Secondary 3 Additional Mathematics Vectors Matrices Quiz

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Secondary 3 Additional Mathematics Quiz - Vectors Matrices

Name: ________________________
Class: ________________________
Date: ________________________
Score: ________ / 60

Duration: 75 minutes
Total Marks: 60

Instructions:

• Answer ALL questions in the spaces provided.
• Show all working clearly. Marks will be awarded for correct reasoning and method, not only for the final answer.
• Non-exact answers should be given correct to 3 significant figures unless otherwise stated.
• The use of a scientific calculator is allowed.
• Vectors may be written in column form $\begin{pmatrix} x \\ y \end{pmatrix}$ or component form $x\mathbf{i} + y\mathbf{j}$.

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Section A: Vectors (Questions 1–10)

Questions 1–5 are multiple-choice. Shade the correct option on your answer sheet. Each question carries 2 marks.

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1. Two vectors are given by $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}$. Find $\mathbf{a} + \mathbf{b}$.

A. $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$
B. $\begin{pmatrix} 4 \\ -7 \end{pmatrix}$
C. $\begin{pmatrix} -3 \\ -10 \end{pmatrix}$
D. $\begin{pmatrix} 2 \\ -7 \end{pmatrix}$

 

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2. Given $\mathbf{p} = 4\mathbf{i} - 3\mathbf{j}$, find the magnitude $|\mathbf{p}|$.

A. $\sqrt{7}$
B. $5$
C. $7$
D. $\sqrt{13}$

 

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3. The vector $\overrightarrow{AB} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}$. Point $A$ has coordinates $(1, 4)$. Find the coordinates of point $B$.

A. $(7, -4)$
B. $(5, 12)$
C. $(-5, 12)$
D. $(7, 12)$

 

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4. Vectors $\mathbf{u} = \begin{pmatrix} 2 \\ k \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 6 \\ 15 \end{pmatrix}$ are parallel. Find the value of $k$.

A. $3$
B. $5$
C. $7$
D. $10$

 

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5. A unit vector in the direction of $\mathbf{a} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$ is:

A. $\begin{pmatrix} \frac{5}{12} \\ 1 \end{pmatrix}$
B. $\begin{pmatrix} \frac{5}{13} \\ \frac{12}{13} \end{pmatrix}$
C. $\begin{pmatrix} \frac{12}{13} \\ \frac{5}{13} \end{pmatrix}$
D. $\begin{pmatrix} 5 \\ 12 \end{pmatrix}$

 

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Questions 6–10 are short-answer. Show your working clearly.

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6. Given $\mathbf{m} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}$ and $\mathbf{n} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$, find:

(a) $3\mathbf{m} - 2\mathbf{n}$
(b) $|3\mathbf{m} - 2\mathbf{n}|$, giving your answer correct to 2 decimal places.

 

 

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7. Points $P$, $Q$, and $R$ have position vectors $\vec{p} = \mathbf{i} + 3\mathbf{j}$, $\vec{q} = 5\mathbf{i} - \mathbf{j}$, and $\vec{r} = -3\mathbf{i} + 7\mathbf{j}$ respectively.

(a) Find $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ as column vectors.
(b) Hence determine whether $P$, $Q$, and $R$ are collinear. Justify your answer.

 

 

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8. A vector $\mathbf{v}$ has magnitude 13 and is in the direction of $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$.

(a) Find a unit vector in the direction of $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$.
(b) Hence find $\mathbf{v}$.

 

 

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9. Given $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$, find the scalar $k$ such that $\mathbf{a} + k\mathbf{b}$ is parallel to $\begin{pmatrix} 5 \\ -6 \end{pmatrix}$.

 

 

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10. The points $A(2, 1)$ and $B(8, 7)$ are given. Point $C$ lies on the line $AB$ such that $\overrightarrow{AC} = \frac{2}{3}\overrightarrow{AB}$.

(a) Find the vector $\overrightarrow{AB}$.
(b) Hence find the coordinates of point $C$.

 

 

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Section B: Matrices (Questions 11–16)

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

(a) $A + B$
(b) $3A - 2B$

 

 

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12. Given $P = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}$ and $Q = \begin{pmatrix} 2 & -1 \\ 5 & 0 \end{pmatrix}$, find the product $PQ$.

 

 

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

 

 

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14. Solve the simultaneous equations using a matrix method:

$3x + 2y = 12$ $5x - y = 7$

Write the equations in the form $A\mathbf{x} = \mathbf{b}$, find $A^{-1}$, and hence solve for $x$ and $y$.

 

 

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15. A transformation is represented by the matrix $T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

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

 

 

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

(a) $\det(A)$
(b) $\det(B)$
(c) $\det(AB)$, and verify that $\det(AB) = \det(A)\det(B)$.

 

 

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Section C: Application Problems (Questions 17–20)

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17. A boat travels with a velocity vector $\mathbf{v}_b = \begin{pmatrix} 8 \\ 6 \end{pmatrix}$ km/h in still water. The river current has a velocity vector $\mathbf{v}_c = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$ km/h.

(a) Find the resultant velocity vector of the boat.
(b) Find the magnitude of the resultant velocity, correct to 2 decimal places.
(c) Find the direction of the resultant velocity as a bearing, correct to the nearest degree.

 

 

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18. In a factory, two products $X$ and $Y$ require different amounts of materials $M_1$ and $M_2$. The requirements are summarised in matrix form:

$R = \begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix}$

where the rows represent materials $M_1$ and $M_2$ (in kg), and the columns represent products $X$ and $Y$ respectively.

An order is placed for 20 units of product $X$ and 15 units of product $Y$, represented by the matrix $O = \begin{pmatrix} 20 \\ 15 \end{pmatrix}$.

(a) Calculate the total amount of each material required using matrix multiplication.
(b) If material $M_1$ costs $4 per kg and material $M_2$ costs $6 per kg, find the total cost of materials for this order.

 

 

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19. The points $A(1, 2)$, $B(4, 6)$, and $C(7, 4)$ form a triangle.

(a) Express $\overrightarrow{AB}$ and $\overrightarrow{AC}$ as column vectors.
(b) Use the scalar product $\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}|\cos\theta$ to find $\angle BAC$, correct to the nearest degree.

 

 

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20. A transformation $T$ maps the point $(x, y)$ to $(x', y')$ where:

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

(a) Find the image of the line $y = 2x + 1$ under this transformation. Express your answer in the form $y' = mx' + c$.
(b) Find the area scale factor of this transformation.
(c) Determine whether the point $(5, 8)$ lies on the image of the line $y = 2x + 1$. Justify your answer.

 

 

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END OF QUIZ

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Secondary 3 Additional Mathematics Quiz - Vectors Matrices

Answer Key

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Section A: Vectors (Questions 1–10)

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1. A. $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$
[2 marks]
$\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 + (-1) \\ -2 + 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$

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2. B. $5$
[2 marks]
$|\mathbf{p}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

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3. A. $(7, -4)$
[2 marks]
$B = A + \overrightarrow{AB} = (1+6, 4+(-8)) = (7, -4)$

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4. B. $5$
[2 marks]
For parallel vectors: $\frac{6}{2} = \frac{15}{k} \Rightarrow 3 = \frac{15}{k} \Rightarrow k = 5$

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5. B. $\begin{pmatrix} \frac{5}{13} \\ \frac{12}{13} \end{pmatrix}$
[2 marks]
$|\mathbf{a}| = \sqrt{25 + 144} = \sqrt{169} = 13$
Unit vector $= \frac{1}{13}\begin{pmatrix} 5 \\ 12 \end{pmatrix} = \begin{pmatrix} \frac{5}{13} \\ \frac{12}{13} \end{pmatrix}$

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6. (a) $3\mathbf{m} - 2\mathbf{n} = \begin{pmatrix} -9 \\ 21 \end{pmatrix} - \begin{pmatrix} 8 \\ -2 \end{pmatrix} = \begin{pmatrix} -17 \\ 23 \end{pmatrix}$
[2 marks]
(b) $|3\mathbf{m} - 2\mathbf{n}| = \sqrt{(-17)^2 + 23^2} = \sqrt{289 + 529} = \sqrt{818} \approx 28.60$
[2 marks]
Marking note: Award 1 mark for correct substitution into magnitude formula; 1 mark for correct final answer.

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7. (a) $\overrightarrow{PQ} = \vec{q} - \vec{p} = \begin{pmatrix} 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}$
$\overrightarrow{QR} = \vec{r} - \vec{q} = \begin{pmatrix} -3 \\ 7 \end{pmatrix} - \begin{pmatrix} 5 \\ -1 \end{pmatrix} = \begin{pmatrix} -8 \\ 8 \end{pmatrix}$
[2 marks]
(b) $\overrightarrow{QR} = -2 \times \overrightarrow{PQ}$ since $\begin{pmatrix} -8 \\ 8 \end{pmatrix} = -2\begin{pmatrix} 4 \\ -4 \end{pmatrix}$.
Since $\overrightarrow{QR}$ is a scalar multiple of $\overrightarrow{PQ}$, the vectors are parallel and share point $Q$, so $P$, $Q$, and $R$ are collinear.
[2 marks]
Marking note: Award 1 mark for showing the scalar multiple relationship; 1 mark for the collinearity conclusion with justification.

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8. (a) Unit vector $= \frac{1}{\sqrt{25+144}}\begin{pmatrix} 5 \\ -12 \end{pmatrix} = \frac{1}{13}\begin{pmatrix} 5 \\ -12 \end{pmatrix} = \begin{pmatrix} \frac{5}{13} \\ -\frac{12}{13} \end{pmatrix}$
[2 marks]
(b) $\mathbf{v} = 13 \times \begin{pmatrix} \frac{5}{13} \\ -\frac{12}{13} \end{pmatrix} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}$
[1 mark]

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9. $\mathbf{a} + k\mathbf{b} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + k\begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1-3k \\ 2+4k \end{pmatrix}$
For this to be parallel to $\begin{pmatrix} 5 \\ -6 \end{pmatrix}$:
$\frac{1-3k}{5} = \frac{2+4k}{-6}$
$-6(1-3k) = 5(2+4k)$
$-6 + 18k = 10 + 20k$
$-16 = 2k$
$k = -8$
[3 marks]
Marking note: Award 1 mark for setting up the vector sum; 1 mark for the proportionality equation; 1 mark for correct solution.

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10. (a) $\overrightarrow{AB} = \begin{pmatrix} 8-2 \\ 7-1 \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}$
[1 mark]
(b) $\overrightarrow{AC} = \frac{2}{3}\begin{pmatrix} 6 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$
$C = A + \overrightarrow{AC} = (2+4, 1+4) = (6, 5)$
[2 marks]
Marking note: Award 1 mark for correct scalar multiplication; 1 mark for correct coordinates.

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Section B: Matrices (Questions 11–16)

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11. (a) $A + B = \begin{pmatrix} 2+5 & -1+0 \\ 3+(-2) & 4+3 \end{pmatrix} = \begin{pmatrix} 7 & -1 \\ 1 & 7 \end{pmatrix}$
[1 mark]
(b) $3A - 2B = \begin{pmatrix} 6 & -3 \\ 9 & 12 \end{pmatrix} - \begin{pmatrix} 10 & 0 \\ -4 & 6 \end{pmatrix} = \begin{pmatrix} -4 & -3 \\ 13 & 6 \end{pmatrix}$
[2 marks]

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12. $PQ = \begin{pmatrix} 1\times2+3\times5 & 1\times(-1)+3\times0 \\ -2\times2+4\times5 & -2\times(-1)+4\times0 \end{pmatrix} = \begin{pmatrix} 2+15 & -1+0 \\ -4+20 & 2+0 \end{pmatrix} = \begin{pmatrix} 17 & -1 \\ 16 & 2 \end{pmatrix}$
[3 marks]
Marking note: Award 1 mark for correct method; 1 mark for correct computation of elements; 1 mark for final matrix.

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13. $\det(M) = (4)(2) - (3)(1) = 8 - 3 = 5 \neq 0$, so the inverse exists.
$M^{-1} = \frac{1}{5}\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} \frac{2}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{4}{5} \end{pmatrix}$
[3 marks]
Marking note: Award 1 mark for determinant; 1 mark for correct formula application; 1 mark for final answer.

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14. $A = \begin{pmatrix} 3 & 2 \\ 5 & -1 \end{pmatrix}$, $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}$, $\mathbf{b} = \begin{pmatrix} 12 \\ 7 \end{pmatrix}$
$\det(A) = (3)(-1) - (2)(5) = -3 - 10 = -13$
$A^{-1} = \frac{1}{-13}\begin{pmatrix} -1 & -2 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{13} & \frac{2}{13} \\ \frac{5}{13} & -\frac{3}{13} \end{pmatrix}$
$\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\mathbf{b} = \begin{pmatrix} \frac{1}{13}\times12 + \frac{2}{13}\times7 \\ \frac{5}{13}\times12 + (-\frac{3}{13})\times7 \end{pmatrix} = \begin{pmatrix} \frac{12+14}{13} \\ \frac{60-21}{13} \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$
$x = 2$, $y = 3$
[5 marks]
Marking note: Award 1 mark for correct matrix setup; 1 mark for determinant; 1 mark for inverse; 1 mark for multiplication; 1 mark for final answer.

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15. (a) $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 5 \end{pmatrix} = \begin{pmatrix} 0\times3+(-1)\times5 \\ 1\times3+0\times5 \end{pmatrix} = \begin{pmatrix} -5 \\ 3 \end{pmatrix}$
Image: $(-5, 3)$
[2 marks]
(b) This is a rotation of $90°$ anticlockwise about the origin.
[1 mark]
Marking note: Accept equivalent descriptions such as "rotation about O through 90° in the anticlockwise direction."

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16. (a) $\det(A) = (2)(3) - (1)(-1) = 6 + 1 = 7$
[1 mark]
(b) $\det(B) = (1)(-1) - (0)(2) = -1$
[1 mark]
(c) $AB = \begin{pmatrix} 2\times1+1\times2 & 2\times0+1\times(-1) \\ -1\times1+3\times2 & -1\times0+3\times(-1) \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ 5 & -3 \end{pmatrix}$
$\det(AB) = (4)(-3) - (-1)(5) = -12 + 5 = -7$
$\det(A)\det(B) = 7 \times (-1) = -7$
Hence $\det(AB) = \det(A)\det(B)$ ✓
[3 marks]
Marking note: Award 1 mark for computing AB; 1 mark for det(AB); 1 mark for verification.

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Section C: Application Problems (Questions 17–20)

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17. (a) $\mathbf{v}_r = \mathbf{v}_b + \mathbf{v}_c = \begin{pmatrix} 8 \\ 6 \end{pmatrix} + \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \end{pmatrix}$ km/h
[2 marks]
(b) $|\mathbf{v}_r| = \sqrt{6^2 + 7^2} = \sqrt{36 + 49} = \sqrt{85} \approx 9.22$ km/h
[1 mark]
(c) $\tan\theta = \frac{7}{6} \Rightarrow \theta = \tan^{-1}\left(\frac{7}{6}\right) \approx 49.4°$
Bearing $= 90° - 49.4° = 40.6° \approx 041°$
[2 marks]
Marking note: Award 1 mark for correct angle; 1 mark for correct bearing. Accept 040° or 041°.

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18. (a) $\begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 20 \\ 15 \end{pmatrix} = \begin{pmatrix} 3\times20+5\times15 \\ 2\times20+4\times15 \end{pmatrix} = \begin{pmatrix} 60+75 \\ 40+60 \end{pmatrix} = \begin{pmatrix} 135 \\ 100 \end{pmatrix}$
Material $M_1$: 135 kg; Material $M_2$: 100 kg
[2 marks]
(b) Total cost = 135 \times 4 + 100 \times 6 = 540 + 600 = \1140$
[2 marks]

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19. (a) $\overrightarrow{AB} = \begin{pmatrix} 4-1 \\ 6-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
$\overrightarrow{AC} = \begin{pmatrix} 7-1 \\ 4-2 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$
[2 marks]
(b) $\overrightarrow{AB} \cdot \overrightarrow{AC} = 3\times6 + 4\times2 = 18 + 8 = 26$
$|\overrightarrow{AB}| = \sqrt{9+16} = 5$
$|\overrightarrow{AC}| = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$
$\cos\angle BAC = \frac{26}{5 \times 2\sqrt{10}} = \frac{26}{10\sqrt{10}} = \frac{13}{5\sqrt{10}} \approx 0.8222$
$\angle BAC = \cos^{-1}(0.8222) \approx 34.7° \approx 35°$
[4 marks]
Marking note: Award 1 mark for scalar product; 1 mark for magnitudes; 1 mark for cos value; 1 mark for angle.

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20. (a) Let $(x, y)$ be a point on $y = 2x + 1$. Then:
$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2x+y \\ x+3y \end{pmatrix}$
Substitute $y = 2x + 1$:
$x' = 2x + (2x+1) = 4x + 1$
$y' = x + 3(2x+1) = x + 6x + 3 = 7x + 3$
From $x' = 4x + 1$: $x = \frac{x'-1}{4}$
Substitute into $y'$: $y' = 7\left(\frac{x'-1}{4}\right) + 3 = \frac{7x'-7}{4} + 3 = \frac{7x'-7+12}{4} = \frac{7x'+5}{4}$
$y' = \frac{7}{4}x' + \frac{5}{4}$
[4 marks]
Marking note: Award 1 mark for substitution; 1 mark for expressing x' and y' in terms of x; 1 mark for eliminating x; 1 mark for final equation.
(b) Area scale factor $= |\det(T)| = |(2)(3)-(1)(1)| = |6-1| = 5$
[1 mark]
(c) Check: $8 = \frac{7}{4}(5) + \frac{5}{4} = \frac{35+5}{4} = \frac{40}{4} = 10$
Since $8 \neq 10$, the point $(5, 8)$ does not lie on the image.
[2 marks]
Marking note: Award 1 mark for substitution; 1 mark for conclusion.

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Total: 60 marks