five years after the survey.
Total population $\approx\tfrac53\times639.84\approx1066$
1 markDetermines the total population, including males.
The totals for years 1 to 5 are approximately 1552, 1382, 1233, 1179 and 1066, all above 1000. The total first falls below 1000 after six years ($\approx977$).
1 markUses mathematical reasoning to justify the decision.
The ranger's claim is not reasonable.
1 markStates an appropriate evaluation of the reasonableness of the claim.
Comparing the females alone with 1000 loses the last three marks, because the total has to include the males. Showing only the five-year total is enough, since the population is falling.
Answer
total after five years $\approx1066$, which is more than 1000the claim is not reasonable
Question 10 of 10
Dominance with an unplayed match; evaluating a claim · Technology-active · Complex familiar · [6 marks]
Teams A, B, C, D and E play each other once in a competition with no draws. All matches have been played except the final match, between teams C and D. The results so far are:
A defeated B, D and E.
B defeated E.
C defeated A and B.
D defeated B and E.
E defeated C.
The organisers will rank the teams using the model $\boldsymbol{N}+0.5\boldsymbol{N}^2$, where $\boldsymbol{N}$ is the dominance matrix with winning teams as rows and losing teams as columns.
The coach of team A claims that team A will be ranked first, regardless of the result of the final match.
Evaluate the reasonableness of the coach's claim. Justify your decision using mathematical reasoning.
Rows and columns in the order A, B, C, D, E. If D defeats C:
1 markCorrectly constructs a dominance matrix for one result of the final match.
Using technology, row sums of $\boldsymbol{N}+0.5\boldsymbol{N}^2$:
A 5.5, B 1.5, C 4, D 5, E 2
Team A is ranked first.
1 markDetermines the model values when D defeats C.
If C defeats D, the elements in row C, column D and row D, column C change to 1 and 0.
Row sums of $\boldsymbol{N}+0.5\boldsymbol{N}^2$: A 5, B 1.5, C 6, D 3, E 2.5
1 markDetermines the model values when C defeats D.
Team C is ranked first, with team A second.
1 markIdentifies the first-placed team in each case.
Team A is ranked first only if D defeats C. If C defeats D, team C is ranked first.
1 markUses mathematical reasoning to justify the decision.
The coach's claim is not reasonable.
1 markStates an appropriate evaluation of the reasonableness of the claim.
Team A has three wins whatever happens, so an argument from one-step dominances alone misses the point and loses the middle marks.
Answer
D beats C: A first (5.5)C beats D: C first (6)the claim is not reasonable
Keep practising
Ready for something harder?
Leslie questions with unknown rates, like 2024 Paper 2, Question 19, are the hardest in this topic so far. Once these feel comfortable, try the complex unfamiliar questions in the collection.
