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Leslie matrices and population models: VCE General Mathematics Unit 4 Matrices

Syllabus dot point

“Set up a Leslie matrix from age-specific birth (fecundity) and survival rates, multiply it by an age-structure state matrix to project a population forward, and interpret the long-term growth and age distribution”

14 min read

Quick answer

A Leslie matrix models the female population of a species in equal-width age groups. Birth rates (average female offspring per female per period) go across the top row; survival rates (the proportion of one age group that lives to reach the next) go on the subdiagonal, in row j+1j + 1, column jj; every other element is 0. Project with Sn+1=LSnS_{n+1} = LS_n, so Sn=LnS0S_n = L^nS_0. In the long run the total population changes by a constant factor each period (found by dividing successive totals for large nn), and the proportions in each age group settle to a stable age distribution.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Leslie matrices and transition matrices
  4. Exam-style questions
  5. Practice questions

What this dot point is asking

The Study Design includes the "use of the matrix recurrence relation: S0=S_0 = initial state matrix, Sn+1=TSnS_{n+1} = TS_n or Sn+1=LSnS_{n+1} = LS_n where TT is a transition matrix, LL is a Leslie matrix, and SS is a column state matrix, to generate a sequence of state matrices (assuming the next state only relies on the current state)". For Leslie matrices that means you should be able to:

  1. build a Leslie matrix from birth and survival rates given in words, a table or a life-cycle diagram;
  2. interpret any element of the matrix in context;
  3. project the age structure forward with Sn+1=LSnS_{n+1} = LS_n (or Sn=LnS0S_n = L^nS_0), including totals, percentages and rounding;
  4. describe the long term: the growth rate of the total population and the proportions in each age group.

The Leslie questions on recent papers have been one-mark multiple-choice items: build the matrix from a life-cycle diagram (2025 Examination 1 Question 29, 84%84\%) and interpret one element (2023 Examination 1 Question 31, 74%74\%). Both are quick if you know where each kind of rate goes. Projection questions use the same working as transition matrices.

The answer

What the model assumes

A Leslie model divides the female population into age groups of equal width, and uses a time step equal to that width (for example, age groups of one year and a time step of one year). In each time step:

  • every female in age group jj produces, on average, bjb_j female offspring, who join age group 1;
  • a proportion sjs_j of the females in age group jj survive and move up to age group j+1j + 1;
  • females in the oldest age group do not move on (in the basic model they die at the end of the step).

Only females are counted because the number of births depends on the number of females.

The structure of LL

For four age groups:

L=[b1b2b3b4s10000s20000s30].L = \begin{bmatrix} b_1 & b_2 & b_3 & b_4 \\ s_1 & 0 & 0 & 0 \\ 0 & s_2 & 0 & 0 \\ 0 & 0 & s_3 & 0 \end{bmatrix}.

  • Row 1 holds the birth rates. b1b_1 is often 0 (the youngest group does not breed). Birth rates can be greater than 1.
  • The subdiagonal holds the survival rates. The element in row j+1j + 1, column jj is the proportion of group jj that survives into group j+1j + 1. Survival rates are between 0 and 1.
  • Everything else is 0, including the leading diagonal (no one stays in the same age group) and the last column below row 1.

Unlike a transition matrix, the columns of a Leslie matrix do not add to 1: births add individuals and deaths remove them, so the total population can grow or shrink.

Definition

A Leslie matrix is a square matrix whose first row contains the birth rates of each age group and whose subdiagonal contains the survival rates from each age group to the next, with all other elements zero. It is used in the recurrence relation Sn+1=LSnS_{n+1} = LS_n, where SnS_n is the column matrix of the number of females in each age group after nn time periods.

Reading a life-cycle diagram

A life-cycle diagram draws each age group as a vertex. Straight arrows from group jj to group j+1j + 1 carry the survival rates; curved arrows back to group 1 carry the birth rates. The diagram below is the one from 2025 Examination 1 Question 29.

Life-cycle diagram for a population with four age groupsFour age groups, 1 to 4, drawn left to right. Straight arrows from each age group to the next are survival rates: 0.9 from group 1 to group 2, 0.7 from group 2 to group 3 and 0.2 from group 3 to group 4. Curved arrows from groups 2, 3 and 4 back to group 1 are birth rates: 2.1, 4.6 and 1.8. Group 1 has no curved arrow, so its birth rate is 0, and group 4 has no arrow to a later group. The birth rates form the top row of the Leslie matrix and the survival rates sit just below the leading diagonal.2.14.61.80.90.70.21234births(top row)survival(subdiagonal)Age group 1 has no birth arrow, so the top-left element of the Leslie matrix is 0.

Reading it: the top row is [02.14.61.8]\begin{bmatrix} 0 & 2.1 & 4.6 & 1.8 \end{bmatrix} (group 1 has no birth arrow) and the survival rates 0.90.9, 0.70.7 and 0.20.2 go in positions (2,1)(2, 1), (3,2)(3, 2) and (4,3)(4, 3).

Projecting forward

With S0S_0 the initial column matrix of the number of females in each age group,

S1=LS0,S2=LS1=L2S0,…,Sn=LnS0.S_1 = LS_0, \quad S_2 = LS_1 = L^2S_0, \quad \dots, \quad S_n = L^nS_0.

Multiplying out one step shows what each row does:

  • row 1 gives the newborns: b1×(group 1)+b2×(group 2)+…b_1 \times (\text{group 1}) + b_2 \times (\text{group 2}) + \dots;
  • row j+1j + 1 gives the survivors moving up: sj×(group j)s_j \times (\text{group } j).

The total population after nn periods is the sum of the elements of SnS_n. Keep full accuracy in the matrix and round only the final answer (numbers of animals are usually rounded to the nearest whole number).

To go backwards, from S1S_1 to S0S_0, use S0=L−1S1S_0 = L^{-1}S_1. This needs det⁡L≠0\det L \neq 0, which happens when the oldest group has a non-zero birth rate and all survival rates are non-zero.

The long term: growth rate and stable age distribution

If you calculate SnS_n for larger and larger nn (CAS: L50S0L^{50}S_0, L51S0L^{51}S_0, and so on), two things usually settle down:

  1. the ratio of successive totals approaches a constant rr. The population then grows (or declines) by the same percentage every period: (r−1)×100%(r - 1) \times 100\%;
  2. the proportion of the population in each age group stops changing. These proportions are the stable age distribution.

If r>1r > 1 the population grows, if r=1r = 1 it is stable, and if r<1r < 1 it declines. Both rr and the stable distribution come from the matrix LL, not from the starting population, although the starting population affects how quickly they appear.

The Study Design asks for this informally: "no noticeable change from one state matrix to the next" once you allow for the growth factor. One caution: when only one age group breeds, the numbers cycle (each individual takes exactly the same number of steps from birth to breeding), so check several consecutive ratios rather than just one.

Formula

LL: birth rates in row 1; survival rate from group jj to group j+1j + 1 in row j+1j + 1, column jj; all else 0. Sn+1=LSnS_{n+1} = LS_n, Sn=LnS0S_n = L^nS_0, S0=L−1S1S_0 = L^{-1}S_1. Long-run growth factor r=total of Sn+1total of Snr = \dfrac{\text{total of } S_{n+1}}{\text{total of } S_n} for large nn; percentage change per period =(r−1)×100%= (r - 1) \times 100\%.

Worked examples: building, projecting and the long term

Building LL from a life-cycle diagram (2025 Examination 1 Question 29)

Step 1: the top row
Births come from groups 2, 3 and 4 at rates 2.1, 4.6 and 1.8. Group 1 has none. Row 1: 0,2.1,4.6,1.80, 2.1, 4.6, 1.8.
Step 2: the subdiagonal
0.90.9 (group 1 to 2) in row 2 column 1; 0.70.7 in row 3 column 2; 0.20.2 in row 4 column 3.
Step 3: eliminate
An option with the survival rates in row 1 has the rates swapped; an option with 2.1 in the top-left gives group 1 babies; an option with survival rates on the leading diagonal keeps animals the same age. Answer: A.

Interpreting one element (2023 Examination 1 Question 31)

L=[0430.20000.40]L = \begin{bmatrix} 0 & 4 & 3 \\ 0.2 & 0 & 0 \\ 0 & 0.4 & 0 \end{bmatrix} for a bird that lives three years.

Step 1: locate the element. Row 2, column 1 is on the subdiagonal, so it is a survival rate from group 1 to group 2.

Step 2: say it in context. On average 20%20\% of first-year birds survive into their second year. Answer: C.

Projecting three years ahead

For the same birds, start with 500 first-year, 100 second-year and 40 third-year females.

Step 1: one step by hand
First-year birds next year: 4×100+3×40=5204 \times 100 + 3 \times 40 = 520. Second-year: 0.2×500=1000.2 \times 500 = 100. Third-year: 0.4×100=400.4 \times 100 = 40. So S1=[52010040]TS_1 = \begin{bmatrix} 520 & 100 & 40 \end{bmatrix}^T.
Step 2: continue on CAS
S2=[52010440]TS_2 = \begin{bmatrix} 520 & 104 & 40 \end{bmatrix}^T and S3=L3S0=[53610441.6]TS_3 = L^3S_0 = \begin{bmatrix} 536 & 104 & 41.6 \end{bmatrix}^T.
Step 3: totals
640→660→664→681.6640 \to 660 \to 664 \to 681.6. After three years there are about 682 female birds, including about 42 in their third year.

Long-run growth and the stable age distribution

Step 1: compute far ahead
S49S_{49} and S50S_{50} on CAS; their totals have ratio 1.0177…1.0177\dots
Step 2: check it has settled
The ratio is also 1.01771.0177 at n=30n = 30 and n=40n = 40, to four decimal places.
Step 3: interpret
In the long term the female bird population grows by about 1.8%1.8\% per year. The proportions in S50S_{50} are about 78.5%78.5\% first-year, 15.4%15.4\% second-year and 6.1%6.1\% third-year birds, and they no longer change: this is the stable age distribution. At this rate the population first exceeds double its starting size (1280) after 40 years.
Common traps
Swapping rows and columns for survival
The survival rate from group 1 to group 2 is in row 2, column 1, not row 1, column 2 (which is the birth rate of group 2).
Putting survival rates on the leading diagonal
That would keep animals in the same age group. They belong one place below the diagonal.
Giving the youngest group a birth rate it does not have
If the question says females do not breed in their first year, the top-left element is 0.
Rounding each step
Keep the decimals in S1S_1, S2S_2, …\dots and round only the final answer.
Using one early ratio for the long-term growth
total(S1)÷total(S0)\text{total}(S_1) \div \text{total}(S_0) is usually not the long-run rate. Use large nn and check two or three consecutive ratios agree.
Expecting the columns to add to 1
That is a transition-matrix property; a Leslie matrix changes the total population.
Exam technique

Write the age groups beside the rows and columns of LL before filling it in, and say each element aloud as "from column group to row group". For projections, store LL and S0S_0 on CAS and calculate LnS0L^nS_0 directly rather than stepping one year at a time, but do the first step by hand if the question asks you to show working. For "explain" questions, name the age groups and use "on average".

Leslie matrices and transition matrices

Both models use a recurrence relation of the form Sn+1=MSnS_{n+1} = MS_n and powers of a matrix to look ahead, and in both the long-term state appears when the state matrix stops changing (for Leslie models, when the proportions stop changing). The differences are in what the matrix means:

  • a transition matrix moves a fixed total between states, so each column adds to 1 and the total stays the same;
  • a Leslie matrix adds newborns and loses animals that die, so its columns do not add to 1 and the total grows or shrinks.

Culling and restocking are modelled by adding a column matrix to each step, Sn+1=MSn+BS_{n+1} = MS_n + B (see the transition matrices page for that form).

Note

A Leslie matrix tracks a population of female animals grouped by age. The top row says how many babies each age group has on average. The numbers just under the diagonal say what fraction of each age group survives to become the next age group. Multiply the matrix by the current numbers in each age group to get next year's numbers, and keep multiplying to look further ahead. After a while the population grows (or shrinks) by the same percentage every year and the share of animals in each age group stays the same.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2025 VCAA Exam 1 Q29Past paper1 mark
The female population of an animal species has been divided into four age groups (1, 2, 3 and 4), with age group 1 being the youngest. The age groups, birth rates and survival rates are presented in the life-cycle diagram below. [Diagram: survival arrows 1→21 \to 2: 0.9, 2→32 \to 3: 0.7, 3→43 \to 4: 0.2; birth arrows to group 1 from groups 2, 3, 4: 2.1, 4.6, 1.8.] Which one of the following is a Leslie matrix that corresponds with this life-cycle diagram? A. [02.14.61.80.900000.700000.20]\begin{bmatrix}0&2.1&4.6&1.8\\0.9&0&0&0\\0&0.7&0&0\\0&0&0.2&0\end{bmatrix} B. [00.90.70.22.100004.600001.80]\begin{bmatrix}0&0.9&0.7&0.2\\2.1&0&0&0\\0&4.6&0&0\\0&0&1.8&0\end{bmatrix} C. [2.14.61.800.900000.700000.20]\begin{bmatrix}2.1&4.6&1.8&0\\0.9&0&0&0\\0&0.7&0&0\\0&0&0.2&0\end{bmatrix} D. [2.14.61.80.800.900000.700000.2]\begin{bmatrix}2.1&4.6&1.8&0.8\\0&0.9&0&0\\0&0&0.7&0\\0&0&0&0.2\end{bmatrix}
Show worked answer →
  • Top row: births. Element jj of row 1 is the birth rate of age group jj. Group 1 has no arrow back to itself, so the top row is 0,2.1,4.6,1.80, 2.1, 4.6, 1.8.
  • Subdiagonal: survival. The rate from group jj to group j+1j + 1 goes in row j+1j + 1, column jj: 0.90.9 in row 2 column 1, 0.70.7 in row 3 column 2, 0.20.2 in row 4 column 3.
  • Everything else is 0.

That is option A (84%84\% correct). B swaps the roles of the birth and survival rates (9%9\%). C shifts the birth rates one column to the left, giving group 1 a birth rate of 2.1 (5%5\%). D puts the survival rates on the leading diagonal, which would keep each group where it is.

Source: VCAA 2025 General Mathematics Examination 1, Question 29, and the 2025 examination report.

2023 VCAA Exam 1 Q31Past paper1 mark
A species of bird has a life span of three years. The females in this species do not reproduce in their first year but produce an average of four female offspring in their second year, and three in their third year. The Leslie matrix, LL, below is used to model the female population distribution of this species of bird. L=[0430.20000.40]L = \begin{bmatrix}0&4&3\\0.2&0&0\\0&0.4&0\end{bmatrix} The element in the second row, first column states that on average 20% of this population will A. be female. B. never reproduce. C. survive into their second year. D. produce offspring in their first year. E. live for the entire lifespan of three years.
Show worked answer →

Row 2, column 1 is on the subdiagonal, so it is a survival rate: the proportion of age group 1 (first-year birds) that survive to become age group 2 (second-year birds).

0.20.2 means that on average 20%20\% of the birds survive into their second year: option C (74%74\% correct).

The other options confuse the two kinds of element. The birth rates are in row 1 (00, 44, 33), and "live for the entire lifespan" would need both survival rates: 0.2×0.4=0.080.2 \times 0.4 = 0.08, so only 8%8\% of birds reach their third year.

Source: VCAA 2023 General Mathematics Examination 1, Question 31, and the 2023 examination report.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks
An insect population has three age groups with birth rates 0, 4 and 3 and survival rates 0.5 (group 1 to group 2) and 0.25 (group 2 to group 3). The initial population is S0=[2006020]S_0 = \begin{bmatrix} 200 \\ 60 \\ 20 \end{bmatrix}. (a) Write the Leslie matrix LL. (b) Find S1S_1 and the total population after one period.
Show worked solution →

(a) Births in the top row, survival on the subdiagonal:

L=[0430.50000.250]L = \begin{bmatrix} 0 & 4 & 3 \\ 0.5 & 0 & 0 \\ 0 & 0.25 & 0 \end{bmatrix} (1 mark)

(b) Row by row:

  • Group 1 (newborns): 0×200+4×60+3×20=3000 \times 200 + 4 \times 60 + 3 \times 20 = 300.
  • Group 2: 0.5×200=1000.5 \times 200 = 100.
  • Group 3: 0.25×60=150.25 \times 60 = 15.

S1=[30010015]S_1 = \begin{bmatrix} 300 \\ 100 \\ 15 \end{bmatrix}, total 415415 (up from 280280). (2 marks)

foundation2 marks
For L=[01.50.80.60000.30]L = \begin{bmatrix} 0 & 1.5 & 0.8 \\ 0.6 & 0 & 0 \\ 0 & 0.3 & 0 \end{bmatrix}, explain in words the meaning of (a) the element in row 1, column 2 and (b) the element in row 3, column 2.
Show worked solution →

(a) 1.51.5: on average, each female in age group 2 produces 1.5 female offspring per time period. (1 mark)

(b) 0.30.3: on average, 30%30\% of the females in age group 2 survive into age group 3. (1 mark)

core3 marks
The bird population in 2023 Examination 1 Question 31 has L=[0430.20000.40]L = \begin{bmatrix} 0 & 4 & 3 \\ 0.2 & 0 & 0 \\ 0 & 0.4 & 0 \end{bmatrix}. At the start there are 500 first-year, 100 second-year and 40 third-year female birds. (a) Find the number of third-year female birds after 3 years, to the nearest whole number. (b) Find the percentage increase in the total female population over the 3 years, correct to one decimal place.
Show worked solution →

Use Sn+1=LSnS_{n+1} = LS_n with S0=[50010040]TS_0 = \begin{bmatrix} 500 & 100 & 40 \end{bmatrix}^T.

S1=[52010040]S_1 = \begin{bmatrix} 520 \\ 100 \\ 40 \end{bmatrix}, S2=[52010440]S_2 = \begin{bmatrix} 520 \\ 104 \\ 40 \end{bmatrix}, S3=[53610441.6]S_3 = \begin{bmatrix} 536 \\ 104 \\ 41.6 \end{bmatrix}.

(a) 41.641.6, so 42 third-year birds. (1 mark) Round only at the end.

(b) Total after 3 years =536+104+41.6=681.6= 536 + 104 + 41.6 = 681.6; initial total =640= 640. Increase =681.6−640640×100%=6.5%= \frac{681.6 - 640}{640} \times 100\% = 6.5\%. (2 marks)

core2 marks
For the same bird population, a later count gives S1=[5506040]S_1 = \begin{bmatrix} 550 \\ 60 \\ 40 \end{bmatrix}. Find S0S_0.
Show worked solution →

S1=LS0S_1 = LS_0, so S0=L−1S1S_0 = L^{-1}S_1. LL has an inverse because its determinant is 0.240.24, not 0 (the last birth rate is not zero).

Using CAS, S0=L−1[5506040]=[30010050]S_0 = L^{-1}\begin{bmatrix} 550 \\ 60 \\ 40 \end{bmatrix} = \begin{bmatrix} 300 \\ 100 \\ 50 \end{bmatrix}. (2 marks)

Check: 4×100+3×50=5504 \times 100 + 3 \times 50 = 550, 0.2×300=600.2 \times 300 = 60, 0.4×100=400.4 \times 100 = 40.

exam3 marks
A population of female animals has four age groups and L=[01.21.60.40.600000.700000.50]L = \begin{bmatrix} 0 & 1.2 & 1.6 & 0.4 \\ 0.6 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.5 & 0 \end{bmatrix}, with S0=[400300200100]TS_0 = \begin{bmatrix} 400 & 300 & 200 & 100 \end{bmatrix}^T. (a) Find the total population after 2 time periods. (b) In the long term the total population increases by a constant percentage each period. Use CAS to find this percentage, correct to one decimal place. (c) Find the long-term percentage of the population in age group 1, to the nearest per cent.
Show worked solution →

(a) S1=[720240210100]TS_1 = \begin{bmatrix} 720 & 240 & 210 & 100 \end{bmatrix}^T and S2=[664432168105]TS_2 = \begin{bmatrix} 664 & 432 & 168 & 105 \end{bmatrix}^T. Total after 2 periods: 13691369. (1 mark)

(b) Calculate S50=L50S0S_{50} = L^{50}S_0 and S51S_{51} and divide their totals: the ratio is 1.1655…1.1655\dots, so the population grows by about 16.6% per period. (1 mark) Earlier ratios are not yet steady (1270÷1000=1.271270 \div 1000 = 1.27, 1369÷1270≈1.0781369 \div 1270 \approx 1.078).

(c) In S50S_{50} the proportions are about 51%51\%, 26%26\%, 16%16\% and 7%7\%, and they do not change for larger nn: about 51% in age group 1. (1 mark)

exam3 marks
An insect has three stages: eggs, juveniles and adults. Each adult lays on average 40 female eggs and then dies. 25%25\% of eggs survive to become juveniles, and 12%12\% of juveniles survive to become adults. (a) Write the Leslie matrix and find S3S_3 if S0=[200040050]TS_0 = \begin{bmatrix} 2000 & 400 & 50 \end{bmatrix}^T (eggs, juveniles, adults). (b) Explain why S3S_3 is a multiple of S0S_0.
Show worked solution →

(a) L=[00400.250000.120]L = \begin{bmatrix} 0 & 0 & 40 \\ 0.25 & 0 & 0 \\ 0 & 0.12 & 0 \end{bmatrix}. (1 mark)

S1=[200050048]TS_1 = \begin{bmatrix} 2000 & 500 & 48 \end{bmatrix}^T, S2=[192050060]TS_2 = \begin{bmatrix} 1920 & 500 & 60 \end{bmatrix}^T, S3=[240048060]TS_3 = \begin{bmatrix} 2400 & 480 & 60 \end{bmatrix}^T. (1 mark)

(b) Only adults breed, so every insect takes exactly three periods to go from egg to egg. Each egg becomes 0.25×0.12=0.030.25 \times 0.12 = 0.03 adults, which lay 0.03×40=1.20.03 \times 40 = 1.2 eggs. So every three periods each stage is multiplied by 1.21.2: S3=1.2S0S_3 = 1.2S_0. (1 mark) The totals (24502450, 25482548, 24802480, 29402940) rise and fall within each cycle rather than growing smoothly.

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