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Secondary 4 Elementary Mathematics Vectors Matrices Quiz

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Secondary 4 Elementary Mathematics Quiz - Vectors Matrices

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50 Instructions: Answer ALL questions. Show all working clearly. Marks are indicated in brackets. Diagrams are not drawn to scale unless stated.

Section A: Vectors (Questions 1–10)

30 marks

1. Given that a = (\begin{pmatrix} 3 \ -2 \end{pmatrix}) and b = (\begin{pmatrix} -1 \ 5 \end{pmatrix}), find:

(a) a + b [1 mark]

(b) 2a – 3b [2 marks]

2. The position vectors of points P and Q are p = (\begin{pmatrix} 4 \ 1 \end{pmatrix}) and q = (\begin{pmatrix} -2 \ 7 \end{pmatrix}).

(a) Find (\overrightarrow{PQ}) as a column vector. [1 mark]

(b) Calculate the magnitude of (\overrightarrow{PQ}), giving your answer in surd form. [2 marks]

3. Given that u = (\begin{pmatrix} 6 \ -8 \end{pmatrix}), find:

(a) |u| [1 mark]

(b) A unit vector in the direction of u. [2 marks]

4. In the diagram, OABC is a parallelogram. (\overrightarrow{OA}) = a and (\overrightarrow{OC}) = c. M is the midpoint of AB.

(a) Express (\overrightarrow{OB}) in terms of a and c. [1 mark]

(b) Express (\overrightarrow{OM}) in terms of a and c. [2 marks]

5. Given that p = (\begin{pmatrix} 2 \ k \end{pmatrix}) and q = (\begin{pmatrix} -6 \ 9 \end{pmatrix}) are parallel, find the value of k. [2 marks]

6. Points A, B, and C have position vectors a = (\begin{pmatrix} 1 \ 3 \end{pmatrix}), b = (\begin{pmatrix} 5 \ -1 \end{pmatrix}), and c = (\begin{pmatrix} 9 \ -5 \end{pmatrix}).

(a) Find (\overrightarrow{AB}) and (\overrightarrow{BC}). [2 marks]

(b) Show that A, B, and C are collinear. [2 marks]

7. The vector v = (\begin{pmatrix} 5 \ 12 \end{pmatrix}).

(a) Find |v|. [1 mark]

(b) A vector w has the same direction as v but magnitude 26. Find w as a column vector. [2 marks]

8. In triangle PQR, (\overrightarrow{PQ}) = u and (\overrightarrow{PR}) = v. S is a point on QR such that QS : SR = 2 : 1.

Express (\overrightarrow{PS}) in terms of u and v. [3 marks]

9. Given that a = (\begin{pmatrix} 4 \ 1 \end{pmatrix}) and b = (\begin{pmatrix} -2 \ 3 \end{pmatrix}), find the value of |2a + b|. [3 marks]

10. A translation T maps the point (2, –3) to the point (5, 1).

(a) Write T as a column vector. [1 mark]

(b) Find the image of the point (–1, 4) under translation T. [1 mark]

Section B: Matrices (Questions 11–20)

20 marks

11. Given that A = (\begin{pmatrix} 2 & -1 \ 0 & 3 \end{pmatrix}) and B = (\begin{pmatrix} 4 & 1 \ -2 & 5 \end{pmatrix}), find:

(a) A + B [1 mark]

(b) 3A [1 mark]

12. A matrix P = (\begin{pmatrix} 5 & -2 \ 3 & 1 \end{pmatrix}) and Q = (\begin{pmatrix} 1 & 4 \ -2 & 0 \end{pmatrix}).

Find the product PQ. [2 marks]

13. Given that M = (\begin{pmatrix} 3 & 1 \ 2 & 4 \end{pmatrix}) and N = (\begin{pmatrix} 0 & -1 \ 5 & 2 \end{pmatrix}), find:

(a) MN [2 marks]

(b) NM [2 marks]

14. A company produces two types of gift hampers, X and Y. Each hamper X contains 3 bottles of juice, 2 boxes of chocolates, and 1 packet of nuts. Each hamper Y contains 2 bottles of juice, 4 boxes of chocolates, and 3 packets of nuts.

(a) Represent this information as a 3 × 2 matrix H. [1 mark]

(b) The company receives an order for 5 hampers of type X and 8 hampers of type Y. Using matrix multiplication, find the total number of each item required. [2 marks]

15. Given that A = (\begin{pmatrix} 2 & 0 \ 1 & -1 \end{pmatrix}), find A². [2 marks]

16. If (\begin{pmatrix} x & 2 \ 3 & y \end{pmatrix}) + (\begin{pmatrix} 4 & -1 \ 0 & 5 \end{pmatrix}) = (\begin{pmatrix} 7 & 1 \ 3 & 8 \end{pmatrix}), find the values of x and y. [2 marks]

17. A matrix R = (\begin{pmatrix} 2 & 1 \ 4 & 3 \end{pmatrix}).

(a) Find the determinant of R. [1 mark]

(b) Hence, find the inverse matrix R⁻¹. [2 marks]

18. Solve the following matrix equation for x:

(\begin{pmatrix} 3 & -1 \ 2 & 0 \end{pmatrix}) x = (\begin{pmatrix} 5 \ 6 \end{pmatrix}) [3 marks]

19. Given that A = (\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}) and B = (\begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}), verify that (A + B)ᵀ = Aᵀ + Bᵀ. [2 marks]

20. A transformation matrix T = (\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}) is applied to the point P(3, 2).

(a) Find the coordinates of the image of P. [1 mark]

(b) Describe fully the geometric transformation represented by T. [1 mark]

END OF QUIZ

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Answers

Secondary 4 Elementary Mathematics Quiz - Vectors Matrices

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

Section A: Vectors (Questions 1–10)

1. a = (\begin{pmatrix} 3 \ -2 \end{pmatrix}), b = (\begin{pmatrix} -1 \ 5 \end{pmatrix})

(a) a + b = (\begin{pmatrix} 3 + (-1) \ -2 + 5 \end{pmatrix}) = (\begin{pmatrix} 2 \ 3 \end{pmatrix}) [M1]

(b) 2a – 3b = 2(\begin{pmatrix} 3 \ -2 \end{pmatrix}) – 3(\begin{pmatrix} -1 \ 5 \end{pmatrix}) = (\begin{pmatrix} 6 \ -4 \end{pmatrix}) – (\begin{pmatrix} -3 \ 15 \end{pmatrix}) = (\begin{pmatrix} 9 \ -19 \end{pmatrix}) [M1, A1]

2. p = (\begin{pmatrix} 4 \ 1 \end{pmatrix}), q = (\begin{pmatrix} -2 \ 7 \end{pmatrix})

(a) (\overrightarrow{PQ}) = q – p = (\begin{pmatrix} -2 - 4 \ 7 - 1 \end{pmatrix}) = (\begin{pmatrix} -6 \ 6 \end{pmatrix}) [M1]

(b) |(\overrightarrow{PQ})| = (\sqrt{(-6)^2 + 6^2}) = (\sqrt{36 + 36}) = (\sqrt{72}) = (6\sqrt{2}) [M1, A1]

3. u = (\begin{pmatrix} 6 \ -8 \end{pmatrix})

(a) |u| = (\sqrt{6^2 + (-8)^2}) = (\sqrt{36 + 64}) = (\sqrt{100}) = 10 [M1]

(b) Unit vector = (\frac{1}{|\mathbf{u}|}) u = (\frac{1}{10}\begin{pmatrix} 6 \ -8 \end{pmatrix}) = (\begin{pmatrix} 0.6 \ -0.8 \end{pmatrix}) or (\begin{pmatrix} \frac{3}{5} \ -\frac{4}{5} \end{pmatrix}) [M1, A1]

4. OABC is a parallelogram, (\overrightarrow{OA}) = a, (\overrightarrow{OC}) = c, M is midpoint of AB.

(a) (\overrightarrow{OB}) = (\overrightarrow{OA}) + (\overrightarrow{OC}) = a + c [B1]

(b) (\overrightarrow{OM}) = (\overrightarrow{OA}) + (\frac{1}{2}\overrightarrow{AB}) = a + (\frac{1}{2})c [M1, A1] (Since (\overrightarrow{AB}) = (\overrightarrow{OC}) = c)

(c) (\overrightarrow{CM}) = (\overrightarrow{OM}) – (\overrightarrow{OC}) = (a + (\frac{1}{2})c) – c = a – (\frac{1}{2})c [M1, A1]

5. p = (\begin{pmatrix} 2 \ k \end{pmatrix}), q = (\begin{pmatrix} -6 \ 9 \end{pmatrix}) are parallel.

For parallel vectors: p = λq for some scalar λ. (\begin{pmatrix} 2 \ k \end{pmatrix}) = λ(\begin{pmatrix} -6 \ 9 \end{pmatrix})

From x-component: 2 = –6λ → λ = –(\frac{1}{3}) [M1] From y-component: k = 9λ = 9(–(\frac{1}{3})) = –3 [A1]

6. a = (\begin{pmatrix} 1 \ 3 \end{pmatrix}), b = (\begin{pmatrix} 5 \ -1 \end{pmatrix}), c = (\begin{pmatrix} 9 \ -5 \end{pmatrix})

(a) (\overrightarrow{AB}) = b – a = (\begin{pmatrix} 5 - 1 \ -1 - 3 \end{pmatrix}) = (\begin{pmatrix} 4 \ -4 \end{pmatrix}) [M1] (\overrightarrow{BC}) = c – b = (\begin{pmatrix} 9 - 5 \ -5 - (-1) \end{pmatrix}) = (\begin{pmatrix} 4 \ -4 \end{pmatrix}) [A1]

(b) Since (\overrightarrow{AB}) = (\overrightarrow{BC}) = (\begin{pmatrix} 4 \ -4 \end{pmatrix}), the vectors are equal. Therefore, A, B, and C are collinear (B is the midpoint of AC). [M1, A1]

7. v = (\begin{pmatrix} 5 \ 12 \end{pmatrix})

(a) |v| = (\sqrt{5^2 + 12^2}) = (\sqrt{25 + 144}) = (\sqrt{169}) = 13 [B1]

(b) Unit vector in direction of v = (\frac{1}{13}\begin{pmatrix} 5 \ 12 \end{pmatrix}) w = 26 × (\frac{1}{13}\begin{pmatrix} 5 \ 12 \end{pmatrix}) = 2(\begin{pmatrix} 5 \ 12 \end{pmatrix}) = (\begin{pmatrix} 10 \ 24 \end{pmatrix}) [M1, A1]

8. (\overrightarrow{PQ}) = u, (\overrightarrow{PR}) = v, QS : SR = 2 : 1

(\overrightarrow{QR}) = (\overrightarrow{PR}) – (\overrightarrow{PQ}) = v – u [M1] (\overrightarrow{QS}) = (\frac{2}{3}\overrightarrow{QR}) = (\frac{2}{3})(v – u) [M1] (\overrightarrow{PS}) = (\overrightarrow{PQ}) + (\overrightarrow{QS}) = u + (\frac{2}{3})(v – u) = u + (\frac{2}{3})v – (\frac{2}{3})u = (\frac{1}{3})u + (\frac{2}{3})v [A1]

9. a = (\begin{pmatrix} 4 \ 1 \end{pmatrix}), b = (\begin{pmatrix} -2 \ 3 \end{pmatrix})

2a + b = 2(\begin{pmatrix} 4 \ 1 \end{pmatrix}) + (\begin{pmatrix} -2 \ 3 \end{pmatrix}) = (\begin{pmatrix} 8 \ 2 \end{pmatrix}) + (\begin{pmatrix} -2 \ 3 \end{pmatrix}) = (\begin{pmatrix} 6 \ 5 \end{pmatrix}) [M1, A1] |2a + b| = (\sqrt{6^2 + 5^2}) = (\sqrt{36 + 25}) = (\sqrt{61}) [A1]

10. Translation T maps (2, –3) → (5, 1).

(a) T = (\begin{pmatrix} 5 - 2 \ 1 - (-3) \end{pmatrix}) = (\begin{pmatrix} 3 \ 4 \end{pmatrix}) [B1]

(b) Image of (–1, 4) = (–1, 4) + T = (\begin{pmatrix} -1 + 3 \ 4 + 4 \end{pmatrix}) = (2, 8) [B1]

Section B: Matrices (Questions 11–20)

11. A = (\begin{pmatrix} 2 & -1 \ 0 & 3 \end{pmatrix}), B = (\begin{pmatrix} 4 & 1 \ -2 & 5 \end{pmatrix})

(a) A + B = (\begin{pmatrix} 2+4 & -1+1 \ 0+(-2) & 3+5 \end{pmatrix}) = (\begin{pmatrix} 6 & 0 \ -2 & 8 \end{pmatrix}) [B1]

(b) 3A = (\begin{pmatrix} 3×2 & 3×(-1) \ 3×0 & 3×3 \end{pmatrix}) = (\begin{pmatrix} 6 & -3 \ 0 & 9 \end{pmatrix}) [B1]

(c) 2A – B = (\begin{pmatrix} 4 & -2 \ 0 & 6 \end{pmatrix}) – (\begin{pmatrix} 4 & 1 \ -2 & 5 \end{pmatrix}) = (\begin{pmatrix} 0 & -3 \ 2 & 1 \end{pmatrix}) [M1, A1]

12. P = (\begin{pmatrix} 5 & -2 \ 3 & 1 \end{pmatrix}), Q = (\begin{pmatrix} 1 & 4 \ -2 & 0 \end{pmatrix})

PQ = (\begin{pmatrix} 5(1)+(-2)(-2) & 5(4)+(-2)(0) \ 3(1)+1(-2) & 3(4)+1(0) \end{pmatrix}) = (\begin{pmatrix} 5+4 & 20+0 \ 3-2 & 12+0 \end{pmatrix}) = (\begin{pmatrix} 9 & 20 \ 1 & 12 \end{pmatrix}) [M1, A1]

13. M = (\begin{pmatrix} 3 & 1 \ 2 & 4 \end{pmatrix}), N = (\begin{pmatrix} 0 & -1 \ 5 & 2 \end{pmatrix})

(a) MN = (\begin{pmatrix} 3(0)+1(5) & 3(-1)+1(2) \ 2(0)+4(5) & 2(-1)+4(2) \end{pmatrix}) = (\begin{pmatrix} 0+5 & -3+2 \ 0+20 & -2+8 \end{pmatrix}) = (\begin{pmatrix} 5 & -1 \ 20 & 6 \end{pmatrix}) [M1, A1]

(b) NM = (\begin{pmatrix} 0(3)+(-1)(2) & 0(1)+(-1)(4) \ 5(3)+2(2) & 5(1)+2(4) \end{pmatrix}) = (\begin{pmatrix} 0-2 & 0-4 \ 15+4 & 5+8 \end{pmatrix}) = (\begin{pmatrix} -2 & -4 \ 19 & 13 \end{pmatrix}) [M1, A1]

14. (a) H = (\begin{pmatrix} 3 & 2 \ 2 & 4 \ 1 & 3 \end{pmatrix}) where rows represent juice, chocolates, nuts; columns represent X, Y. [B1]

(b) Order vector = (\begin{pmatrix} 5 \ 8 \end{pmatrix}) Total items = H × (\begin{pmatrix} 5 \ 8 \end{pmatrix}) = (\begin{pmatrix} 3(5)+2(8) \ 2(5)+4(8) \ 1(5)+3(8) \end{pmatrix}) = (\begin{pmatrix} 15+16 \ 10+32 \ 5+24 \end{pmatrix}) = (\begin{pmatrix} 31 \ 42 \ 29 \end{pmatrix}) [M1, A1] 31 bottles of juice, 42 boxes of chocolates, 29 packets of nuts.

15. A = (\begin{pmatrix} 2 & 0 \ 1 & -1 \end{pmatrix})

A² = (\begin{pmatrix} 2 & 0 \ 1 & -1 \end{pmatrix}) (\begin{pmatrix} 2 & 0 \ 1 & -1 \end{pmatrix}) = (\begin{pmatrix} 2(2)+0(1) & 2(0)+0(-1) \ 1(2)+(-1)(1) & 1(0)+(-1)(-1) \end{pmatrix}) = (\begin{pmatrix} 4 & 0 \ 1 & 1 \end{pmatrix}) [M1, A1]

16. (\begin{pmatrix} x & 2 \ 3 & y \end{pmatrix}) + (\begin{pmatrix} 4 & -1 \ 0 & 5 \end{pmatrix}) = (\begin{pmatrix} 7 & 1 \ 3 & 8 \end{pmatrix})

x + 4 = 7 → x = 3 [M1] y + 5 = 8 → y = 3 [A1]

17. R = (\begin{pmatrix} 2 & 1 \ 4 & 3 \end{pmatrix})

(a) det(R) = (2)(3) – (1)(4) = 6 – 4 = 2 [B1]

(b) R⁻¹ = (\frac{1}{\det(R)}\begin{pmatrix} 3 & -1 \ -4 & 2 \end{pmatrix}) = (\frac{1}{2}\begin{pmatrix} 3 & -1 \ -4 & 2 \end{pmatrix}) = (\begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \ -2 & 1 \end{pmatrix}) [M1, A1]

18. (\begin{pmatrix} 3 & -1 \ 2 & 0 \end{pmatrix}) x = (\begin{pmatrix} 5 \ 6 \end{pmatrix})

Let x = (\begin{pmatrix} x \ y \end{pmatrix}) 3x – y = 5 ...(1) 2x + 0y = 6 ...(2) [M1]

From (2): 2x = 6 → x = 3 [M1] Substitute into (1): 3(3) – y = 5 → 9 – y = 5 → y = 4 x = (\begin{pmatrix} 3 \ 4 \end{pmatrix}) [A1]

19. A = (\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}), B = (\begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix})

A + B = (\begin{pmatrix} 6 & 8 \ 10 & 12 \end{pmatrix}) (A + B)ᵀ = (\begin{pmatrix} 6 & 10 \ 8 & 12 \end{pmatrix}) [M1]

Aᵀ = (\begin{pmatrix} 1 & 3 \ 2 & 4 \end{pmatrix}), Bᵀ = (\begin{pmatrix} 5 & 7 \ 6 & 8 \end{pmatrix}) Aᵀ + Bᵀ = (\begin{pmatrix} 6 & 10 \ 8 & 12 \end{pmatrix}) [M1]

Since (A + B)ᵀ = Aᵀ + Bᵀ, the property is verified. [A1]

20. T = (\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}), P(3, 2)

(a) Image = T(\begin{pmatrix} 3 \ 2 \end{pmatrix}) = (\begin{pmatrix} 0(3)+(-1)(2) \ 1(3)+0(2) \end{pmatrix}) = (\begin{pmatrix} -2 \ 3 \end{pmatrix}) [B1] Coordinates: (–2, 3)

(b) Rotation of 90° anticlockwise about the origin. [B1]

END OF ANSWER KEY