Find the exact eigenvalues of the matrix $A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$.

1. Set up the characteristic equation by setting the determinant of $(A - \lambda I)$ to zero: $\det\begin{pmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{pmatrix} = 0$. 2. Expand the determinant: $(4-\lambda)(3-\lambda) - (2)(1) = 0$. 3. Simplify the resulting quadratic equation: $12 - 4\lambda - 3\lambda + \lambda^2 - 2 = 0 \implies \lambda^2 - 7\lambda + 10 = 0$. 4. Factorise the quadratic: $(\lambda - 5)(\lambda - 2) = 0$. 5. The exact eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = 2$ .

Matrix $M = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$. Show algebraically that $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ is an eigenvector of $M$, and state its corresponding eigenvalue.

1. Multiply the matrix $M$ by the vector $\mathbf{v}$: $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. 2. Calculate the top row: $3(1) + 1(1) = 4$. 3. Calculate the bottom row: $1(1) + 3(1) = 4$. 4. Express the resulting vector as a scalar multiple of the original vector: $\begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. 5. Since $M\mathbf{v} = \lambda\mathbf{v}$, it is verified as an eigenvector with an eigenvalue of $\lambda = 4$ .

A system is modelled by a Markov chain with the transition matrix $T = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}$. Find the exact steady state probability vector $\mathbf{s} = \begin{pmatrix} x \\ y \end{pmatrix}$.

1. Recall that at the steady state, multiplying by the transition matrix leaves the state unchanged: $T\mathbf{s} = \mathbf{s}$. 2. Set up the matrix equation: $\begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$. 3. Extract the first linear equation: $0.8x + 0.3y = x \implies 0.3y = 0.2x \implies 3y = 2x$. 4. Apply the fundamental property of probability vectors, which states that the sum of the states must equal $1$: $x + y = 1 \implies y = 1 - x$. 5. Substitute and solve: $3(1 - x) = 2x \implies 3 - 3x = 2x \implies 5x = 3 \implies x = 0.6$. Consequently, $y = 0.4$. The steady state vector is $\begin{pmatrix} 0.6 \\ 0.4 \end{pmatrix}$ .

Eigenvalues & Markov Chains Questions

What's examined in AI HL eigenvalues, eigenvectors and Markov chains

The question bank covers these eigenvalues, eigenvectors and Markov chains question types (number of questions in brackets):

Markov Chain Modelling (20)
Eigenvalue/Eigenvector Calculation (5)
Matrix Diagonalisation (2)

Eigenvalues, Eigenvectors and Markov Chains worked examples

Worked example 1: Finding Eigenvalues of a 2x2 Matrix · easy

Find the exact eigenvalues of the matrix $A = (\begin{matrix} 4 & 2 \\ 1 & 3 \end{matrix})$ .

Solution

5 steps

Examiner tip: You can quickly verify your eigenvalues are correct by checking that their sum equals the trace (sum of the main diagonal elements) of the original matrix.

Worked example 2: Verifying an Eigenvector · medium

Matrix $M = (\begin{matrix} 3 & 1 \\ 1 & 3 \end{matrix})$ . Show algebraically that $v = (\begin{matrix} 1 \\ 1 \end{matrix})$ is an eigenvector of $M$ , and state its corresponding eigenvalue.

Solution

5 steps · 1 to fill in

Examiner tip: Eigenvectors represent the invariant directions of a linear transformation, where the matrix multiplication merely scales the vector by the eigenvalue

λ

.

Worked example 3: Finding the Steady State of a Markov Chain · hard

A system is modelled by a Markov chain with the transition matrix $T = (\begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix})$ . Find the exact steady state probability vector $s = (\begin{matrix} x \\ y \end{matrix})$ .

Solution

5 steps

Examiner tip: The long-term steady state of a regular Markov chain effectively corresponds to the normalised eigenvector associated with the eigenvalue

λ = 1

.

Try these IB Maths AI HL eigenvalues, eigenvectors and Markov chains questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area. One of them has its full worked solution here, open to everyone.

Question 1 · easy · 4 marks · Paper 1

The weather in a specific town can only be Sunny (S) or Rainy (R). If it is Sunny today, there is an $80 %$ chance it will be Sunny tomorrow. If it is Rainy today, there is a $60 %$ chance it will be Rainy tomorrow.
(a) Construct a $2 \times 2$ transition matrix $T$ to represent this information, ordering the columns and rows as S, then R. [2 marks]
(b) If it is Rainy on Monday, calculate the exact probability that it will be Sunny on Wednesday. [2 marks]

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Question 2 · medium · 5 marks · Paper 1

A biologist is modelling a population of insects using a Leslie matrix. The population is divided into two age groups: Juveniles (J) and Adults (A). Each month, $40 %$ of Juveniles survive to become Adults. Each month, every Adult produces an average of 3 new Juveniles, and $50 %$ of the Adults survive to the next month.
(a) Write down the $2 \times 2$ Leslie matrix $L$ for this population. [2 marks]
(b) The initial population consists of $100$ Juveniles and $50$ Adults. Calculate the exact population of each age group after 2 months. [3 marks]

Full worked solution Free, no sign-in

(a) The Leslie Matrix $L$ takes the form $(\begin{matrix} F_{J} & F_{A} \\ S_{J} & S_{A} \end{matrix})$ where $F$ is fecundity and $S$ is survival.
Juveniles don’t reproduce: $F_{J} = 0$ . Adults produce 3: $F_{A} = 3$ .
Juvenile survival: $S_{J} = 0.4$ . Adult survival: $S_{A} = 0.5$ .
$L = (\begin{matrix} 0 & 3 \\ 0.4 & 0.5 \end{matrix})$ . [A1][A1]
(b) Evaluate $V_{2} = L^{2} V_{0}$ .
$V_{0} = (\begin{matrix} 100 \\ 50 \end{matrix})$ .
$V_{1} = L V_{0} = (\begin{matrix} 0 & 3 \\ 0.4 & 0.5 \end{matrix}) (\begin{matrix} 100 \\ 50 \end{matrix}) = (\begin{matrix} 150 \\ 65 \end{matrix})$ . [M1][A1]
$V_{2} = L V_{1} = (\begin{matrix} 0 & 3 \\ 0.4 & 0.5 \end{matrix}) (\begin{matrix} 150 \\ 65 \end{matrix}) = (\begin{matrix} 195 \\ 92.5 \end{matrix})$ .
Population: $195$ Juveniles, $92.5$ Adults. [A1]

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Question 3 · hard · 7 marks · Paper 1

A system's state $(x_{n}, y_{n})$ evolves over discrete time steps according to the transformation $T (x_{n}, y_{n}) = (0.8 x_{n} + 0.1 y_{n}, 0.2 x_{n} + 0.9 y_{n})$ . Let $M$ be the matrix associated with this transformation.

Write down the matrix $M$ . [1 mark]
Find the eigenvalues of $M$ . [2 marks]
Given the initial state $(x_{0}, y_{0}) = (300, 0)$ , determine the state of the system after a very long time. [4 marks]

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FAQ

How many IB Maths AI HL eigenvalues, eigenvectors and Markov chains questions are there?

There are 27 exam-style eigenvalues, eigenvectors and Markov chains questions in the AI HL question bank (Paper 1: 27), graded 6 easy, 11 medium, 10 hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is eigenvalues, eigenvectors and Markov chains on Paper 1 or Paper 2?

In the question bank these questions are set as Paper 1 questions.

Where can I get the mark schemes?

Open the AI HL Unit 1 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

IB Maths AI HL Eigenvalues, Eigenvectors and Markov Chains Questions

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What's examined in AI HL eigenvalues, eigenvectors and Markov chains

Eigenvalues, Eigenvectors and Markov Chains worked examples

Worked example 1: Finding Eigenvalues of a 2x2 Matrix · easy

Solution

Worked example 2: Verifying an Eigenvector · medium

Solution

Worked example 3: Finding the Steady State of a Markov Chain · hard

Solution

Try these IB Maths AI HL eigenvalues, eigenvectors and Markov chains questions

Question 1 · easy · 4 marks · Paper 1

Question 2 · medium · 5 marks · Paper 1

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Question 3 · hard · 7 marks · Paper 1

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