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HKDSE Mathematics Extended Module 2 Practice Paper: Matrices, Determinants and Vectors (12 Questions, 100 Marks, 75 Minutes, with Answers)
HKDSE Mathematics Module 2: Matrices, Determinants and Vectors
HKDSE Mathematics Module 2 practice paper: 12 original questions on determinants, inverses, linear systems, transformations, vectors and planes. 100 marks, 75 minutes.
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Paper details Level: S6 | Subject: Mathematics (Extended Module 2) | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (Q1-Q4): Determinants, Inverse Matrices and Linear Systems (32 marks)
Matrices and determinants: expansion of a 3 by 3 determinant, cofactors, inverses, the Cramer rule and uniqueness of solutions.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (Determinants) Let A be the 3 by 3 matrix with rows 1 2 3, 0 4 5 and 1 0 6. (a) Expand along the first row to find det A. (b) Find the cofactor of the (2,3) entry of A. (c) Find det(3A). | Answer: (a) det A = 1x(4×6 – 5×0) – 2x(0x6 – 5×1) + 3x(0x0 – 4×1) = 24 + 10 – 12 = 22. [4 marks] (b) Deleting row 2 and column 3 leaves rows 1 2 and 1 0, whose determinant is 1×0 – 2×1 = -2, so the cofactor is (-1)^5 x (-2) = 2. [2 marks] (c) det(3A) = 3^3 x 22 = 594. [2 marks] |
| 2 | (Inverse matrices) Let B be the 2 by 2 matrix with rows 3 5 and 1 2. (a) Find det B and the inverse of B. (b) Use the inverse to solve 3x + 5y = 4 and x + 2y = 1. (c) Explain why a square matrix with zero determinant has no inverse. | Answer: (a) det B = 3×2 – 5×1 = 1, so the inverse has rows 2 -5 and -1 3. [4 marks] (b) x = 2×4 – 5×1 = 3 and y = -1×4 + 3×1 = -1. [3 marks] (c) The inverse formula carries the factor 1 over det B; when det B is zero that factor is undefined, and the matrix collapses the plane onto a line so the effect cannot be undone. [1 mark] |
| 3 | (Linear systems) Solve x + 2y + z = 8, 2x – y + 3z = 9 and 3x + y – z = 2. (a) Find the determinant of the coefficient matrix. (b) Use the Cramer rule to find x, y and z. (c) Explain how a non-zero determinant guarantees a unique solution. | Answer: (a) The determinant is 1x((-1)x(-1) – 3×1) – 2x(2x(-1) – 3×3) + 1x(2×1 – (-1)x3) = -2 + 22 + 5 = 25. [3 marks] (b) Replacing each column in turn by the constants gives determinants 25, 50 and 75, so x = 1, y = 2 and z = 3. [3 marks] (c) A non-zero determinant means the three equations are independent, so the denominator in the Cramer rule never vanishes and the solution is unique. [2 marks] |
| 4 | (Matrix multiplication) Let A be the 2 by 2 matrix with rows 1 2 and 3 4, and B the 2 by 2 matrix with rows 0 1 and 1 0. (a) Find AB and BA, and explain why matrix multiplication is not commutative. (b) Find A squared. (c) Verify that det(AB) equals det A times det B. | Answer: (a) AB has rows 2 1 and 4 3, while BA has rows 3 4 and 1 2; they differ, so matrix multiplication is not commutative. [4 marks] (b) A squared has rows 7 10 and 15 22. [2 marks] (c) det A = -2, det B = -1 and det(AB) = 2×3 – 1×4 = 2, which equals (-2)x(-1). [2 marks] |
Section B (Q5-Q8): Matrix Properties, Transformations and Vector Products (36 marks)
Matrix algebra and transformations: non-commutative products, determinant identities, rotation and reflection matrices, scalar and vector products.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (Determinant properties) Let M be the 3 by 3 matrix with rows 2 0 1, 1 3 2 and 0 1 1. (a) Find det M. (b) Hence find det(2M), det(M squared), det(M inverse) and det(M transpose). (c) If N is a 3 by 3 matrix with det N = 5, find det(MN) and det(N inverse). | Answer: (a) det M = 2x(3×1 – 2×1) – 0x(1×1 – 2×0) + 1x(1×1 – 3×0) = 2 + 0 + 1 = 3. [3 marks] (b) det(2M) = 2^3 x 3 = 24; det(M squared) = 3^2 = 9; det(M inverse) = 1/3; det(M transpose) = 3. [3 marks] (c) det(MN) = 3×5 = 15 and det(N inverse) = 1/5. [2 marks] |
| 6 | (Transformation matrices) R is the matrix with rows 0 -1 and 1 0, a rotation of 90 degrees anticlockwise. S is the matrix with rows -1 0 and 0 1, a reflection in the y-axis. (a) Find RS and describe the geometric effect of the composite transformation. (b) Find the image of (3, 4) after S and then R. (c) Find det(RS) and explain its link with area. | Answer: (a) RS has rows 0 -1 and -1 0, which is a reflection in the line y = -x. [4 marks] (b) S(3, 4) = (-3, 4) and then R(-3, 4) = (-4, -3). [2 marks] (c) det(RS) = 0x0 – (-1)x(-1) = -1; the size 1 means areas are unchanged and the negative sign means orientation is reversed. [2 marks] |
| 7 | (Scalar product and projection) Let a = (1, 2, 2), b = (2, -2, 1) and c = (3, 1, 5). (a) Show that a is perpendicular to b, and find the lengths of a and b. (b) Find the projection of c onto a, both as a vector and as a length. (c) Find the shortest distance from c to the line r = t a, to 3 decimal places. | Answer: (a) a.b = 1×2 + 2x(-2) + 2×1 = 0, so the vectors are perpendicular; both have length 3. [4 marks] (b) c.a = 3 + 2 + 10 = 15, so the projection vector is (15/9)(1, 2, 2) = (5/3, 10/3, 10/3) with length 15/3 = 5. [3 marks] (c) The squared length of c is 35 and the squared projection is 25, so the distance is the square root of 10, about 3.162. [3 marks] |
| 8 | (Vector product and planes) Let P(1, 0, 2), Q(3, 2, 1) and R(0, 1, 4). (a) Find the vectors PQ and PR, then their vector product. (b) Find the area of triangle PQR to 3 decimal places. (c) Find the equation of the plane through P, Q and R. (d) Find the distance from the origin to that plane, to 3 decimal places. | Answer: (a) PQ = (2, 2, -1) and PR = (-1, 1, 2); their vector product is (5, -3, 4). [4 marks] (b) The magnitude is the square root of 50, so the area is half of it, about 3.536. [2 marks] (c) Taking (5, -3, 4) as normal and substituting P gives 5x – 3y + 4z = 13. [2 marks] (d) The distance is 13 divided by the square root of 50, about 1.838. [2 marks] |
Section C (Q9-Q12): Lines in Space, Planes and Parametric Discussion (32 marks)
Solid geometry: intersecting lines and their angle, the plane through three points, point-to-plane distance, and discussion of parametric systems.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | (Lines in space) L1: r = (1, 2, 3) + t(2, -1, 1) and L2: r = (1, -1, 0) + s(1, 1, 2). (a) Show that L1 and L2 intersect, and find the point of intersection with the values of t and s. (b) Find the angle between the two lines, to 1 decimal place. | Answer: (a) Comparing components gives 1 + 2t = 1 + s, 2 – t = -1 + s and 3 + t = 2s, so t = 1 and s = 2, with intersection (3, 1, 4). [4 marks] (b) The scalar product of the directions is 3 and each has length the square root of 6, so the cosine is 0.5 and the angle is 60.0 degrees. [4 marks] |
| 10 | (Point and plane) The plane pi has equation 2x – y + 2z = 10 and A is the point (3, 1, 2). (a) Write down a unit normal to pi. (b) Find the distance from A to pi. (c) Find the foot of the perpendicular from A to pi. | Answer: (a) The normal (2, -1, 2) has length 3, so a unit normal is (2/3, -1/3, 2/3). [2 marks] (b) The distance is |6 – 1 + 4 – 10|/3 = 1/3, about 0.333. [3 marks] (c) The foot is A plus (1/9)(2, -1, 2) = (29/9, 8/9, 20/9), about (3.222, 0.889, 2.222). [3 marks] |
| 11 | (Vector product) Let a = 2i + 3j – k and b = i – j + 2k. (a) Find a x b and its magnitude. (b) Find a unit vector perpendicular to both a and b. (c) Find the area of the triangle with sides a and b, to 3 decimal places. (d) Explain why the scalar product of a with a x b must be zero. | Answer: (a) a x b = (3×2 – (-1)x(-1), (-1)x1 – 2×2, 2x(-1) – 3×1) = (5, -5, -5), with magnitude 5 times the square root of 3, about 8.660. [3 marks] (b) The unit vector is (1, -1, -1) divided by the square root of 3. [2 marks] (c) The area is half that magnitude, about 4.330. [2 marks] (d) The vector product is perpendicular to a, so its scalar product with a must be zero. [1 mark] |
| 12 | (Parameter discussion) Consider x + y + kz = 1, x + ky + z = 1 and kx + y + z = 1. (a) Show that the determinant of the coefficient matrix is -(k-1)^2 (k+2). (b) Show that the system has no solution when k = -2. (c) Find the general solution when k = 1. (d) State the values of k that give a unique solution. | Answer: (a) Expanding gives -k^3 + 3k – 2, which factorises as -(k-1)^2 (k+2). [3 marks] (b) When k = -2 the three equations add to 0 = 3, a contradiction, so there is no solution. [2 marks] (c) When k = 1 all three reduce to x + y + z = 1, so the solution is (1 – s – t, s, t) for real s and t. [2 marks] (d) A unique solution needs a non-zero determinant, so k is neither 1 nor -2. [1 mark] |
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