OxfordAQA International GCSE Mathematics: Matrices and matrix transformations (9260) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards), section 3.3.3, statements G23 to G26. All of this content is Extension only, so every question is marked (Extension only). Read the study guide or the revision notes first if you need to. Course pages: the OxfordAQA IGCSE Mathematics hub and the printable checklist. In text, a slash separates the rows of a matrix: (0 −1 / 1 0) has top row 0, −1.

Questions

1. (Extension only) (a) Work out (4 −1 / 2 5)(3 / −2). [2] (b) Work out −2(3 −1 / 0 5). [1]

2. (Extension only) A = (1 3 / −2 4) and B = (2 0 / 5 −1). (a) Work out AB. [2] (b) Work out BA. [2] (c) State whether AB = BA. [1]

3. (Extension only) (p 3 / 2 q)(2 / 5) = (19 / 24). Find p and q. [4]

4. (Extension only) M = (5 −2 / 3 1). (a) Write down the 2 × 2 identity matrix I. [1] (b) Work out MI. [2] (c) Explain what the matrix I does to every point. [1]

5. (Extension only) The unit square has vertices O(0, 0), A(1, 0), B(1, 1) and C(0, 1). It is transformed by the matrix (0 −1 / −1 0). (a) Find the images of A, B and C. [3] (b) Describe fully the single transformation. [2]

6. (Extension only) Describe fully the single transformation represented by each matrix. (a) (−1 0 / 0 1) [2] (b) (0 1 / −1 0) [2] (c) (½ 0 / 0 ½) [2]

7. (Extension only) A transformation sends A(1, 0) to (−1, 0) and C(0, 1) to (0, 1). (a) Write down its matrix. [2] (b) Describe the transformation. [1] (c) Find the image of P(4, −3). [2]

8. (Extension only) (a) Write down the matrix for reflection in the line y = x. [1] (b) Use it to find the images of the vertices (1, 2), (4, 2) and (4, 5) of a triangle. [3] (c) Show that the point (3, 3) is unchanged by the transformation. [2]

9. (Extension only) X is reflection in x = 0. Y is reflection in y = x. (a) Write down the matrices for X and Y. [2] (b) Find the single matrix for X followed by Y. [2] (c) Describe the single transformation in part (b). [1] (d) Find the matrix for Y followed by X. What does your answer show? [2]

10. (Extension only) Enlargement with scale factor 4, centre the origin, is followed by a rotation of 180° about the origin. (a) Write down the matrix for each transformation. [2] (b) Find the single matrix for the combination. [2] (c) Find the image of (−3, 2). [2] (d) Show that doing the rotation first gives the same matrix. [2]

11. (Extension only) R = (0 −1 / 1 0). (a) Work out R². [2] (b) Work out R³. [2] (c) Work out R⁴ and say what it represents. [2] (d) Use your answers to find the image of (3, −5) after R is applied seven times. [2]

12. (Extension only) A is a rotation of 270° anticlockwise about the origin. B is reflection in y = −x. C is enlargement with scale factor 3, centre the origin. (a) Find the single matrix for A followed by B. [3] (b) Find the single matrix for A, then B, then C. [2] (c) The point P is transformed by A, then B, then C, and ends at (12, 9). Find P. [3]

Answers

1. (a) Top: 4×3 + (−1)×(−2) = 14 [1]. Bottom: 2×3 + 5×(−2) = −4. Answer (14 / −4) [1] (b) (−6 2 / 0 −10) [1] Examiner insight: Put brackets round negative numbers when you multiply, such as (−1)×(−2), so a sign slip is easy to spot.

2. (a) Top row: 1×2 + 3×5 = 17 and 1×0 + 3×(−1) = −3 [1]. Bottom row: −2×2 + 4×5 = 16 and −2×0 + 4×(−1) = −4 [1]. AB = (17 −3 / 16 −4) (b) Top row: 2×1 + 0×(−2) = 2 and 2×3 + 0×4 = 6 [1]. Bottom row: 5×1 + (−1)×(−2) = 7 and 5×3 + (−1)×4 = 11 [1]. BA = (2 6 / 7 11) (c) No, AB is not equal to BA [1] Examiner insight: Writing out each entry as a sum, as above, shows your method even if one number is wrong.

3. 2p + 3×5 = 19 [1], so 2p = 4 and p = 2 [1]. 2×2 + 5q = 24 [1], so 5q = 20 and q = 4 [1] Examiner insight: Form two equations from the two rows first. Checking the answer by multiplying back is quick and catches errors.

4. (a) I = (1 0 / 0 1) [1] (b) Top row: 5×1 + (−2)×0 = 5 and 5×0 + (−2)×1 = −2 [1]. Bottom row: 3 and 1. MI = (5 −2 / 3 1) = M [1] (c) It leaves every point unchanged; the image is the same as the original [1] Examiner insight: A statement that the matrix “does nothing” is clearer if you also say the point stays where it is.

5. (a) The matrix times (1 / 0) gives the first column: A′ = (0, −1) [1]. B′: 0×1 + (−1)×1 = −1 and (−1)×1 + 0×1 = −1, so (−1, −1) [1]. The second column gives C′ = (−1, 0) [1] (b) Reflection [1] in the line y = −x [1] Examiner insight: A full description of a reflection includes the equation of the mirror line. “Reflection” on its own is incomplete.

6. (a) Reflection [1] in x = 0 (the y-axis) [1] (b) Rotation about the origin [1] of 270° anticlockwise (or 90° clockwise) [1] (c) Enlargement [1] scale factor ½, centre the origin [1] Examiner insight: A rotation needs three things: the angle, the direction and the centre. Give all three.

7. (a) The first column is the image of A and the second is the image of C [1]. Matrix (−1 0 / 0 1) [1] (b) Reflection in x = 0 (the y-axis) [1] (c) (−1)×4 + 0×(−3) = −4 and 0×4 + 1×(−3) = −3 [1]. Image (−4, −3) [1] Examiner insight: To build a matrix, write the images of (1, 0) and (0, 1) as columns, not rows.

8. (a) (0 1 / 1 0) [1] (b) (1, 2) → (0×1 + 1×2, 1×1 + 0×2) = (2, 1) [1]. (4, 2) → (2, 4) [1]. (4, 5) → (5, 4) [1] (c) (0 1 / 1 0)(3 / 3) = (0×3 + 1×3 / 1×3 + 0×3) [1] = (3 / 3), the same point [1] Examiner insight: On a “show that” question, show the multiplication and then state the conclusion; a bare statement that the point is fixed does not show the working.

9. (a) X = (−1 0 / 0 1) [1], Y = (0 1 / 1 0) [1] (b) X is first, so the matrix is YX [1]: (0 1 / 1 0)(−1 0 / 0 1) = (0 1 / −1 0) [1] (c) Rotation of 270° anticlockwise about the origin (90° clockwise) [1] (d) XY = (−1 0 / 0 1)(0 1 / 1 0) = (0 −1 / 1 0) [1]. The two answers differ, so the order of the transformations matters [1] Examiner insight: Explain which matrix goes first in your product. Writing YX with no comment on order leaves the reader guessing.

10. (a) Enlargement (4 0 / 0 4) [1], rotation (−1 0 / 0 −1) [1] (b) Rotation is second, so the product is (−1 0 / 0 −1)(4 0 / 0 4) [1] = (−4 0 / 0 −4) [1] (c) (−4)×(−3) + 0×2 = 12 [1] and 0×(−3) + (−4)×2 = −8, so (12, −8) [1] (d) (4 0 / 0 4)(−1 0 / 0 −1) = (−4 0 / 0 −4) [1], which is the same matrix [1] Examiner insight: Part (d) asks for evidence. Work out the reversed product in full rather than saying “scalars commute”.

11. (a) R² = (0 −1 / 1 0)(0 −1 / 1 0): top row 0×0 + (−1)×1 = −1 and 0 [1]; bottom row 0 and 1×(−1) = −1. R² = (−1 0 / 0 −1) [1] (b) R³ = R²R = (−1 0 / 0 −1)(0 −1 / 1 0) [1] = (0 1 / −1 0) [1] (c) R⁴ = R²R² = (1 0 / 0 1) [1], the identity matrix (a full turn, which leaves every point unchanged) [1] (d) R⁷ = R⁴R³ = IR³ = R³ [1]. (0 1 / −1 0)(3 / −5) = (−5, −3) [1] Examiner insight: Look for a pattern before multiplying seven times. Using R⁴ = I saves a lot of working and gives fewer chances to slip.

12. (a) A = (0 1 / −1 0) and B = (0 −1 / −1 0) [1]. A is first, so BA = (0 −1 / −1 0)(0 1 / −1 0) [1] = (1 0 / 0 −1) [1] (b) C = (3 0 / 0 3) is last, so CBA = 3(1 0 / 0 −1) [1] = (3 0 / 0 −3) [1] (c) (3 0 / 0 −3)(x / y) = (3x / −3y) = (12 / 9) [1]. So 3x = 12, giving x = 4 [1], and −3y = 9, giving y = −3. P = (4, −3) [1] Examiner insight: Check by applying A, B and C to P one at a time: (4, −3) goes to (−3, −4), then (4, 3), then (12, 9).

Where marks are usually lost

• Putting matrices in the wrong order for a combined transformation. The first transformation is the right-hand matrix.
• Slips with negative signs in row-times-column sums.
• Writing the images of A and C as rows when they should be columns.
• Giving an incomplete description, such as “rotation” with no angle, direction or centre.
• Mixing up reflection in y = x and reflection in y = −x.
• Multiplying a column by a matrix from the left, which is not defined here.
• Skipping the check on one point, which would show an error in the matrix.

Next steps

• Return to the revision notes or the study guide.
• Course pages: the OxfordAQA IGCSE Mathematics hub and the printable checklist.
• Find your weak spots with the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1 (for exams May/June 2018 onwards), section 3.3.3, statements G23 to G26.