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Date	May 2022	Marks available	2	Reference code	22M.1.AHL.TZ1.11
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	11	Adapted from	N/A

Question

The matrix $M = (\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix})$ has eigenvalues $- 5$ and $1$ .

A switch has two states, $A$ and $B$ . Each second it either remains in the same state or moves according to the following rule: If it is in state $A$ it will move to state $B$ with a probability of $0.8$ and if it is in state $B$ it will move to state $A$ with a probability of $0.7$ .

Find an eigenvector corresponding to the eigenvalue of $1$ . Give your answer in the form $(\begin{matrix} a \\ b \end{matrix})$ , where $a, b \in ℤ$ .

[3]

Using your answer to (a), or otherwise, find the long-term probability of the switch being in state $A$ . Give your answer in the form $\frac{c}{d}$ , where $c, d \in ℤ^{+}$ .

[2]

Markscheme

$λ = 1$

$(\begin{matrix} - 0.8 & 0.7 \\ 0.8 & - 0.7 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ OR $(\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ y \end{matrix})$ (M1)

$0.8 x = 0.7 y$ (A1)

an eigenvector is $(\begin{matrix} 7 \\ 8 \end{matrix})$ (or equivalent with integer values) A1

[3 marks]

EITHER

(the long-term probability matrix is given by the eigenvector corresponding to the eigenvalue equal to $1$ , scaled so that the sum of the entries is $1$ )

$8 + 7 = 15$ (M1)

$(\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix}) (\begin{matrix} p \\ 1 - p \end{matrix}) = (\begin{matrix} p \\ 1 - p \end{matrix})$ (M1)

considering high powers of the matrix e.g. ${(\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix})}^{50}$ (M1)

$(\begin{matrix} \frac{7}{15} & \frac{7}{15} \\ \frac{8}{15} & \frac{8}{15} \end{matrix})$

THEN

probability of being in state $A$ is $\frac{7}{15}$ A1

[2 marks]

Examiners report

In part (a), some candidates could correctly use either $(A - λ I) x = 0$ or $A x = λ x$ to find an eigenvector but many did not pay attention to the fact that integer values of the eigenvector were required. Some candidates used the method of finding the steady state by finding $A^{n}$ for some high value of n in part (b) but ignored the fact that they needed to express their answer in rational form. Some did try to convert their calculated answer of 0.467 to $\frac{467}{1000}$ but this could only receive partial credit as an exact answer was required.

[N/A]

Syllabus sections

Topic 4—Statistics and probability » AHL 4.19—Transition matrices – Markov chains

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EXM.3.AHL.TZ0.4f:
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