IB DP Mathematics: Applications and Interpretation – Matrices, eigenvalues and eigenvectors (HL) Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the matrices unit of IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 1.14 and 1.15, and every question is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

All three HL papers require technology, so every question here is calculator allowed. Questions labelled “(calculator allowed, by hand)” test the 2 × 2 skills the guide expects you to do without the GDC, so show every line. Give answers exactly or to 3 significant figures. Matrices are written row by row: “rows (a, b), (c, d)”.

Learn the methods first in the matrices and eigenvalues study guide and the revision notes. The IB DP Maths AI course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator allowed) Matrix A has rows (2, 0, −1) and (4, 3, 5).

(a) State the order of A. [1] (b) Write down the element a₂₃. [1]

2. (calculator allowed, by hand) A has rows (3, −2), (1, 4) and B has rows (0, 5), (−1, 2). Find 2A + B and AB. [3]

3. (calculator allowed, by hand) C has rows (6, 4), (2, 3).

(a) Find C⁻¹. [2] (b) The matrix with rows (k, 6), (2, 3) is singular. Find k. [2]

4. (calculator allowed, by hand) The matrix with rows (x + 2y, 3), (4, x − y) is equal to the matrix with rows (7, 3), (4, 1). Find x and y. [3]

5. (calculator allowed, by hand) P has rows (1, 2), (0, 1) and Q has rows (1, 0), (3, 1).

(a) Find PQ and QP. [2] (b) Show that (P + Q)² ≠ P² + 2PQ + Q², and explain why. [2]

6. (calculator allowed) A bakery has two shops, X and Y. In one day, X sells 40 loaves, 65 pastries and 30 cakes; Y sells 55 loaves, 50 pastries and 20 cakes. The selling prices are 3.20, 1.90 and 12.50 dollars, and the costs to make are 1.40, 0.75 and 5.00 dollars.

(a) Write the sales as a 2 × 3 matrix S and the prices as a column p. State the order of Sp. [2] (b) Find the revenue of each shop. [2] (c) Use a matrix product to find the profit of each shop. [2]

7. (calculator allowed) Notebooks cost n dollars, pens p dollars and folders f dollars. Three orders are: 3 notebooks, 4 pens, 2 folders for 19.60 dollars; 2 notebooks, 6 pens, 1 folder for 16.35 dollars; 1 notebook, 2 pens, 3 folders for 12.35 dollars.

(a) Write the system in the form Ax = b. [2] (b) Find det A and explain why the system has a unique solution. [2] (c) Use the inverse matrix to find the three prices. [2]

8. (calculator allowed, by hand) A message is coded with A = 1, B = 2, …, Z = 26, in pairs, using the encoding matrix E with rows (2, 3), (1, 2). The coded message is 27, 14, 37, 21.

(a) Find E⁻¹. [2] (b) Decode the message. [3]

9. (calculator allowed, by hand) A has rows (1, 4), (2, 3).

(a) Find the characteristic polynomial of A. [2] (b) Find the eigenvalues of A. [1] (c) Find an eigenvector for each eigenvalue. [2]

10. (calculator allowed, by hand) B has rows (5, −2), (1, 2).

(a) Show that the characteristic polynomial of B is λ² − 7λ + 12. [2] (b) Find the eigenvalues of B and a corresponding eigenvector for each. [3] (c) Write down matrices P and D such that B = PDP⁻¹, and find P⁻¹. [2] (d) Find an expression for Bⁿ, and hence find the top-left element of B⁵. [3]

11. (calculator allowed) Two towns, Northby and Southam, have 12 000 and 8000 people. Each year 15% of Northby’s population moves to Southam and 5% of Southam’s population moves to Northby. There are no other changes. Let xₙ be the column (Nₙ, Sₙ) after n years, with xₙ₊₁ = Txₙ.

(a) Write down T. [2] (b) Show that the eigenvalues of T are 1 and 0.8. [2] (c) Find an eigenvector for each eigenvalue. [2] (d) Express x₀ in terms of the eigenvectors, and hence find N₃ and S₃. [3] (e) State the long-term population of Northby. [1] (f) Find the first year in which Northby has fewer than 6000 people. [2]

12. (calculator allowed, by hand) The matrix A has rows (k, 2), (3, 1), and 4 is an eigenvalue of A.

(a) Find k. [3] (b) Find the other eigenvalue of A. [2]

Answers

1. (a) 2 × 3 [1] (b) 5 [1] Examiner insight: The order is rows × columns; “3 × 2” scores nothing, and there is no follow-through on a one-mark answer.

2. 2A + B has rows (6, 1), (1, 10) [1]. AB: row-by-column products, e.g. top-left 3(0) + (−2)(−1) [1], giving rows (2, 11), (−4, 13) [1] Examiner insight: The method mark for AB needs visible row-by-column working; multiplying corresponding elements earns nothing.

3. (a) det C = 18 − 8 = 10 [1]; C⁻¹ = (1/10) × rows (3, −4), (−2, 6) = rows (0.3, −0.4), (−0.2, 0.6) [1] (b) 3k − 12 = 0 [1], so k = 4 [1] Examiner insight: An inverse with the leading diagonal negated instead of swapped loses the accuracy mark; multiply back to I to catch it.

4. Equate elements: x + 2y = 7 and x − y = 1 [1]. Subtracting gives 3y = 6 [1], so x = 3, y = 2 [1] Examiner insight: The first mark is for equating corresponding elements; writing only the answers without the equations can lose it.

5. (a) PQ has rows (7, 2), (3, 1); QP has rows (1, 2), (3, 7) [1] [1] (b) (P + Q)² has rows (10, 8), (12, 10), but P² + 2PQ + Q² has rows (16, 8), (12, 4) [1]. (P + Q)² = P² + PQ + QP + Q², which equals P² + 2PQ + Q² only if PQ = QP; here PQ ≠ QP because matrix multiplication is not commutative [1] Examiner insight: A “show that” needs both matrices written out; stating “they are different” without the elements earns no marks.

6. (a) S has rows (40, 65, 30), (55, 50, 20) and p is the column (3.20, 1.90, 12.50) [1]. Sp is 2 × 1 [1] (b) Sp: X = 128 + 123.50 + 375 [1] = 626.50 dollars; Y = 521 dollars [1] (c) Profit = S(p − c), where p − c = (1.80, 1.15, 7.50) [1]. X: 371.75 dollars; Y: 306.50 dollars [1] Examiner insight: The product must be Sp, not pS; pS is (3 × 1)(2 × 3), which is not defined, so writing it earns no method mark.

7. (a) A has rows (3, 4, 2), (2, 6, 1), (1, 2, 3) [1]; x is the column (n, p, f) and b is the column (19.60, 16.35, 12.35) [1] (b) det A = 24 [1]. It is non-zero, so A⁻¹ exists and x = A⁻¹b is unique [1] (c) x = A⁻¹b [1]: n = 3.50, p = 1.20, f = 2.15 dollars [1] Examiner insight: The question says “use the inverse matrix”, so write x = A⁻¹b before the GDC result; a bare answer risks the method mark.

8. (a) det E = 4 − 3 = 1 [1]; E⁻¹ has rows (2, −3), (−1, 2) [1] (b) E⁻¹(27, 14) = (54 − 42, −27 + 28) = (12, 1) [1]; E⁻¹(37, 21) = (74 − 63, −37 + 42) = (11, 5) [1]. 12, 1, 11, 5 is LAKE [1] Examiner insight: Decoding with E instead of E⁻¹ gives numbers above 26; that check alone tells you the method is wrong.

9. (a) det(A − λI) = (1 − λ)(3 − λ) − 8 [1] = λ² − 4λ − 5 [1] (b) (λ − 5)(λ + 1) = 0, so λ = 5, λ = −1 [1] (c) λ = 5: −4x + 4y = 0, so (1, 1) [1]. λ = −1: 2x + 4y = 0, so (2, −1) [1] Examiner insight: Any non-zero multiple of an eigenvector scores full marks; (0, 0) never does.

10. (a) det(B − λI) = (5 − λ)(2 − λ) − (−2)(1) [1] = 10 − 7λ + λ² + 2 = λ² − 7λ + 12 [1] (b) (λ − 3)(λ − 4) = 0, so λ = 3, 4 [1]. λ = 3: 2x − 2y = 0, giving (1, 1) [1]. λ = 4: x − 2y = 0, giving (2, 1) [1] (c) P has rows (1, 2), (1, 1) and D has rows (3, 0), (0, 4) [1]. det P = −1, so P⁻¹ has rows (−1, 2), (1, −1) [1] (d) Bⁿ = PDⁿP⁻¹ with Dⁿ having diagonal 3ⁿ, 4ⁿ [1]. Bⁿ has rows (2·4ⁿ − 3ⁿ, 2·3ⁿ − 2·4ⁿ), (4ⁿ − 3ⁿ, 2·3ⁿ − 4ⁿ) [1]. Top-left of B⁵: 2(1024) − 243 = 1805 [1] Examiner insight: In (a) the “show that” mark needs the unexpanded determinant line; starting from the given quadratic earns nothing.

11. (a) Columns are “from”: T has rows (0.85, 0.05), (0.15, 0.95) [1] [1] (b) (0.85 − λ)(0.95 − λ) − 0.0075 = λ² − 1.8λ + 0.8 [1] = (λ − 1)(λ − 0.8), so λ = 1 and 0.8 [1] (c) λ = 1: −0.15N + 0.05S = 0, giving (1, 3) [1]. λ = 0.8: 0.05N + 0.05S = 0, giving (1, −1) [1] (d) x₀ = 5000(1, 3) + 7000(1, −1) [1]. So xₙ = 5000(1, 3) + 7000(0.8)ⁿ(1, −1) [1]. With 0.8³ = 0.512: N₃ = 8584, S₃ = 11 416 [1] (e) 0.8ⁿ → 0, so Northby tends to 5000 [1] (f) 5000 + 7000(0.8)ⁿ < 6000 gives 0.8ⁿ < 1/7 [1]; N₈ ≈ 6174 and N₉ ≈ 5940, so year 9 [1] Examiner insight: In (f) the answer must be a whole number of years; writing n = 8.72 without interpreting it loses the final mark.

12. (a) det(A − 4I) = 0 [1]: (k − 4)(−3) − 6 = 0 [1], so k = 2 [1] (b) det(A − λI) = (2 − λ)(1 − λ) − 6 = λ² − 3λ − 4 [1] = (λ − 4)(λ + 1), so the other eigenvalue is −1 [1] Examiner insight: Substituting λ = 4 into det(A − λI) = 0 is the method mark; guessing k and checking earns nothing unless the check is shown in full.

Where marks are usually lost

• Stating an order as columns × rows, or reading aᵢⱼ with i as the column.
• Multiplying matrices element by element instead of row by column.
• Negating the leading diagonal of a 2 × 2 inverse instead of swapping it.
• Giving a GDC answer to “use the inverse matrix” without writing x = A⁻¹b.
• Writing a context product in an order that is not defined, such as pS instead of Sp.
• Starting a “show that” from the given result instead of the determinant line.
• Using (0, 0), or one eigenvector for both eigenvalues.
• Ordering the eigenvalues in D differently from the eigenvector columns in P.
• Writing a transition matrix with “to” and “from” mixed up.
• Leaving a year as a decimal instead of the first whole year that satisfies the condition.

Next steps

• Recap with the revision notes.
• Re-learn any weak method in the study guide.
• See the whole course on the IB DP Maths AI course hub.
• Tick off topics with the printable syllabus checklist.
• Read AI exam preparation for paper-by-paper advice.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021. Sections AHL 1.14 and AHL 1.15.