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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: 45 minutes
Total Marks: 50

Instructions:

Answer all questions.
Show all working clearly. Marks are awarded for method.
Unless otherwise stated, give non-exact answers correct to 3 significant figures.
You may use an approved calculator.

Section A: Matrices (Questions 1–5)

Each question carries 2 marks.

1. A matrix $P$ is given by $P = \begin{pmatrix} 3 & -1 \\ 2 & 0 \\ 5 & 4 \end{pmatrix}$ .

(a) State the order of matrix $P$ .
(b) Write down the element in the second row and first column of $P$ .

2. Given $A = \begin{pmatrix} 4 & 1 \\ -2 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 5 \\ 1 & -2 \end{pmatrix}$ , find $A + B$ .

3. Given $M = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix}$ , find $3M$ .

4. Given $X = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $Y = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$ , find $XY$ .

5. A shop sells two types of gift hampers, A and B. The table below shows the number of items in each hamper.

Item	Hamper A	Hamper B
Chocolates	3	2
Biscuits	1	4
Candies	5	3

The cost of each chocolate is $2, each biscuit is$ 1, and each candy is $0.50.

Represent the hamper contents as a $3 \times 2$ matrix $H$ , and the costs as a $1 \times 3$ matrix $C$ . Hence, or otherwise, find the total cost of the items in each hamper.

Section B: Vectors – Basic Operations (Questions 6–10)

Each question carries 2 marks.

6. Given $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ , find $\mathbf{a} + \mathbf{b}$ .

7. Given $\mathbf{u} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ , find $\mathbf{u} - \mathbf{v}$ .

8. Given $\mathbf{p} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ , find $|\mathbf{p}|$ , the magnitude of $\mathbf{p}$ .

9. Given $\mathbf{r} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$ , find $-2\mathbf{r}$ .

10. The position vectors of points $A$ and $B$ are $\vec{OA} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}$ . Find $\vec{AB}$ .

Section C: Vectors – Geometry and Applications (Questions 11–15)

Each question carries 3 marks.

11. Given $\mathbf{a} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$ , find the vector $\mathbf{c}$ such that $2\mathbf{a} + \mathbf{c} = \mathbf{b}$ .

12. Points $P$ and $Q$ have position vectors $\vec{OP} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\vec{OQ} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ . $M$ is the midpoint of $PQ$ . Find the position vector of $M$ .

13. Given $\mathbf{u} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}$ , find a vector that is parallel to $\mathbf{u}$ and has magnitude 5.

14. In triangle $ABC$ , $\vec{AB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\vec{AC} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ . Find $\vec{BC}$ .

15. The points $A$ , $B$ , and $C$ have position vectors $\mathbf{a} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ , $\mathbf{b} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ , and $\mathbf{c} = \begin{pmatrix} 8 \\ 5 \end{pmatrix}$ . Show that $A$ , $B$ , and $C$ are collinear.

Section D: Vectors – Problem Solving (Questions 16–20)

Each question carries 4 marks.

16. In the diagram, $OABC$ is a parallelogram. $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$ . $M$ is the midpoint of $AB$ , and $N$ is the point on $BC$ such that $BN : NC = 1 : 2$ .

(a) Express $\vec{OM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ .
(b) Express $\vec{ON}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ .
(c) Hence, find $\vec{MN}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ .

17. A particle moves from point $P$ to point $Q$ with velocity vector $\mathbf{v} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ m/s. The particle takes 5 seconds to travel from $P$ to $Q$ .

(a) Find the displacement vector $\vec{PQ}$ .
(b) Given that the position vector of $P$ is $\begin{pmatrix} 2 \\ 7 \end{pmatrix}$ , find the position vector of $Q$ .
(c) Find the distance $PQ$ .

18. Given $\mathbf{p} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ .

(a) Find $|\mathbf{p} + \mathbf{q}|$ .
(b) Find the value of $k$ such that $\mathbf{p} + k\mathbf{q}$ is parallel to the $x$ -axis.

19. In triangle $PQR$ , $\vec{PQ} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\vec{PR} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ .

(a) Find $\vec{QR}$ .
(b) Calculate $|\vec{PQ}|$ , $|\vec{PR}|$ , and $|\vec{QR}|$ .
(c) Hence, determine whether triangle $PQR$ is right-angled. Justify your answer.

20. A quadrilateral $ABCD$ has vertices with position vectors: $\vec{OA} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ , $\vec{OB} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ , $\vec{OC} = \begin{pmatrix} 7 \\ 7 \end{pmatrix}$ , $\vec{OD} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$ .

(a) Find $\vec{AB}$ and $\vec{DC}$ .
(b) What type of quadrilateral is $ABCD$ ? Justify your answer.
(c) Find the position vector of the intersection point of the diagonals $AC$ and $BD$ .

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: Matrices (2 marks each)

1. $P = \begin{pmatrix} 3 & -1 \\ 2 & 0 \\ 5 & 4 \end{pmatrix}$

(a) Order of $P$ is $3 \times 2$ . [1 mark]
(b) Element in second row, first column is $2$ . [1 mark]

2. $A + B = \begin{pmatrix} 4 & 1 \\ -2 & 3 \end{pmatrix} + \begin{pmatrix} 0 & 5 \\ 1 & -2 \end{pmatrix} = \begin{pmatrix} 4+0 & 1+5 \\ -2+1 & 3+(-2) \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ -1 & 1 \end{pmatrix}$ [2 marks]

Award 1 mark for correct method, 1 mark for correct answer.

3. $3M = 3 \times \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 6 & -3 \\ 0 & 9 \end{pmatrix}$ [2 marks]

Award 1 mark for multiplying each element, 1 mark for correct answer.

4. $XY = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} = \begin{pmatrix} 1(5)+2(7) & 1(6)+2(8) \\ 3(5)+4(7) & 3(6)+4(8) \end{pmatrix} = \begin{pmatrix} 5+14 & 6+16 \\ 15+28 & 18+32 \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}$ [2 marks]

Award 1 mark for correct row-column multiplication setup, 1 mark for correct answer.

5. $H = \begin{pmatrix} 3 & 2 \\ 1 & 4 \\ 5 & 3 \end{pmatrix}$ , $C = \begin{pmatrix} 2 & 1 & 0.5 \end{pmatrix}$ [1 mark]

Total cost matrix $= CH = \begin{pmatrix} 2 & 1 & 0.5 \end{pmatrix} \begin{pmatrix} 3 & 2 \\ 1 & 4 \\ 5 & 3 \end{pmatrix}$

$= \begin{pmatrix} 2(3)+1(1)+0.5(5) & 2(2)+1(4)+0.5(3) \end{pmatrix} = \begin{pmatrix} 6+1+2.5 & 4+4+1.5 \end{pmatrix} = \begin{pmatrix} 9.5 & 9.5 \end{pmatrix}$ [1 mark]

Total cost of Hamper A = $9.50; Total cost of Hamper B = $9.50.

Section B: Vectors – Basic Operations (2 marks each)

6. $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} + \begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$ [2 marks]

7. $\mathbf{u} - \mathbf{v} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ [2 marks]

8. $|\mathbf{p}| = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \approx 7.21$ [2 marks]

Award 1 mark for correct formula, 1 mark for correct simplified or decimal answer.

9. $-2\mathbf{r} = -2 \times \begin{pmatrix} 2 \\ -5 \end{pmatrix} = \begin{pmatrix} -4 \\ 10 \end{pmatrix}$ [2 marks]

10. $\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 7 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ [2 marks]

Section C: Vectors – Geometry and Applications (3 marks each)

11. $2\mathbf{a} + \mathbf{c} = \mathbf{b}$
$2\begin{pmatrix} 4 \\ 1 \end{pmatrix} + \mathbf{c} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$
$\begin{pmatrix} 8 \\ 2 \end{pmatrix} + \mathbf{c} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$ [1 mark]
$\mathbf{c} = \begin{pmatrix} -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 8 \\ 2 \end{pmatrix} = \begin{pmatrix} -10 \\ 3 \end{pmatrix}$ [2 marks]

Award 1 mark for correct substitution, 2 marks for correct answer.

12. $\vec{OM} = \frac{1}{2}(\vec{OP} + \vec{OQ}) = \frac{1}{2}\left( \begin{pmatrix} 3 \\ -2 \end{pmatrix} + \begin{pmatrix} -1 \\ 4 \end{pmatrix} \right) = \frac{1}{2}\begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ [3 marks]

Award 1 mark for midpoint formula, 1 mark for correct addition, 1 mark for correct answer.

13. $|\mathbf{u}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ [1 mark]

Unit vector in direction of $\mathbf{u} = \frac{1}{10}\begin{pmatrix} 6 \\ -8 \end{pmatrix} = \begin{pmatrix} 0.6 \\ -0.8 \end{pmatrix}$ [1 mark]

Vector parallel to $\mathbf{u}$ with magnitude 5: $5 \times \begin{pmatrix} 0.6 \\ -0.8 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ [1 mark]

Also accept $\begin{pmatrix} -3 \\ 4 \end{pmatrix}$ (opposite direction).

14. $\vec{BC} = \vec{BA} + \vec{AC} = -\vec{AB} + \vec{AC} = -\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ [3 marks]

Alternatively: $\vec{BC} = \vec{AC} - \vec{AB} = \begin{pmatrix} -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ .

Award 1 mark for correct vector relationship, 2 marks for correct answer.

15. $\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ [1 mark]

$\vec{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} 8 \\ 5 \end{pmatrix} - \begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ [1 mark]

Since $\vec{AB} = \vec{BC}$ , the vectors are parallel and share point $B$ . Therefore, $A$ , $B$ , and $C$ are collinear. [1 mark]

Section D: Vectors – Problem Solving (4 marks each)

16. (a) $\vec{OB} = \vec{OA} + \vec{OC} = \mathbf{a} + \mathbf{c}$ (parallelogram)
$\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2}(\mathbf{a} + \mathbf{a} + \mathbf{c}) = \frac{1}{2}(2\mathbf{a} + \mathbf{c}) = \mathbf{a} + \frac{1}{2}\mathbf{c}$ [1 mark]

(b) $\vec{ON} = \vec{OC} + \vec{CN} = \mathbf{c} + \frac{2}{3}\vec{CB}$
$\vec{CB} = \vec{OB} - \vec{OC} = (\mathbf{a} + \mathbf{c}) - \mathbf{c} = \mathbf{a}$
$\vec{ON} = \mathbf{c} + \frac{2}{3}\mathbf{a}$ [1 mark]

(c) $\vec{MN} = \vec{ON} - \vec{OM} = (\mathbf{c} + \frac{2}{3}\mathbf{a}) - (\mathbf{a} + \frac{1}{2}\mathbf{c}) = \frac{2}{3}\mathbf{a} - \mathbf{a} + \mathbf{c} - \frac{1}{2}\mathbf{c} = -\frac{1}{3}\mathbf{a} + \frac{1}{2}\mathbf{c}$ [2 marks]

Award 1 mark for correct expression, 1 mark for correct simplification.

17. (a) Displacement $\vec{PQ} = \mathbf{v} \times t = \begin{pmatrix} 3 \\ -4 \end{pmatrix} \times 5 = \begin{pmatrix} 15 \\ -20 \end{pmatrix}$ m [1 mark]

(b) $\vec{OQ} = \vec{OP} + \vec{PQ} = \begin{pmatrix} 2 \\ 7 \end{pmatrix} + \begin{pmatrix} 15 \\ -20 \end{pmatrix} = \begin{pmatrix} 17 \\ -13 \end{pmatrix}$ [1 mark]

Award 1 mark for correct magnitude formula, 1 mark for correct answer.

18. (a) $\mathbf{p} + \mathbf{q} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} -3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ [1 mark]
$|\mathbf{p} + \mathbf{q}| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$ [1 mark]

(b) $\mathbf{p} + k\mathbf{q} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + k\begin{pmatrix} -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 - 3k \\ 2 + k \end{pmatrix}$ [1 mark]
For the vector to be parallel to the $x$ -axis, its $y$ -component must be zero:
$2 + k = 0 \implies k = -2$ [1 mark]

19. (a) $\vec{QR} = \vec{PR} - \vec{PQ} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$ [1 mark]

(b) $|\vec{PQ}| = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}$
$|\vec{PR}| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$
$|\vec{QR}| = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$ [1 mark]

(c) Check Pythagoras: $|\vec{PQ}|^2 = 17$ , $|\vec{PR}|^2 = 10$ , $|\vec{QR}|^2 = 13$
$10 + 13 = 23 \neq 17$ ; $17 + 13 = 30 \neq 10$ ; $17 + 10 = 27 \neq 13$
None of the sums of squares of two sides equals the square of the third side. Therefore, triangle $PQR$ is not right-angled. [2 marks]

Award 1 mark for checking Pythagoras, 1 mark for correct conclusion with justification.

20. (a) $\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$
$\vec{DC} = \vec{OC} - \vec{OD} = \begin{pmatrix} 7 \\ 7 \end{pmatrix} - \begin{pmatrix} 3 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ [1 mark]

(b) Since $\vec{AB} = \vec{DC}$ , one pair of opposite sides are equal and parallel.
Check the other pair: $\vec{BC} = \vec{OC} - \vec{OB} = \begin{pmatrix} 7 \\ 7 \end{pmatrix} - \begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$
$\vec{AD} = \vec{OD} - \vec{OA} = \begin{pmatrix} 3 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$
Since $\vec{BC} = \vec{AD}$ , both pairs of opposite sides are equal and parallel. Therefore, $ABCD$ is a parallelogram. [1 mark]

(c) The diagonals of a parallelogram bisect each other. The intersection point $M$ is the midpoint of $AC$ (or $BD$ ).
$\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OC}) = \frac{1}{2}\left( \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 7 \\ 7 \end{pmatrix} \right) = \frac{1}{2}\begin{pmatrix} 8 \\ 9 \end{pmatrix} = \begin{pmatrix} 4 \\ 4.5 \end{pmatrix}$ [2 marks]

Award 1 mark for using midpoint of diagonal, 1 mark for correct answer.

END OF ANSWER KEY