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Date	May 2021	Marks available	2	Reference code	21M.2.AHL.TZ2.4
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Hence or otherwise and Determine	Question number	4	Adapted from	N/A

Question

In a small village there are two doctors’ clinics, one owned by Doctor Black and the other owned by Doctor Green. It was noted after each year that $3.5 %$ of Doctor Black’s patients moved to Doctor Green’s clinic and $5 %$ of Doctor Green’s patients moved to Doctor Black’s clinic. All additional losses and gains of patients by the clinics may be ignored.

At the start of a particular year, it was noted that Doctor Black had $2100$ patients on their register, compared to Doctor Green’s $3500$ patients.

Write down a transition matrix $T$ indicating the annual population movement between clinics.

[2]

Find a prediction for the ratio of the number of patients Doctor Black will have, compared to Doctor Green, after two years.

[2]

Find a matrix $P$ , with integer elements, such that $T = P D P^{- 1}$ , where $D$ is a diagonal matrix.

[6]

Hence, show that the long-term transition matrix $T^{\infty}$ is given by $T^{\infty} = (\begin{matrix} \frac{10}{17} & \frac{10}{17} \\ \frac{7}{17} & \frac{7}{17} \end{matrix})$ .

[6]

Hence, or otherwise, determine the expected ratio of the number of patients Doctor Black would have compared to Doctor Green in the long term.

[2]

Markscheme

$T = (\begin{matrix} 0.965 & 0.05 \\ 0.035 & 0.95 \end{matrix})$ M1A1

Note: Award M1A1 for $T = (\begin{matrix} 0.95 & 0.035 \\ 0.05 & 0.965 \end{matrix})$ .
Award the A1 for a transposed $T$ if used correctly in part (b) i.e. preceded by $1 \times 2$ matrix $(2100 3500)$ rather than followed by a $2 \times 1$ matrix.

[2 marks]

${(\begin{matrix} 0.965 & 0.05 \\ 0.035 & 0.95 \end{matrix})}^{2} (\begin{matrix} 2100 \\ 3500 \end{matrix})$ (M1)

$= (\begin{matrix} 2294 \\ 3306 \end{matrix})$

so ratio is $2294 : 3306 (= 1147 : 1653, 0.693889 \dots)$ A1

[2 marks]

to solve $A x = λ x :$

$|\begin{matrix} 0.965 - λ & 0.05 \\ 0.035 & 0.95 - λ \end{matrix}| = 0$ (M1)

$(0.965 - λ) (0.95 - λ) - 0.05 \times 0.035 = 0$

$λ = 0.915$ OR $λ = 1$ (A1)

attempt to find eigenvectors for at least one eigenvalue (M1)

when $λ = 0.915, x = (\begin{matrix} 1 \\ - 1 \end{matrix})$ (or any real multiple) (A1)

when $λ = 1, x = (\begin{matrix} 10 \\ 7 \end{matrix})$ (or any real multiple) (A1)

therefore $P = (\begin{matrix} 1 & 10 \\ - 1 & 7 \end{matrix})$ (accept integer valued multiples of their eigenvectors and columns in either order) A1

[6 marks]

$P^{- 1} = {(\begin{matrix} 1 & 10 \\ - 1 & 7 \end{matrix})}^{- 1} = \frac{1}{17} (\begin{matrix} 7 & - 10 \\ 1 & 1 \end{matrix})$ (A1)

Note: This mark is independent, and may be seen anywhere in part (d).

$D = (\begin{matrix} 0.915 & 0 \\ 0 & 1 \end{matrix})$ (A1)

$T^{n} = P D^{n} P^{- 1} = (\begin{matrix} 1 & 10 \\ - 1 & 7 \end{matrix}) (\begin{matrix} 0 . 915^{n} & 0 \\ 0 & 1^{n} \end{matrix}) \frac{1}{17} (\begin{matrix} 7 & - 10 \\ 1 & 1 \end{matrix})$ (M1)A1

Note: Award (M1)A0 for finding $P^{- 1} D^{n} P$ correctly.

as $n \to \infty, D^{n} = (\begin{matrix} 0 . 915^{n} & 0 \\ 0 & 1^{n} \end{matrix}) \to (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ R1

so $T^{n} \to \frac{1}{17} (\begin{matrix} 1 & 10 \\ - 1 & 7 \end{matrix}) (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} 7 & - 10 \\ 1 & 1 \end{matrix})$ A1

$= (\begin{matrix} \frac{10}{17} & \frac{10}{17} \\ \frac{7}{17} & \frac{7}{17} \end{matrix})$ AG

Note: The AG line must be seen for the final A1 to be awarded.

[6 marks]

METHOD ONE

$(\begin{matrix} \frac{10}{17} & \frac{10}{17} \\ \frac{7}{17} & \frac{7}{17} \end{matrix}) (\begin{matrix} 2100 \\ 3500 \end{matrix}) = (\begin{matrix} 3294 \\ 2306 \end{matrix})$ (M1)