IB Mathematics AHL 1.15 Eigenvalues and eigenvectors AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question investigates methods for predicting future international football match outcomes using historical data.
Two friends, Somsak and Naree, discuss approaches to predicting outcomes of international football matches involving Argentina.
Somsak proposes using historical data to predict outcomes, analyzing the results of Argentina’s most recent 240 matches, grouped into 60 blocks of four matches each. He counts the number of wins per block, with results shown below.
| Number of wins for Argentina (per block of four matches) | Frequency |
|---|---|
| 0 | 0 |
| 1 | 11 |
| 2 | 21 |
| 3 | 21 |
| 4 | 7 |
(a) Calculate the mean number of wins per block of four matches for Argentina. [2]
Somsak believes the data follows a binomial distribution with \( n = 4 \) and conducts a \(\chi^2\) goodness of fit test.
(b) Estimate the probability \( p \) for the binomial model using Somsak’s data. [1]
(c)(i) Calculate the probability that Argentina wins zero matches in a block of four matches using the binomial model. [1]
(c)(ii) Calculate the expected frequency for zero wins in 60 blocks. [2]
Since some expected frequencies are below 5, Somsak combines categories to produce the following observed and expected frequencies, with expected values rounded to one decimal place.
| Number of wins for Argentina (per block of four matches) | Observed frequency | Expected frequency |
|---|---|---|
| 0 or 1 | 11 | 10.8 |
| 2 | 21 | 20.7 |
| 3 | 21 | 20.7 |
| 4 | 7 | 7.8 |
Somsak performs a \(\chi^2\) goodness of fit test at the 5% significance level to test if the data follows a binomial distribution with \( n = 4 \).
(d) For this test, state:
(i) the null hypothesis; [1]
(ii) the number of degrees of freedom; [1]
(iii) the \( p \)-value; [2]
(iv) the conclusion, justifying your answer. [2]
(i) the null hypothesis; [1]
(ii) the number of degrees of freedom; [1]
(iii) the \( p \)-value; [2]
(iv) the conclusion, justifying your answer. [2]
(e) Using Somsak’s binomial model, calculate the probability that Argentina wins at least one of their next four matches. [2]
Naree suggests a prediction model based on transitions between match outcomes, using two states: Argentina won (A) or did not win (B). She counts transitions in Somsak’s data: AA (win followed by win), AB (win followed by no win), BA (no win followed by win), BB (no win followed by no win). Her results are shown below.
| Transition | Frequency |
|---|---|
| AA | 85 |
| AB | 60 |
| BA | 62 |
| BB | 32 |
Naree uses relative frequencies to estimate probabilities for a transition matrix.
(f)(i) Show that Naree’s estimate for the probability of Argentina winning the next match, given a previous win, is \( \frac{17}{29} \). [2]
(f)(ii) Write down the transition matrix \( T \) for Naree’s model. [2]
(g)(i) Demonstrate that the characteristic polynomial of \( T \) is \( 1363\lambda^2 – 1263\lambda – 100 = 0 \). [3]
(g)(ii) Find the eigenvalues of \( T \). [1]
(g)(iii) Determine the eigenvectors corresponding to the eigenvalues. [3]
In the distant future, Naree plans to attend three consecutive Argentina matches. Use Naree’s model to calculate the probability that Argentina wins all three matches.
(h) Calculate the probability that Argentina wins all three matches. [4]
▶️ Answer/Explanation
Markscheme
(a)
Calculate the mean number of wins per block (4 matches):
\( \bar{x} = \frac{0 \cdot 0 + 1 \cdot 11 + 2 \cdot 21 + 3 \cdot 21 + 4 \cdot 7}{60} = \frac{0 + 11 + 42 + 63 + 28}{60} = \frac{144}{60} = 2.4 \).
Mean = 2.4 M1A1
[2 marks]
Calculate the mean number of wins per block (4 matches):
\( \bar{x} = \frac{0 \cdot 0 + 1 \cdot 11 + 2 \cdot 21 + 3 \cdot 21 + 4 \cdot 7}{60} = \frac{0 + 11 + 42 + 63 + 28}{60} = \frac{144}{60} = 2.4 \).
Mean = 2.4 M1A1
[2 marks]
(b)
For \( X \sim \text{Bin}(4, p) \), the expected value is \( \mathbb{E}[X] = np \). Given \( \bar{x} = 2.4 \), \( n = 4 \):
\( \hat{p} = \frac{2.4}{4} = 0.6 = \frac{3}{5} \).
\( p = 0.6 \) A1
[1 mark]
For \( X \sim \text{Bin}(4, p) \), the expected value is \( \mathbb{E}[X] = np \). Given \( \bar{x} = 2.4 \), \( n = 4 \):
\( \hat{p} = \frac{2.4}{4} = 0.6 = \frac{3}{5} \).
\( p = 0.6 \) A1
[1 mark]
(c)(i)
For \( X \sim \text{Bin}(4, 0.6) \), probability of zero wins:
\( P(X = 0) = (1 – 0.6)^4 = 0.4^4 = 0.0256 \).
\( P(X = 0) = 0.0256 \) A1
[1 mark]
For \( X \sim \text{Bin}(4, 0.6) \), probability of zero wins:
\( P(X = 0) = (1 – 0.6)^4 = 0.4^4 = 0.0256 \).
\( P(X = 0) = 0.0256 \) A1
[1 mark]
(c)(ii)
Expected frequency for zero wins over 60 blocks:
\( E_0 = 60 \times 0.0256 = 1.536 \).
Expected frequency = 1.536 M1A1
[2 marks]
Expected frequency for zero wins over 60 blocks:
\( E_0 = 60 \times 0.0256 = 1.536 \).
Expected frequency = 1.536 M1A1
[2 marks]
(d)
(i) Null hypothesis:
\( H_0 \): Data follows \( \text{Bin}(4, 0.6) \), \( H_1 \): Data does not follow \( \text{Bin}(4, 0.6) \).
\( H_0 \): Binomial distribution with \( n = 4 \), \( p = 0.6 \) A1
\( H_0 \): Data follows \( \text{Bin}(4, 0.6) \), \( H_1 \): Data does not follow \( \text{Bin}(4, 0.6) \).
\( H_0 \): Binomial distribution with \( n = 4 \), \( p = 0.6 \) A1
(ii) Degrees of freedom: 4 categories (0 or 1, 2, 3, 4), 1 parameter estimated (\( p \)):
\( \nu = 4 – 1 – 1 = 2 \).
Degrees of freedom = 2 A1
\( \nu = 4 – 1 – 1 = 2 \).
Degrees of freedom = 2 A1
(iii) \(\chi^2\) statistic:
Observed: \( (11, 21, 21, 7) \), Expected: \( (10.8, 20.7, 20.7, 7.8) \).
\( \chi^2 = \sum \frac{(O – E)^2}{E} = \frac{(11 – 10.8)^2}{10.8} + \frac{(21 – 20.7)^2}{20.7} + \frac{(21 – 20.7)^2}{20.7} + \frac{(7 – 7.8)^2}{7.8} \approx 0.003704 + 0.004348 + 0.004348 + 0.082051 \approx 0.0945 \).
For \( \nu = 2 \), \( p \)-value: \( P(\chi^2_{(2)} \geq 0.0945) \approx 0.954 \).
\( p \)-value \( \approx 0.954 \) M1A1
Observed: \( (11, 21, 21, 7) \), Expected: \( (10.8, 20.7, 20.7, 7.8) \).
\( \chi^2 = \sum \frac{(O – E)^2}{E} = \frac{(11 – 10.8)^2}{10.8} + \frac{(21 – 20.7)^2}{20.7} + \frac{(21 – 20.7)^2}{20.7} + \frac{(7 – 7.8)^2}{7.8} \approx 0.003704 + 0.004348 + 0.004348 + 0.082051 \approx 0.0945 \).
For \( \nu = 2 \), \( p \)-value: \( P(\chi^2_{(2)} \geq 0.0945) \approx 0.954 \).
\( p \)-value \( \approx 0.954 \) M1A1
(iv) Conclusion: Since \( p \approx 0.954 > 0.05 \), do not reject \( H_0 \). The data is consistent with a binomial distribution with \( n = 4 \), \( p = 0.6 \).
Do not reject \( H_0 \); binomial model fits A1A1
[6 marks]Do not reject \( H_0 \); binomial model fits A1A1
(e)
For \( X \sim \text{Bin}(4, 0.6) \), probability of at least one win:
\( P(X \geq 1) = 1 – P(X = 0) = 1 – (0.4)^4 = 1 – 0.0256 = 0.9744 \approx 0.974 \).
Probability \( \approx 0.974 \) M1A1
[2 marks]
For \( X \sim \text{Bin}(4, 0.6) \), probability of at least one win:
\( P(X \geq 1) = 1 – P(X = 0) = 1 – (0.4)^4 = 1 – 0.0256 = 0.9744 \approx 0.974 \).
Probability \( \approx 0.974 \) M1A1
[2 marks]
(f)(i)
Transitions from A: \( AA = 85 \), \( AB = 60 \), total = \( 85 + 60 = 145 \).
Probability of winning given previous win:
\( P(A \to A) = \frac{AA}{AA + AB} = \frac{85}{145} = \frac{85 \div 5}{145 \div 5} = \frac{17}{29} \approx 0.586 \).
\( P(A \to A) = \frac{17}{29} \) M1A1
[2 marks]
Transitions from A: \( AA = 85 \), \( AB = 60 \), total = \( 85 + 60 = 145 \).
Probability of winning given previous win:
\( P(A \to A) = \frac{AA}{AA + AB} = \frac{85}{145} = \frac{85 \div 5}{145 \div 5} = \frac{17}{29} \approx 0.586 \).
\( P(A \to A) = \frac{17}{29} \) M1A1
[2 marks]
(f)(ii)
Transitions from B: \( BA = 62 \), \( BB = 32 \), total = \( 62 + 32 = 94 \).
\( P(B \to A) = \frac{BA}{BA + BB} = \frac{62}{94} = \frac{31}{47} \approx 0.660 \).
\( P(A \to B) = \frac{12}{29} \approx 0.414 \), \( P(B \to B) = \frac{16}{47} \approx 0.340 \).
Column-stochastic transition matrix (current state: A, B; next state: A, B):
\( T = \begin{bmatrix} \frac{17}{29} & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} \end{bmatrix} \approx \begin{bmatrix} 0.586 & 0.660 \\ 0.414 & 0.340 \end{bmatrix} \).
\( T = \begin{bmatrix} \frac{17}{29} & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} \end{bmatrix} \) A1A1
[2 marks]
Transitions from B: \( BA = 62 \), \( BB = 32 \), total = \( 62 + 32 = 94 \).
\( P(B \to A) = \frac{BA}{BA + BB} = \frac{62}{94} = \frac{31}{47} \approx 0.660 \).
\( P(A \to B) = \frac{12}{29} \approx 0.414 \), \( P(B \to B) = \frac{16}{47} \approx 0.340 \).
Column-stochastic transition matrix (current state: A, B; next state: A, B):
\( T = \begin{bmatrix} \frac{17}{29} & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} \end{bmatrix} \approx \begin{bmatrix} 0.586 & 0.660 \\ 0.414 & 0.340 \end{bmatrix} \).
\( T = \begin{bmatrix} \frac{17}{29} & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} \end{bmatrix} \) A1A1
[2 marks]
(g)(i)
For \( T = \begin{bmatrix} \frac{17}{29} & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} \end{bmatrix} \), compute characteristic polynomial:
\( \det(T – \lambda I) = \det \begin{bmatrix} \frac{17}{29} – \lambda & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} – \lambda \end{bmatrix} = \left( \frac{17}{29} – \lambda \right) \left( \frac{16}{47} – \lambda \right) – \frac{12}{29} \cdot \frac{31}{47} \).
Compute: \( \left( \frac{17}{29} – \lambda \right) \left( \frac{16}{47} – \lambda \right) = \lambda^2 – \left( \frac{17}{29} + \frac{16}{47} \right) \lambda + \frac{17 \cdot 16}{29 \cdot 47} \).
\( \frac{17}{29} + \frac{16}{47} = \frac{17 \cdot 47 + 16 \cdot 29}{29 \cdot 47} = \frac{799 + 464}{1363} = \frac{1263}{1363} \).
\( \frac{17 \cdot 16}{29 \cdot 47} = \frac{272}{1363} \), \( \frac{12 \cdot 31}{29 \cdot 47} = \frac{372}{1363} \).
\( \det(T – \lambda I) = \lambda^2 – \frac{1263}{1363} \lambda + \frac{272}{1363} – \frac{372}{1363} = \lambda^2 – \frac{1263}{1363} \lambda – \frac{100}{1363} \).
Multiply by 1363: \( 1363 \lambda^2 – 1263 \lambda – 100 = 0 \).
Characteristic polynomial: \( 1363 \lambda^2 – 1263 \lambda – 100 = 0 \) M1A1A1
For \( T = \begin{bmatrix} \frac{17}{29} & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} \end{bmatrix} \), compute characteristic polynomial:
\( \det(T – \lambda I) = \det \begin{bmatrix} \frac{17}{29} – \lambda & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} – \lambda \end{bmatrix} = \left( \frac{17}{29} – \lambda \right) \left( \frac{16}{47} – \lambda \right) – \frac{12}{29} \cdot \frac{31}{47} \).
Compute: \( \left( \frac{17}{29} – \lambda \right) \left( \frac{16}{47} – \lambda \right) = \lambda^2 – \left( \frac{17}{29} + \frac{16}{47} \right) \lambda + \frac{17 \cdot 16}{29 \cdot 47} \).
\( \frac{17}{29} + \frac{16}{47} = \frac{17 \cdot 47 + 16 \cdot 29}{29 \cdot 47} = \frac{799 + 464}{1363} = \frac{1263}{1363} \).
\( \frac{17 \cdot 16}{29 \cdot 47} = \frac{272}{1363} \), \( \frac{12 \cdot 31}{29 \cdot 47} = \frac{372}{1363} \).
\( \det(T – \lambda I) = \lambda^2 – \frac{1263}{1363} \lambda + \frac{272}{1363} – \frac{372}{1363} = \lambda^2 – \frac{1263}{1363} \lambda – \frac{100}{1363} \).
Multiply by 1363: \( 1363 \lambda^2 – 1263 \lambda – 100 = 0 \).
Characteristic polynomial: \( 1363 \lambda^2 – 1263 \lambda – 100 = 0 \) M1A1A1
(g)(ii)
Solve: \( 1363 \lambda^2 – 1263 \lambda – 100 = 0 \).
\( \lambda = \frac{1263 \pm \sqrt{1263^2 + 4 \cdot 1363 \cdot 100}}{2 \cdot 1363} \).
Discriminant: \( 1263^2 + 4 \cdot 1363 \cdot 100 = 1595169 + 545200 = 2140369 = 1463^2 \).
\( \lambda = \frac{1263 \pm 1463}{2726} \).
\( \lambda_1 = \frac{1263 + 1463}{2726} = \frac{2726}{2726} = 1 \), \( \lambda_2 = \frac{1263 – 1463}{2726} = -\frac{200}{2726} = -\frac{100}{1363} \approx -0.07337 \).
Eigenvalues: \( \lambda_1 = 1 \), \( \lambda_2 = -\frac{100}{1363} \) A1
Solve: \( 1363 \lambda^2 – 1263 \lambda – 100 = 0 \).
\( \lambda = \frac{1263 \pm \sqrt{1263^2 + 4 \cdot 1363 \cdot 100}}{2 \cdot 1363} \).
Discriminant: \( 1263^2 + 4 \cdot 1363 \cdot 100 = 1595169 + 545200 = 2140369 = 1463^2 \).
\( \lambda = \frac{1263 \pm 1463}{2726} \).
\( \lambda_1 = \frac{1263 + 1463}{2726} = \frac{2726}{2726} = 1 \), \( \lambda_2 = \frac{1263 – 1463}{2726} = -\frac{200}{2726} = -\frac{100}{1363} \approx -0.07337 \).
Eigenvalues: \( \lambda_1 = 1 \), \( \lambda_2 = -\frac{100}{1363} \) A1
(g)(iii)
For \( \lambda_1 = 1 \): Solve \( (T – I)v = 0 \).
\( T – I = \begin{bmatrix} \frac{17}{29} – 1 & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} – 1 \end{bmatrix} = \begin{bmatrix} -\frac{12}{29} & \frac{31}{47} \\ \frac{12}{29} & -\frac{31}{47} \end{bmatrix} \).
Row reduce: \( \frac{12}{29} v_1 – \frac{31}{47} v_2 = 0 \Rightarrow v_1 = \frac{31 \cdot 29}{47 \cdot 12} v_2 = \frac{899}{564} v_2 \).
Choose \( v_2 = 564 \): \( v_1 = 899 \). Eigenvector: \( \begin{bmatrix} 899 \\ 564 \end{bmatrix} \sim \begin{bmatrix} 1 \\ \frac{564}{899} \end{bmatrix} \approx \begin{bmatrix} 1 \\ 0.627 \end{bmatrix} \).
For \( \lambda_2 = -\frac{100}{1363} \): Solve \( (T – \lambda_2 I)v = 0 \).
Choose simple eigenvector: \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \).
Eigenvectors: \( \begin{bmatrix} 1 \\ \frac{564}{899} \end{bmatrix} \), \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) M1A1A1
[7 marks]
For \( \lambda_1 = 1 \): Solve \( (T – I)v = 0 \).
\( T – I = \begin{bmatrix} \frac{17}{29} – 1 & \frac{31}{47} \\ \frac{12}{29} & \frac{16}{47} – 1 \end{bmatrix} = \begin{bmatrix} -\frac{12}{29} & \frac{31}{47} \\ \frac{12}{29} & -\frac{31}{47} \end{bmatrix} \).
Row reduce: \( \frac{12}{29} v_1 – \frac{31}{47} v_2 = 0 \Rightarrow v_1 = \frac{31 \cdot 29}{47 \cdot 12} v_2 = \frac{899}{564} v_2 \).
Choose \( v_2 = 564 \): \( v_1 = 899 \). Eigenvector: \( \begin{bmatrix} 899 \\ 564 \end{bmatrix} \sim \begin{bmatrix} 1 \\ \frac{564}{899} \end{bmatrix} \approx \begin{bmatrix} 1 \\ 0.627 \end{bmatrix} \).
For \( \lambda_2 = -\frac{100}{1363} \): Solve \( (T – \lambda_2 I)v = 0 \).
Choose simple eigenvector: \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \).
Eigenvectors: \( \begin{bmatrix} 1 \\ \frac{564}{899} \end{bmatrix} \), \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) M1A1A1
[7 marks]
(h)
In steady state, \( \pi = T \pi \), \( \pi_A + \pi_B = 1 \). Use eigenvector for \( \lambda = 1 \): \( \begin{bmatrix} 1 \\ \frac{564}{899} \end{bmatrix} \).
Normalize: \( \pi_A = \frac{1}{1 + \frac{564}{899}} = \frac{899}{899 + 564} = \frac{899}{1463} \approx 0.614 \).
\( \pi_B = 1 – \frac{899}{1463} = \frac{564}{1463} \approx 0.386 \).
Probability of three consecutive wins: Start in A (\( \pi_A \)) and two \( A \to A \) transitions:
\( P(\text{3 wins}) = \pi_A \cdot \left( \frac{17}{29} \right)^2 = \frac{899}{1463} \cdot \frac{289}{841} = \frac{899 \cdot 289}{1463 \cdot 841} \approx 0.211 \).
Probability \( \approx 0.211 \) M1A1A1A1
[4 marks]
In steady state, \( \pi = T \pi \), \( \pi_A + \pi_B = 1 \). Use eigenvector for \( \lambda = 1 \): \( \begin{bmatrix} 1 \\ \frac{564}{899} \end{bmatrix} \).
Normalize: \( \pi_A = \frac{1}{1 + \frac{564}{899}} = \frac{899}{899 + 564} = \frac{899}{1463} \approx 0.614 \).
\( \pi_B = 1 – \frac{899}{1463} = \frac{564}{1463} \approx 0.386 \).
Probability of three consecutive wins: Start in A (\( \pi_A \)) and two \( A \to A \) transitions:
\( P(\text{3 wins}) = \pi_A \cdot \left( \frac{17}{29} \right)^2 = \frac{899}{1463} \cdot \frac{289}{841} = \frac{899 \cdot 289}{1463 \cdot 841} \approx 0.211 \).
Probability \( \approx 0.211 \) M1A1A1A1
[4 marks]
Total Marks: 28