IB · MATH AI HL

Mathematics: Applications & Interpretation HL

Matrices and Markov chains — Topic 1 HL

Name: ____________________Date: October 2, 2026

1.
Given A = [[2, 3], [−1, 4]] and B = [[5, −2], [0, 3]], find (a) A + B and (b) A − B.
[2 marks]
Marking points
- Adds corresponding entries to obtain A + B = [[7, 1], [−1, 7]].
- Subtracts corresponding entries to obtain A − B = [[−3, 5], [−1, 1]].
Examiner tip: Matrix addition and subtraction only combine entries in the same position — the matrices must have identical dimensions for this to be defined.
2.
Marking analysis: A learner attempts the following task: “Given A = [[2, 3], [−1, 4]] and B = [[5, −2], [0, 3]], find (a) A + B and (b) A − B.” Their response addresses only this point: “Adds corresponding entries to obtain A + B = [[7, 1], [−1, 7]].” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]
Marking points
- Recognises credit for the stated point: Adds corresponding entries to obtain A + B = [[7, 1], [−1, 7]].
- Identifies the missing requirement: Subtracts corresponding entries to obtain A − B = [[−3, 5], [−1, 1]].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
Given A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 3]], find the matrix product AB.
[3 marks]
Marking points
- Multiplies row 1 of A by each column of B: (1×2 + 2×1, 1×0 + 2×3) = (4, 6).
- Multiplies row 2 of A by each column of B: (3×2 + 4×1, 3×0 + 4×3) = (10, 12).
- States AB = [[4, 6], [10, 12]].
Examiner tip: Remember that matrix multiplication is generally not commutative — AB and BA can give different results, so always keep track of the order.
4.
Marking analysis: A learner attempts the following task: “Given A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 3]], find the matrix product AB.” Their response addresses only this point: “Multiplies row 1 of A by each column of B: (1×2 + 2×1, 1×0 + 2×3) = (4, 6).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Multiplies row 1 of A by each column of B: (1×2 + 2×1, 1×0 + 2×3) = (4, 6).
- Identifies the missing requirement: Multiplies row 2 of A by each column of B: (3×2 + 4×1, 3×0 + 4×3) = (10, 12).
- Identifies the missing requirement: States AB = [[4, 6], [10, 12]].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
Find the determinant of the matrix A = [[4, 7], [2, 6]].
[2 marks]
Marking points
- Uses det(A) = ad − bc = (4)(6) − (7)(2).
- Obtains det(A) = 10.
Examiner tip: For a 2×2 matrix [[a, b], [c, d]], the determinant ad − bc is always computed as the product of the leading diagonal minus the product of the other diagonal.
6.
Marking analysis: A learner attempts the following task: “Find the determinant of the matrix A = [[4, 7], [2, 6]].” Their response addresses only this point: “Uses det(A) = ad − bc = (4)(6) − (7)(2).” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]
Marking points
- Recognises credit for the stated point: Uses det(A) = ad − bc = (4)(6) − (7)(2).
- Identifies the missing requirement: Obtains det(A) = 10.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
Find the inverse of the matrix A = [[3, 5], [1, 2]].
[3 marks]
Marking points
- Calculates det(A) = (3)(2) − (5)(1) = 1.
- Uses the inverse formula A⁻¹ = (1/det(A))[[d, −b], [−c, a]].
- Obtains A⁻¹ = [[2, −5], [−1, 3]].
Examiner tip: Swap the leading diagonal entries, negate the other two entries, then divide every entry by the determinant — a matrix only has an inverse when its determinant is non-zero.
8.
Marking analysis: A learner attempts the following task: “Find the inverse of the matrix A = [[3, 5], [1, 2]].” Their response addresses only this point: “Calculates det(A) = (3)(2) − (5)(1) = 1.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Calculates det(A) = (3)(2) − (5)(1) = 1.
- Identifies the missing requirement: Uses the inverse formula A⁻¹ = (1/det(A))[[d, −b], [−c, a]].
- Identifies the missing requirement: Obtains A⁻¹ = [[2, −5], [−1, 3]].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
Use a matrix method to solve the simultaneous equations 2x + 3y = 12 and x − y = 1.
[5 marks]
Marking points
- Writes the system in matrix form [[2, 3], [1, −1]][x, y]ᵀ = [12, 1]ᵀ.
- Calculates the determinant of the coefficient matrix: (2)(−1) − (3)(1) = −5.
- Finds the inverse of the coefficient matrix.
- Multiplies the inverse by the constants vector [12, 1]ᵀ.
- Obtains x = 3, y = 2.
Examiner tip: The matrix method for solving simultaneous equations works regardless of how many variables there are, as long as the coefficient matrix has a non-zero determinant — a GDC can perform the inverse and multiplication directly.
10.
Marking analysis: A learner attempts the following task: “Use a matrix method to solve the simultaneous equations 2x + 3y = 12 and x − y = 1.” Their response addresses only this point: “Writes the system in matrix form [[2, 3], [1, −1]][x, y]ᵀ = [12, 1]ᵀ.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Writes the system in matrix form [[2, 3], [1, −1]][x, y]ᵀ = [12, 1]ᵀ.
- Identifies the missing requirement: Calculates the determinant of the coefficient matrix: (2)(−1) − (3)(1) = −5.
- Identifies the missing requirement: Finds the inverse of the coefficient matrix.
- Identifies the missing requirement: Multiplies the inverse by the constants vector [12, 1]ᵀ.
- Identifies the missing requirement: Obtains x = 3, y = 2.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
Find the determinant of the 3×3 matrix A = [[1, 2, 3], [0, 1, 4], [5, 6, 0]].
[4 marks]
Marking points
- Expands along the first row: det(A) = 1(1×0 − 4×6) − 2(0×0 − 4×5) + 3(0×6 − 1×5).
- Evaluates each 2×2 minor: 1(−24) − 2(−20) + 3(−5).
- Simplifies to −24 + 40 − 15.
- Obtains det(A) = 1.
Examiner tip: Cofactor expansion along a row or column with the most zeros minimises the arithmetic — here the first row has no zeros, but a GDC's determinant function avoids this choice entirely.
12.
Marking analysis: A learner attempts the following task: “Find the determinant of the 3×3 matrix A = [[1, 2, 3], [0, 1, 4], [5, 6, 0]].” Their response addresses only this point: “Expands along the first row: det(A) = 1(1×0 − 4×6) − 2(0×0 − 4×5) + 3(0×6 − 1×5).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Expands along the first row: det(A) = 1(1×0 − 4×6) − 2(0×0 − 4×5) + 3(0×6 − 1×5).
- Identifies the missing requirement: Evaluates each 2×2 minor: 1(−24) − 2(−20) + 3(−5).
- Identifies the missing requirement: Simplifies to −24 + 40 − 15.
- Identifies the missing requirement: Obtains det(A) = 1.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
Given A = [[1, 2, 3], [4, 5, 6]] and B = [[7, 8], [9, 10], [11, 12]], find the matrix product AB, stating its dimensions.
[4 marks]
Marking points
- Confirms the product is defined since A is 2×3 and B is 3×2, so AB will be 2×2.
- Calculates the first row of AB: (1×7+2×9+3×11, 1×8+2×10+3×12) = (58, 64).
- Calculates the second row of AB: (4×7+5×9+6×11, 4×8+5×10+6×12) = (139, 154).
- States AB = [[58, 64], [139, 154]].
Examiner tip: Two matrices can only be multiplied if the number of columns in the first equals the number of rows in the second — always check this before attempting the multiplication.
14.
Marking analysis: A learner attempts the following task: “Given A = [[1, 2, 3], [4, 5, 6]] and B = [[7, 8], [9, 10], [11, 12]], find the matrix product AB, stating its dimensions.” Their response addresses only this point: “Confirms the product is defined since A is 2×3 and B is 3×2, so AB will be 2×2.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Confirms the product is defined since A is 2×3 and B is 3×2, so AB will be 2×2.
- Identifies the missing requirement: Calculates the first row of AB: (1×7+2×9+3×11, 1×8+2×10+3×12) = (58, 64).
- Identifies the missing requirement: Calculates the second row of AB: (4×7+5×9+6×11, 4×8+5×10+6×12) = (139, 154).
- Identifies the missing requirement: States AB = [[58, 64], [139, 154]].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
A machine is always in one of two states: Working (W) or Broken (B). If it is Working on a given day, the probability it is Working the next day is 0.8. If it is Broken, the probability it is Working the next day is 0.3. The machine is Working today. Use the transition matrix P = [[0.8, 0.2], [0.3, 0.7]] and the state vector [1, 0] to find the probability distribution for tomorrow.
[3 marks]
Marking points
- Multiplies the row state vector by the transition matrix: [1, 0] × P.
- Obtains the new state vector [0.8, 0.2].
- Interprets this as an 80% chance the machine is Working tomorrow and a 20% chance it is Broken.
Examiner tip: The row of the transition matrix corresponding to the current state gives next period's probabilities directly — multiplying by a state vector with a single 1 simply selects that row.
16.
Marking analysis: A learner attempts the following task: “A machine is always in one of two states: Working (W) or Broken (B). If it is Working on a given day, the probability it is Working the next day is 0.8. If it is Broken, the probability it is Working the next day is 0.3. The machine is Working today. Use the transition matrix P = [[0.8, 0.2], [0.3, 0.7]] and the state vector [1, 0] to find the probability distribution for tomorrow.” Their response addresses only this point: “Multiplies the row state vector by the transition matrix: [1, 0] × P.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Multiplies the row state vector by the transition matrix: [1, 0] × P.
- Identifies the missing requirement: Obtains the new state vector [0.8, 0.2].
- Identifies the missing requirement: Interprets this as an 80% chance the machine is Working tomorrow and a 20% chance it is Broken.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
Using the same machine from the previous question (transition matrix P = [[0.8, 0.2], [0.3, 0.7]], starting state [1, 0] for Working today), find the probability distribution for the day after tomorrow (two days ahead).
[4 marks]
Marking points
- Uses tomorrow's state vector [0.8, 0.2] found previously.
- Multiplies this state vector by P again: [0.8, 0.2] × P.
- Calculates the first entry: 0.8(0.8) + 0.2(0.3) = 0.70.
- Obtains the state vector for two days ahead as [0.70, 0.30].
Examiner tip: Applying the transition matrix repeatedly from the current state is equivalent to multiplying the original state vector by P raised to that power — both approaches give the same result.
18.
Marking analysis: A learner attempts the following task: “Using the same machine from the previous question (transition matrix P = [[0.8, 0.2], [0.3, 0.7]], starting state [1, 0] for Working today), find the probability distribution for the day after tomorrow (two days ahead).” Their response addresses only this point: “Uses tomorrow's state vector [0.8, 0.2] found previously.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Uses tomorrow's state vector [0.8, 0.2] found previously.
- Identifies the missing requirement: Multiplies this state vector by P again: [0.8, 0.2] × P.
- Identifies the missing requirement: Calculates the first entry: 0.8(0.8) + 0.2(0.3) = 0.70.
- Identifies the missing requirement: Obtains the state vector for two days ahead as [0.70, 0.30].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
For the transition matrix P = [[0.8, 0.2], [0.3, 0.7]], find the steady-state (long-run) probability distribution π = [π₁, π₂], using πP = π and π₁ + π₂ = 1.
[5 marks]
Marking points
- Sets up the steady-state equations from πP = π: 0.8π₁ + 0.3π₂ = π₁ and 0.2π₁ + 0.7π₂ = π₂.
- Simplifies the first equation to −0.2π₁ + 0.3π₂ = 0, giving π₂ = (2/3)π₁.
- Substitutes into π₁ + π₂ = 1 to obtain π₁(1 + 2/3) = 1.
- Solves to obtain π₁ = 0.6.
- Obtains π₂ = 0.4, giving the steady-state distribution [0.6, 0.4].
Examiner tip: The steady-state vector is the same regardless of the starting state, for a regular Markov chain — it depends only on the transition matrix itself.
20.
Marking analysis: A learner attempts the following task: “For the transition matrix P = [[0.8, 0.2], [0.3, 0.7]], find the steady-state (long-run) probability distribution π = [π₁, π₂], using πP = π and π₁ + π₂ = 1.” Their response addresses only this point: “Sets up the steady-state equations from πP = π: 0.8π₁ + 0.3π₂ = π₁ and 0.2π₁ + 0.7π₂ = π₂.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Sets up the steady-state equations from πP = π: 0.8π₁ + 0.3π₂ = π₁ and 0.2π₁ + 0.7π₂ = π₂.
- Identifies the missing requirement: Simplifies the first equation to −0.2π₁ + 0.3π₂ = 0, giving π₂ = (2/3)π₁.
- Identifies the missing requirement: Substitutes into π₁ + π₂ = 1 to obtain π₁(1 + 2/3) = 1.
- Identifies the missing requirement: Solves to obtain π₁ = 0.6.
- Identifies the missing requirement: Obtains π₂ = 0.4, giving the steady-state distribution [0.6, 0.4].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
For the transition matrix P = [[0.8, 0.2], [0.3, 0.7]], calculate P², and explain what the entries of P² represent in the context of the Markov chain.
[4 marks]
Marking points
- Multiplies P by itself to obtain P² = [[0.70, 0.30], [0.45, 0.55]].
- Explains that each entry (i, j) of P² gives the probability of moving from state i to state j in exactly two steps.
- Notes that the rows of P² still sum to 1, since the matrix still represents a valid probability distribution after two steps.
- Cross-checks that the top row of P² matches the two-step state vector found starting from [1, 0].
Examiner tip: Raising a transition matrix to the nth power always gives the n-step transition probabilities directly — this is often faster than repeatedly multiplying a state vector when you need several different starting states.
22.
Marking analysis: A learner attempts the following task: “For the transition matrix P = [[0.8, 0.2], [0.3, 0.7]], calculate P², and explain what the entries of P² represent in the context of the Markov chain.” Their response addresses only this point: “Multiplies P by itself to obtain P² = [[0.70, 0.30], [0.45, 0.55]].” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Multiplies P by itself to obtain P² = [[0.70, 0.30], [0.45, 0.55]].
- Identifies the missing requirement: Explains that each entry (i, j) of P² gives the probability of moving from state i to state j in exactly two steps.
- Identifies the missing requirement: Notes that the rows of P² still sum to 1, since the matrix still represents a valid probability distribution after two steps.
- Identifies the missing requirement: Cross-checks that the top row of P² matches the two-step state vector found starting from [1, 0].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
Solve the matrix equation AX = B for the column vector X, where A = [[2, 1], [1, 1]] and B = [[8], [5]], by finding A⁻¹ and using X = A⁻¹B.
[5 marks]
Marking points
- Calculates det(A) = (2)(1) − (1)(1) = 1.
- Finds A⁻¹ = [[1, −1], [−1, 2]].
- Multiplies X = A⁻¹B.
- Calculates the first entry: 1(8) + (−1)(5) = 3.
- Obtains X = [[3], [2]], so x = 3 and y = 2.
Examiner tip: Solving AX = B using the inverse matrix is mathematically identical to the elimination method for simultaneous equations — it is simply a more systematic, scalable way to organise the same process.
24.
Marking analysis: A learner attempts the following task: “Solve the matrix equation AX = B for the column vector X, where A = [[2, 1], [1, 1]] and B = [[8], [5]], by finding A⁻¹ and using X = A⁻¹B.” Their response addresses only this point: “Calculates det(A) = (2)(1) − (1)(1) = 1.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Calculates det(A) = (2)(1) − (1)(1) = 1.
- Identifies the missing requirement: Finds A⁻¹ = [[1, −1], [−1, 2]].
- Identifies the missing requirement: Multiplies X = A⁻¹B.
- Identifies the missing requirement: Calculates the first entry: 1(8) + (−1)(5) = 3.
- Identifies the missing requirement: Obtains X = [[3], [2]], so x = 3 and y = 2.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
A three-state weather model has transition matrix P = [[0.7, 0.2, 0.1], [0.1, 0.6, 0.3], [0.2, 0.2, 0.6]] for states (Sunny, Cloudy, Rainy) respectively. Today's distribution is [0.5, 0.3, 0.2]. Find tomorrow's probability distribution.
[5 marks]
Marking points
- Multiplies the row state vector [0.5, 0.3, 0.2] by the transition matrix P.
- Calculates the first entry: 0.5(0.7) + 0.3(0.1) + 0.2(0.2) = 0.42.
- Calculates the second entry: 0.5(0.2) + 0.3(0.6) + 0.2(0.2) = 0.32.
- Calculates the third entry: 0.5(0.1) + 0.3(0.3) + 0.2(0.6) = 0.26.
- States tomorrow's distribution as [0.42, 0.32, 0.26].
Examiner tip: With more than two states, a GDC's matrix multiplication function avoids tedious hand calculation — enter the state vector and transition matrix directly rather than computing each entry manually.
26.
Marking analysis: A learner attempts the following task: “A three-state weather model has transition matrix P = [[0.7, 0.2, 0.1], [0.1, 0.6, 0.3], [0.2, 0.2, 0.6]] for states (Sunny, Cloudy, Rainy) respectively. Today's distribution is [0.5, 0.3, 0.2]. Find tomorrow's probability distribution.” Their response addresses only this point: “Multiplies the row state vector [0.5, 0.3, 0.2] by the transition matrix P.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Multiplies the row state vector [0.5, 0.3, 0.2] by the transition matrix P.
- Identifies the missing requirement: Calculates the first entry: 0.5(0.7) + 0.3(0.1) + 0.2(0.2) = 0.42.
- Identifies the missing requirement: Calculates the second entry: 0.5(0.2) + 0.3(0.6) + 0.2(0.2) = 0.32.
- Identifies the missing requirement: Calculates the third entry: 0.5(0.1) + 0.3(0.3) + 0.2(0.6) = 0.26.
- Identifies the missing requirement: States tomorrow's distribution as [0.42, 0.32, 0.26].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
A city's residents are classified as living Downtown (D) or in the Suburbs (S). Each year, 40% of Downtown residents move to the Suburbs, and 50% of Suburb residents move Downtown. The transition matrix is P = [[0.6, 0.4], [0.5, 0.5]]. If the population starts entirely Downtown, [1, 0], find the distribution after 2 years.
[4 marks]
Marking points
- Multiplies [1, 0] by P to obtain the distribution after 1 year: [0.6, 0.4].
- Multiplies [0.6, 0.4] by P to obtain the distribution after 2 years.
- Calculates the first entry: 0.6(0.6) + 0.4(0.5) = 0.56.
- Obtains the distribution after 2 years as [0.56, 0.44].
Examiner tip: Each year's distribution depends only on the previous year's distribution, not on how the population reached that state — this 'memoryless' property is the defining feature of a Markov chain.
28.
Marking analysis: A learner attempts the following task: “A city's residents are classified as living Downtown (D) or in the Suburbs (S). Each year, 40% of Downtown residents move to the Suburbs, and 50% of Suburb residents move Downtown. The transition matrix is P = [[0.6, 0.4], [0.5, 0.5]]. If the population starts entirely Downtown, [1, 0], find the distribution after 2 years.” Their response addresses only this point: “Multiplies [1, 0] by P to obtain the distribution after 1 year: [0.6, 0.4].” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Multiplies [1, 0] by P to obtain the distribution after 1 year: [0.6, 0.4].
- Identifies the missing requirement: Multiplies [0.6, 0.4] by P to obtain the distribution after 2 years.
- Identifies the missing requirement: Calculates the first entry: 0.6(0.6) + 0.4(0.5) = 0.56.
- Identifies the missing requirement: Obtains the distribution after 2 years as [0.56, 0.44].
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
Show that the uniform distribution π = [1/3, 1/3, 1/3] is the steady-state distribution for the transition matrix P = [[0.7, 0.2, 0.1], [0.1, 0.6, 0.3], [0.2, 0.2, 0.6]] from an earlier question, by verifying that πP = π.
[4 marks]
Marking points
- Multiplies π = [1/3, 1/3, 1/3] by P.
- Calculates the first entry: (1/3)(0.7 + 0.1 + 0.2) = (1/3)(1) = 1/3.
- Notes that each column of P sums to 1 (the matrix is doubly stochastic), so every entry of πP equals (1/3)(1) = 1/3 by the same reasoning.
- Concludes that πP = [1/3, 1/3, 1/3] = π, confirming π is the steady-state distribution.
Examiner tip: Whenever every column of a transition matrix sums to 1 (a doubly stochastic matrix), the uniform distribution is always its steady state — this is a useful shortcut that avoids solving simultaneous equations.
30.
Marking analysis: A learner attempts the following task: “Show that the uniform distribution π = [1/3, 1/3, 1/3] is the steady-state distribution for the transition matrix P = [[0.7, 0.2, 0.1], [0.1, 0.6, 0.3], [0.2, 0.2, 0.6]] from an earlier question, by verifying that πP = π.” Their response addresses only this point: “Multiplies π = [1/3, 1/3, 1/3] by P.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Multiplies π = [1/3, 1/3, 1/3] by P.
- Identifies the missing requirement: Calculates the first entry: (1/3)(0.7 + 0.1 + 0.2) = (1/3)(1) = 1/3.
- Identifies the missing requirement: Notes that each column of P sums to 1 (the matrix is doubly stochastic), so every entry of πP equals (1/3)(1) = 1/3 by the same reasoning.
- Identifies the missing requirement: Concludes that πP = [1/3, 1/3, 1/3] = π, confirming π is the steady-state distribution.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
31.
Explain what it means for a Markov chain to be 'regular', and explain why a regular Markov chain always converges to a unique steady-state distribution regardless of its starting state.
[3 marks] · no calculator
Marking points
- Explains that a Markov chain is regular if some power of its transition matrix has all strictly positive entries, meaning every state can eventually be reached from every other state.
- Explains that this property guarantees the chain 'mixes' thoroughly over time, washing out the influence of the starting state.
- Concludes that as a result, the probability distribution converges to the same unique steady-state vector no matter which state the chain began in.
Examiner tip: You do not need to prove convergence formally for this course — describing the intuition that a regular chain 'forgets' its starting point over many steps is sufficient for full marks.
32.
Marking analysis: A learner attempts the following task: “Explain what it means for a Markov chain to be 'regular', and explain why a regular Markov chain always converges to a unique steady-state distribution regardless of its starting state.” Their response addresses only this point: “Explains that a Markov chain is regular if some power of its transition matrix has all strictly positive entries, meaning every state can eventually be reached from every other state.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: Explains that a Markov chain is regular if some power of its transition matrix has all strictly positive entries, meaning every state can eventually be reached from every other state.
- Identifies the missing requirement: Explains that this property guarantees the chain 'mixes' thoroughly over time, washing out the influence of the starting state.
- Identifies the missing requirement: Concludes that as a result, the probability distribution converges to the same unique steady-state vector no matter which state the chain began in.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.