A2 June 2023 Paper 1 Q8
8. A colony of small mammals is being studied.
In the study, the mammals are divided into 3 categories
| \(N\) (newborns) | 0 to less than 1 month old |
| \(J\) (juveniles) | 1 to 3 months old |
| \(B\) (breeders) | over 3 months old |
A model for the population of the colony is given by the matrix equation
\[\begin{pmatrix}N_{n+1}\\ J_{n+1}\\ B_{n+1}\end{pmatrix} = \begin{pmatrix}0 & 0 & 2\\ a & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix}\begin{pmatrix}N_n\\ J_n\\ B_n\end{pmatrix}\]where \(a\) and \(b\) are constants, and \(N_n\), \(J_n\) and \(B_n\) are the respective numbers of the mammals in each category \(n\) months after the start of the study.
At the start of the study the colony has breeders only, with no newborns or juveniles.
According to the model, after 2 months the number of newborns is 48 and the number of juveniles is 40
Given that the model predicts approximately 1015 mammals in total at the start of a particular month, and approximately 596 newborns, 464 juveniles and 437 breeders at the start of the next month,
It is decided to monitor the number of newborn males and females as a part of the study.
Assuming that 42% of newborns are male,
(There is no need to estimate any unknown values for the refined model, but any known values should be made clear.) (2)
| Scheme | Marks | AO |
|---|---|---|
Accept E.g.
| B1 | 3.5b |
| (1) |
Notes
B1: Any valid limitation – see scheme for some examples. Must refer to a feature of the categories given.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\begin{pmatrix}0 & 0 & 2\\ a & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix}^2\begin{pmatrix}0\\ 0\\ k\end{pmatrix} = \begin{pmatrix}0 & 0 & 2\\ a & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix}\begin{pmatrix}2k\\ 0\\ 0.96k\end{pmatrix}\) or \(\begin{pmatrix}0 & 0.96 & 1.92\\ ab & b^2 & 2a\\ 0.48a & 0.48b + 0.96 \times 0.48 & 0.96^2\end{pmatrix}\begin{pmatrix}0\\ 0\\ k\end{pmatrix} = \begin{pmatrix}1.92k\\ 2ak\\ 0.9216k\end{pmatrix}\) | M1 | 3.4 |
| \(48 = 2 \times 0.96k \Rightarrow k = \ldots\) | dM1 | 1.1b |
| \(k = 25\) so 25 mammals at the start of the study | A1 | 3.2a |
| (ii) \(40 = 2ka \Rightarrow a = 0.8\,*\) | A1* | 1.1b |
| (4) |
Notes
(b)(i)
M1: Attempts to use the given information to set up a matrix equation and find the numbers of mammals after one month e.g.
or attempts to square the matrix to find the number of mammals after two months e.g.
\[\begin{pmatrix}0 & 0.96 & 1.92\\ ab & b^2 & 2a\\ 0.48a & 0.48b + 0.96 \times 0.48 & 0.96^2\end{pmatrix}\begin{pmatrix}N_0\\ J_0\\ B_0\end{pmatrix} = \ldots\]dM1: Forms an equation, in their variable for number of breeders at the start, setting their number of newborns after 2 months equal to 48 and solves for their variable to find the initial number of breeders.
A1: For identifying 25 mammals at the start of the study. Allow 25 mammals or just 25 or e.g. \(B_0 = 25\) so ignore how they label it just look for 25
Note that in some cases work may be minimal e.g.
\[\begin{pmatrix}0 & 0 & 2\\ a & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix}\begin{pmatrix}0\\ 0\\ B_0\end{pmatrix} = \begin{pmatrix}N_1\\ J_1\\ B_1\end{pmatrix} \Rightarrow 0.96B_0 = B_1,\ \begin{pmatrix}0 & 0 & 2\\ a & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix}\begin{pmatrix}N_1\\ J_1\\ B_1\end{pmatrix} = \begin{pmatrix}48\\ 40\\ B_2\end{pmatrix} \Rightarrow 2B_1 = 48\]\[B_1 = 24 = 0.96B_0 \Rightarrow B_0 = 25\](ii)
A1*: For correctly showing \(a = 0.8\). Must see the correct work to establish the correct value or equivalent by verification with a minimal conclusion e.g.
| Scheme | Marks | AO |
|---|---|---|
| \(\det\begin{pmatrix}0 & 0 & 2\\ 0.8 & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix} = 0 - 0 + 2(0.48 \times 0.8 - 0) = 0.768\) | B1 | 2.2a |
| \(\operatorname{adj}\begin{pmatrix}0 & 0 & 2\\ 0.8 & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix} = \begin{pmatrix}0.96b & 0.96 & -2b\\ -0.768 & 0 & 1.6\\ 0.384 & 0 & 0\end{pmatrix}\) | M1 | 1.1b |
| \(= \begin{pmatrix}1.25b & 1.25 & -\frac{125}{48}b\\ -1 & 0 & \frac{25}{12}\\ 0.5 & 0 & 0\end{pmatrix}\) oe e.g. \(\dfrac{1}{0.768}\begin{pmatrix}0.96b & 0.96 & -2b\\ -0.768 & 0 & 1.6\\ 0.384 & 0 & 0\end{pmatrix}\) or e.g. \(\dfrac{125}{96}\begin{pmatrix}0.96b & 0.96 & -2b\\ -0.768 & 0 & 1.6\\ 0.384 & 0 & 0\end{pmatrix}\) | A1 | 1.1b |
| (3) |
Notes
B1: Deduces correct determinant for the matrix. Allow equivalents e.g. \(\dfrac{96}{125}\) May be implied.
M1: Recognisable attempt at the adjoint matrix. Look for at least 3 non-zero entries correct.
A1: Correct inverse. Accept awrt \(-2.6b\) for the upper right entry and awrt 2.08 for middle right entry, or accept with determinant still outside. Apply isw once a correct answer is seen.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}1.25b & 1.25 & -\frac{125}{48}b\\ -1 & 0 & \frac{25}{12}\\ 0.5 & 0 & 0\end{pmatrix}\begin{pmatrix}596\\ 464\\ 437\end{pmatrix}\) Total \(= 1.25b \times 596 + 1.25 \times 464 - \frac{125}{48}b \times 437 - 596 + \frac{25}{12} \times 437 + 0.5 \times 596\) | M1 | 3.1b |
| \(\Rightarrow 1015 = x + y + z = 745b + 580 - 1138b - 596 + 910.4 + 298 \Rightarrow b = \ldots\) | dM1 | 3.4 |
| \(b =\) awrt 0.45 | A1 | 1.1b |
| (3) |
Notes
(d) Alternative:
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}0 & 0 & 2\\ 0.8 & b & 0\\ 0 & 0.48 & 0.96\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}596\\ 464\\ 437\end{pmatrix} \Rightarrow \begin{aligned}2z &= 596\\ 0.8x + by &= 464\\ 0.48y + 0.96z &= 437\end{aligned}\) \(\Rightarrow z = 298,\ y = \dfrac{3773}{12}\,(314.4\ldots)\) \(x + y + z = 1015 \Rightarrow x = \ldots\dfrac{4831}{12}\,(402.5\ldots)\) | M1 | 3.1b |
| \(0.8x + by = 464 \Rightarrow 0.8 \times \dfrac{4831}{12} + b \times \dfrac{3773}{12} = 464 \Rightarrow b = \ldots\) | dM1 | 3.4 |
| \(b =\) awrt 0.45 | A1 | 1.1b |
| (3) |
Notes
M1: Attempts (their inverse matrix) \(\times \begin{pmatrix}596\\ 464\\ 437\end{pmatrix}\) correctly and adds the 3 expressions together to find the total in terms of \(b\).
M1: Sets their total = 1015 and solves for \(b\).
A1: awrt 0.45
Alternative:
M1: Uses the original matrix with \(a = 0.8\) and \(\begin{pmatrix}596\\ 464\\ 437\end{pmatrix}\) to form 3 equations in their variables and \(b\) and uses these and the 1015 to find the number of Newborns.
M1: Uses their values in the \(y\) component and solves for \(b\).
A1: awrt 0.45
| Scheme | Marks | AO |
|---|---|---|
| Let \(NM_n\) be newborn males and \(NF_n\) be newborn females in month \(n\) \(\begin{pmatrix}NM_{n+1}\\ NF_{n+1}\\ J_{n+1}\\ B_{n+1}\end{pmatrix} = \begin{pmatrix}0 & 0 & 0 & 0.84\\ 0 & 0 & 0 & 1.16\\ ? & ? & 0.45 & 0\\ 0 & 0 & 0.48 & 0.96\end{pmatrix}\begin{pmatrix}NM_n\\ NF_n\\ J_n\\ B_n\end{pmatrix}\) or e.g. \(\begin{pmatrix}NF_{n+1}\\ NM_{n+1}\\ J_{n+1}\\ B_{n+1}\end{pmatrix} = \begin{pmatrix}0 & 0 & 0 & 1.16\\ 0 & 0 & 0 & 0.84\\ ? & ? & 0.45 & 0\\ 0 & 0 & 0.48 & 0.96\end{pmatrix}\begin{pmatrix}NF_n\\ NM_n\\ J_n\\ B_n\end{pmatrix}\) | M1 A1ft | 3.5c 3.3 |
| (2) | ||
| (13 marks) |
Notes
M1: Defines new variables for male and female newborns (accept if a clear notation is used if not defined) and sets up a 4×4 matrix with structure shown, or male and female rows swapped, with the correct 0 entries in at least 4 places.
A1ft: Fully correct matrix system shown, accepting anything (including 0) for the unknown spaces shown – but must have all the 0’s and upper right entries correct. Accept \(b\) or their value of \(b\) in place of 0.45