Secondary 4 Elementary Mathematics Vectors Matrices Quiz

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Secondary 4 Elementary Mathematics Quiz - Vectors Matrices

Name: ___________________________

Class: ___________________________

Date: ___________________________

Score: ________ / 50

Duration: 60 minutes

Total Marks: 50

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Instructions

• Answer all questions in the spaces provided.
• Show all working clearly. Marks are awarded for correct method as well as final answer.
• Use a pencil for any diagrams or sketches.
• The use of a calculator is allowed unless otherwise stated.
• Write your final answer clearly and include appropriate units where required.
• Each question is worth between 1 and 4 marks as indicated.

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Section A: Vectors — Fundamentals and Operations (Questions 1–10)

Answer all questions. Each question is worth 2–3 marks.

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1. Given that $\mathbf{p} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}$, find the vector $2\mathbf{p} - \mathbf{q}$.

[2 marks]

 

 

 

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2. The position vector of point $A$ is $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and the position vector of point $B$ is $\begin{pmatrix} -2 \\ 7 \end{pmatrix}$.

(a) Find the vector $\overrightarrow{AB}$.

[1 mark]

 

 

(b) Hence find $|\overrightarrow{AB}|$, giving your answer correct to 2 decimal places.

[2 marks]

 

 

 

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3. Given that $\mathbf{u} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$, find the unit vector in the direction of $\mathbf{u}$.

[2 marks]

 

 

 

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4. Vectors $\mathbf{a}$ and $\mathbf{b}$ are such that $\mathbf{a} = \begin{pmatrix} 2 \\ k \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$. Given that $\mathbf{a}$ is parallel to $\mathbf{b}$, find the value of $k$.

[2 marks]

 

 

 

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5. In the diagram below (not drawn to scale), $OABC$ is a parallelogram. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $AB$.

Express the following in terms of $\mathbf{a}$ and $\mathbf{c}$:

(a) $\overrightarrow{OB}$

[1 mark]

 

 

(b) $\overrightarrow{OM}$

[2 marks]

 

 

 

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6. A particle moves from point $P$ with position vector $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ to point $Q$ with position vector $\begin{pmatrix} 7 \\ 5 \end{pmatrix}$ in 4 seconds.

(a) Find the displacement vector $\overrightarrow{PQ}$.

[1 mark]

 

 

(b) Find the average velocity vector of the particle.

[2 marks]

 

 

 

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7. Given $\mathbf{m} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}$ and $\mathbf{n} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, calculate the scalar product $\mathbf{m} \cdot \mathbf{n}$.

[2 marks]

 

 

 

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8. Vectors $\mathbf{p} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} x \\ 3 \end{pmatrix}$ are perpendicular. Find the value of $x$.

[2 marks]

 

 

 

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9. The points $A$, $B$, and $C$ have position vectors $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$, $\begin{pmatrix} 3 \\ 8 \end{pmatrix}$, and $\begin{pmatrix} 6 \\ 14 \end{pmatrix}$ respectively. Show that $A$, $B$, and $C$ are collinear.

[3 marks]

 

 

 

 

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10. In triangle $PQR$, $\overrightarrow{PQ} = 3\mathbf{i} + 2\mathbf{j}$ and $\overrightarrow{PR} = 5\mathbf{i} - 4\mathbf{j}$. $S$ lies on $QR$ such that $QS : SR = 1 : 2$.

(a) Express $\overrightarrow{QR}$ in terms of $\mathbf{i}$ and $\mathbf{j}$.

[1 mark]

 

 

(b) Hence find $\overrightarrow{PS}$.

[3 marks]

 

 

 

 

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Section B: Matrices — Operations and Applications (Questions 11–17)

Answer all questions. Each question is worth 2–4 marks.

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

[2 marks]

 

 

 

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12. Given $M = \begin{pmatrix} 3 & 1 \\ -2 & 4 \end{pmatrix}$ and $N = \begin{pmatrix} 2 & -3 \\ 5 & 1 \end{pmatrix}$, find the matrix product $MN$.

[3 marks]

 

 

 

 

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13. Find the inverse of the matrix $P = \begin{pmatrix} 4 & 3 \\ 1 & 2 \end{pmatrix}$, if it exists.

[3 marks]

 

 

 

 

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14. Solve the following simultaneous equations using the matrix method:

$$\begin{aligned} 3x + 2y &= 12 \\ 5x - y &= 7 \end{aligned}$$

[4 marks]

 

 

 

 

 

 

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15. A shop sells two types of items. The table below shows the number of items sold on two days and the total revenue.

	Item X	Item Y	Total Revenue ($)
Monday	4	6	62
Tuesday	3	5	49

Let the price of one Item X be $x$ dollars and the price of one Item Y be $y$ dollars.

(a) Write down two simultaneous equations in $x$ and $y$ to represent the information.

[1 mark]

 

 

(b) Express the system in matrix form and solve for $x$ and $y$.

[3 marks]

 

 

 

 

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16. Given $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$, find $A^2$.

[2 marks]

 

 

 

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17. The transformation matrix $T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is applied to the point $(3, 5)$.

(a) Find the image of the point $(3, 5)$ under this transformation.

[2 marks]

 

 

(b) Describe the geometric effect of this transformation.

[1 mark]

 

 

 

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Section C: Mixed Application Problems (Questions 18–20)

Answer all questions. Each question is worth 3–4 marks.

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18. A boat travels with a velocity vector $\mathbf{v} = \begin{pmatrix} 12 \\ 5 \end{pmatrix}$ km/h in still water. A current flows with velocity $\mathbf{c} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ km/h.

(a) Find the resultant velocity vector of the boat.

[1 mark]

 

 

(b) Calculate the magnitude of the resultant velocity, correct to 1 decimal place.

[2 marks]

 

 

 

(c) Find the bearing on which the boat actually travels, correct to the nearest degree.

[2 marks]

 

 

 

 

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19. The matrix $R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ represents a rotation about the origin.

(a) Write down the matrix $R$ when $\theta = 90°$.

[1 mark]

 

 

(b) The point $A(4, 1)$ is rotated $90°$ anticlockwise about the origin. Find the coordinates of the image point $A'$.

[2 marks]

 

 

(c) The point $B(x, y)$ is rotated $90°$ anticlockwise about the origin to $B'(-2, 6)$. Find the coordinates of $B$.

[2 marks]

 

 

 

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20. In the figure (not drawn to scale), $OAB$ is a triangle. $\overrightarrow{OA} = 2\mathbf{a}$ and $\overrightarrow{OB} = 3\mathbf{b}$. Point $M$ lies on $OA$ such that $OM : MA = 1 : 1$. Point $N$ lies on $AB$ such that $AN : NB = 2 : 1$. Line $ON$ is extended to meet line $MB$ produced at point $X$.

(a) Express $\overrightarrow{ON}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

[2 marks]

 

 

 

(b) Given that $\overrightarrow{OX} = k\overrightarrow{ON}$ and $\overrightarrow{MX} = h\overrightarrow{MB}$, where $k$ and $h$ are scalars, find the values of $k$ and $h$.

[4 marks]

 

 

 

 

 

 

 

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End of Quiz

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Secondary 4 Elementary Mathematics Quiz - Vectors Matrices

Answer Key

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Question 1 [2 marks]

$$2\mathbf{p} - \mathbf{q} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} - \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} - \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ -9 \end{pmatrix}$$

Answer: $\begin{pmatrix} 7 \\ -9 \end{pmatrix}$

Marking: 1 mark for correct scalar multiplication, 1 mark for correct subtraction.

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Question 2 [3 marks]

(a) [1 mark]

$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} -2 \\ 7 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} -6 \\ 6 \end{pmatrix}$$

Answer: $\begin{pmatrix} -6 \\ 6 \end{pmatrix}$

(b) [2 marks]

$$|\overrightarrow{AB}| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \approx 8.49$$

Answer: 8.49 units

Marking: 1 mark for correct formula, 1 mark for correct evaluation to 2 d.p.

Common mistake: Forgetting to subtract in the correct order ($\mathbf{b} - \mathbf{a}$, not $\mathbf{a} - \mathbf{b}$).

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Question 3 [2 marks]

$$|\mathbf{u}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

Unit vector:

$$\hat{\mathbf{u}} = \frac{1}{10}\begin{pmatrix} 6 \\ 8 \end{pmatrix} = \begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$$

Answer: $\begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$

Marking: 1 mark for correct magnitude, 1 mark for correct unit vector.

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Question 4 [2 marks]

Since $\mathbf{a}$ is parallel to $\mathbf{b}$, $\mathbf{a} = \lambda\mathbf{b}$ for some scalar $\lambda$.

$$\begin{pmatrix} 2 \\ k \end{pmatrix} = \lambda\begin{pmatrix} -3 \\ 4 \end{pmatrix}$$

From the first component: $2 = -3\lambda \Rightarrow \lambda = -\frac{2}{3}$

From the second component: $k = 4\lambda = 4 \times \left(-\frac{2}{3}\right) = -\frac{8}{3}$

Answer: $k = -\dfrac{8}{3}$

Marking: 1 mark for setting up proportionality, 1 mark for correct value.

Common mistake: Setting $\frac{2}{-3} = \frac{k}{4}$ and cross-multiplying incorrectly.

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Question 5 [3 marks]

(a) [1 mark]

In parallelogram $OABC$:

$$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = \mathbf{a} + \mathbf{c}$$

Answer: $\mathbf{a} + \mathbf{c}$

(b) [2 marks]

Since $M$ is the midpoint of $AB$:

$$\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}(\mathbf{a} + \mathbf{c} - \mathbf{a}) = \frac{1}{2}\mathbf{c}$$ $$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \frac{1}{2}\mathbf{c}$$

Answer: $\mathbf{a} + \dfrac{1}{2}\mathbf{c}$

Marking: 1 mark for finding $\overrightarrow{AB} = \mathbf{c}$, 1 mark for correct $\overrightarrow{OM}$.

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Question 6 [3 marks]

(a) [1 mark]

$$\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 7 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$$

Answer: $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$

(b) [2 marks]

Average velocity = displacement ÷ time:

$$\mathbf{v}_{avg} = \frac{1}{4}\begin{pmatrix} 6 \\ 8 \end{pmatrix} = \begin{pmatrix} 1.5 \\ 2 \end{pmatrix}$$

Answer: $\begin{pmatrix} 1.5 \\ 2 \end{pmatrix}$ units/s

Marking: 1 mark for correct formula, 1 mark for correct answer.

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Question 7 [2 marks]

$$\mathbf{m} \cdot \mathbf{n} = (5)(2) + (-3)(1) = 10 - 3 = 7$$

Answer: 7

Marking: 1 mark for correct formula, 1 mark for correct answer.

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Question 8 [2 marks]

For perpendicular vectors, $\mathbf{p} \cdot \mathbf{q} = 0$:

$$(4)(x) + (-1)(3) = 0$$ $$4x - 3 = 0$$ $$x = \frac{3}{4}$$

Answer: $x = \dfrac{3}{4}$

Marking: 1 mark for setting dot product = 0, 1 mark for correct value.

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Question 9 [3 marks]

$$\overrightarrow{AB} = \begin{pmatrix} 3 \\ 8 \end{pmatrix} - \begin{pmatrix} 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$$ $$\overrightarrow{BC} = \begin{pmatrix} 6 \\ 14 \end{pmatrix} - \begin{pmatrix} 3 \\ 8 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$$

Since $\overrightarrow{AB} = \overrightarrow{BC}$, the vectors are equal (same magnitude and direction), so $A$, $B$, and $C$ are collinear.

Answer: $\overrightarrow{AB} = \overrightarrow{BC} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$, hence $A$, $B$, $C$ are collinear.

Marking: 1 mark for $\overrightarrow{AB}$, 1 mark for $\overrightarrow{BC}$, 1 mark for conclusion with reasoning.

Common mistake: Students may find $\overrightarrow{AB}$ and $\overrightarrow{AC}$ instead; $\overrightarrow{AC} = 2\overrightarrow{AB}$ also proves collinearity and should be accepted.

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Question 10 [4 marks]

(a) [1 mark]

$$\overrightarrow{QR} = \overrightarrow{PR} - \overrightarrow{PQ} = (5\mathbf{i} - 4\mathbf{j}) - (3\mathbf{i} + 2\mathbf{j}) = 2\mathbf{i} - 6\mathbf{j}$$

Answer: $2\mathbf{i} - 6\mathbf{j}$

(b) [3 marks]

$$\overrightarrow{QS} = \frac{1}{3}\overrightarrow{QR} = \frac{1}{3}(2\mathbf{i} - 6\mathbf{j}) = \frac{2}{3}\mathbf{i} - 2\mathbf{j}$$ $$\overrightarrow{PS} = \overrightarrow{PQ} + \overrightarrow{QS} = (3\mathbf{i} + 2\mathbf{j}) + \left(\frac{2}{3}\mathbf{i} - 2\mathbf{j}\right) = \frac{11}{3}\mathbf{i} + 0\mathbf{j}$$

Answer: $\dfrac{11}{3}\mathbf{i}$

Marking: 1 mark for $\overrightarrow{QS}$, 1 mark for correct addition, 1 mark for simplified answer.

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Question 11 [2 marks]

$$A + B = \begin{pmatrix} 2+5 & -1+0 \\ 3+(-2) & 4+1 \end{pmatrix} = \begin{pmatrix} 7 & -1 \\ 1 & 5 \end{pmatrix}$$

Answer: $\begin{pmatrix} 7 & -1 \\ 1 & 5 \end{pmatrix}$

Marking: 1 mark for correct addition process, 1 mark for correct final matrix.

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Question 12 [3 marks]

$$MN = \begin{pmatrix} 3 & 1 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} 2 & -3 \\ 5 & 1 \end{pmatrix}$$ $$= \begin{pmatrix} (3)(2)+(1)(5) & (3)(-3)+(1)(1) \\ (-2)(2)+(4)(5) & (-2)(-3)+(4)(1) \end{pmatrix}$$ $$= \begin{pmatrix} 6+5 & -9+1 \\ -4+20 & 6+4 \end{pmatrix} = \begin{pmatrix} 11 & -8 \\ 16 & 10 \end{pmatrix}$$

Answer: $\begin{pmatrix} 11 & -8 \\ 16 & 10 \end{pmatrix}$

Marking: 1 mark for correct method (row × column), 1 mark for correct entries, 1 mark for final matrix.

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Question 13 [3 marks]

$$\det(P) = (4)(2) - (3)(1) = 8 - 3 = 5$$

Since $\det(P) \neq 0$, the inverse exists.

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

Answer: $\begin{pmatrix} \dfrac{2}{5} & -\dfrac{3}{5} \\ -\dfrac{1}{5} & \dfrac{4}{5} \end{pmatrix}$

Marking: 1 mark for determinant, 1 mark for correct formula application, 1 mark for correct final answer.

Common mistake: Swapping the wrong elements or getting the signs wrong on the cofactor matrix.

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Question 14 [4 marks]

In matrix form:

$$\begin{pmatrix} 3 & 2 \\ 5 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 12 \\ 7 \end{pmatrix}$$

Determinant: $(3)(-1) - (2)(5) = -3 - 10 = -13$

$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{-13}\begin{pmatrix} -1 & -2 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 12 \\ 7 \end{pmatrix}$$ $$= \frac{1}{-13}\begin{pmatrix} (-1)(12)+(-2)(7) \\ (-5)(12)+(3)(7) \end{pmatrix} = \frac{1}{-13}\begin{pmatrix} -12-14 \\ -60+21 \end{pmatrix} = \frac{1}{-13}\begin{pmatrix} -26 \\ -39 \end{pmatrix}$$ $$= \begin{pmatrix} 2 \\ 3 \end{pmatrix}$$

Answer: $x = 2$, $y = 3$

Marking: 1 mark for correct matrix form, 1 mark for determinant, 1 mark for inverse, 1 mark for correct solution.

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Question 15 [4 marks]

(a) [1 mark]

$$4x + 6y = 62$$ $$3x + 5y = 49$$

(b) [3 marks]

Matrix form:

$$\begin{pmatrix} 4 & 6 \\ 3 & 5 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 62 \\ 49 \end{pmatrix}$$

Determinant: $(4)(5) - (6)(3) = 20 - 18 = 2$

$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 5 & -6 \\ -3 & 4 \end{pmatrix}\begin{pmatrix} 62 \\ 49 \end{pmatrix}$$ $$= \frac{1}{2}\begin{pmatrix} (5)(62)+(-6)(49) \\ (-3)(62)+(4)(49) \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 310-294 \\ -186+196 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 16 \\ 10 \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \end{pmatrix}$$

Answer: $x = 8$, $y = 5$. Item X costs $8, Item Y costs $5.

Marking: 1 mark for matrix form, 1 mark for determinant and inverse, 1 mark for correct values.

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Question 16 [2 marks]

$$A^2 = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} (2)(2)+(1)(0) & (2)(1)+(1)(3) \\ (0)(2)+(3)(0) & (0)(1)+(3)(3) \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}$$

Answer: $\begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}$

Marking: 1 mark for correct method, 1 mark for correct answer.

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Question 17 [3 marks]

(a) [2 marks]

\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 5 \end{pmatrix} = \begin{pmatrix} (0)(3)+(-1)(5) \\ (1)(3)+(0)(5) \end{pmatrix} = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ** **Answer:** $(-5, 3)$ **(b) [1 mark]** **Answer:** Rotation of $90°$ anticlockwise about the origin. **Marking:** 1 mark for matrix multiplication, 1 mark for correct image, 1 mark for correct description. --- ### Question 18 [5 marks] **(a) [1 mark]**

\mathbf{v}_{resultant} = \mathbf{v} + \mathbf{c} = \begin{pmatrix} 12 \ 5 \end{pmatrix} + \begin{pmatrix} -2 \ 3 \end{pmatrix} = \begin{pmatrix} 10 \ 8 \end{pmatrix} \text{ km/h}

**Answer:** $\begin{pmatrix} 10 \\ 8 \end{pmatrix}$ km/h **(b) [2 marks]**

|\mathbf{v}_{resultant}| = \sqrt{10^2 + 8^2} = \sqrt{100 + 64} = \sqrt{164} \approx 12.8 \text{ km/h}

**Answer:** 12.8 km/h **(c) [2 marks]** The resultant vector is $\begin{pmatrix} 10 \\ 8 \end{pmatrix}$, i.e., 10 km/h east and 8 km/h north. Angle east of north: $\theta = \tan^{-1}\left(\frac{10}{8}\right) = \tan^{-1}(1.25) \approx 51.3°$ Bearing = $090° - 51.3° = 038.7° \approx 039°$ Alternatively, bearing measured clockwise from north: $\tan^{-1}\left(\frac{10}{8}\right) \approx 51°$ east of north, so bearing $\approx 051°$. *Note: Convention — bearing is measured clockwise from north. The vector $\begin{pmatrix} 10 \\ 8 \end{pmatrix}$ has east component 10 and north component 8.* Angle from north towards east: $\alpha = \tan^{-1}\left(\dfrac{10}{8}\right) \approx 51.3°$ Bearing $\approx 051°$ (to nearest degree). **Answer:** Bearing $051°$ **Marking:** 1 mark for resultant, 1 mark for magnitude, 1 mark for angle calculation, 1 mark for bearing. --- ### Question 19 [5 marks] **(a) [1 mark]** When $\theta = 90°$: $\cos 90° = 0$, $\sin 90° = 1$

R = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}

**Answer:** $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ **(b) [2 marks]**

\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}\begin{pmatrix} 4 \ 1 \end{pmatrix} = \begin{pmatrix} -1 \ 4 \end{pmatrix}

**Answer:** $A'(-1, 4)$ **(c) [2 marks]** For a $90°$ anticlockwise rotation: $(x, y) \rightarrow (-y, x)$ So $B'(-2, 6)$ means: $-y = -2 \Rightarrow y = 2$ and $x = 6$. **Answer:** $B(6, 2)$ **Marking:** 1 mark for matrix, 1 mark for multiplication, 1 mark for $A'$, 1 mark for setting up equations, 1 mark for $B$. --- ### Question 20 [6 marks] **(a) [2 marks]** $\overrightarrow{OA} = 2\mathbf{a}$, so $M$ is the midpoint of $OA$: $\overrightarrow{OM} = \mathbf{a}$. $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = 3\mathbf{b} - 2\mathbf{a}$ Since $AN : NB = 2 : 1$, point $N$ divides $AB$ in the ratio $2:1$:

\overrightarrow{ON} = \overrightarrow{OA} + \frac{2}{3}\overrightarrow{AB} = 2\mathbf{a} + \frac{2}{3}(3\mathbf{b} - 2\mathbf{a}) = 2\mathbf{a} + 2\mathbf{b} - \frac{4}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} + 2\mathbf{b}

**Answer:** $\dfrac{2}{3}\mathbf{a} + 2\mathbf{b}$ **(b) [4 marks]** $\overrightarrow{OX} = k\overrightarrow{ON} = k\left(\dfrac{2}{3}\mathbf{a} + 2\mathbf{b}\right) = \dfrac{2k}{3}\mathbf{a} + 2k\mathbf{b}$ $\overrightarrow{MB} = \overrightarrow{OB} - \overrightarrow{OM} = 3\mathbf{b} - \mathbf{a}$ $\overrightarrow{MX} = h\overrightarrow{MB} = h(3\mathbf{b} - \mathbf{a}) = -h\mathbf{a} + 3h\mathbf{b}$ Also: $\overrightarrow{OX} = \overrightarrow{OM} + \overrightarrow{MX} = \mathbf{a} + (-h\mathbf{a} + 3h\mathbf{b}) = (1-h)\mathbf{a} + 3h\mathbf{b}$ Equating the two expressions for $\overrightarrow{OX}$:

\dfrac{2k}{3}\mathbf{a} + 2k\mathbf{b} = (1-h)\mathbf{a} + 3h\mathbf{b}

Comparing coefficients of $\mathbf{a}$: $\dfrac{2k}{3} = 1 - h$ ... (i) Comparing coefficients of $\mathbf{b}$: $2k = 3h$ ... (ii) From (ii): $k = \dfrac{3h}{2}$ Substitute into (i): $\dfrac{2}{3} \cdot \dfrac{3h}{2} = 1 - h \Rightarrow h = 1 - h \Rightarrow 2h = 1 \Rightarrow h = \dfrac{1}{2}$ From (ii): $2k = 3 \times \dfrac{1}{2} = \dfrac{3}{2} \Rightarrow k = \dfrac{3}{4}$ **Answer:** $k = \dfrac{3}{4}$, $h = \dfrac{1}{2}$ **Marking:** 1 mark for $\overrightarrow{ON}$, 1 mark for $\overrightarrow{OX} = k\overrightarrow{ON}$, 1 mark for $\overrightarrow{MX} = h\overrightarrow{MB}$, 1 mark for equating and solving. **Common mistake:** Incorrect ratio division for point $N$ on $AB$. Students should use section formula carefully. --- **Total: 50 marks**