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Secondary 3 Additional Mathematics Vectors Matrices Quiz

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Questions

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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 (xy)\begin{pmatrix} x \\ y \end{pmatrix} or component form xi+yjx\mathbf{i} + y\mathbf{j}.

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.

1. Two vectors are given by a=(3−2)\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} and b=(−15)\mathbf{b} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}. Find a+b\mathbf{a} + \mathbf{b}.

A. (23)\begin{pmatrix} 2 \\ 3 \end{pmatrix}
B. (4−7)\begin{pmatrix} 4 \\ -7 \end{pmatrix}
C. (−3−10)\begin{pmatrix} -3 \\ -10 \end{pmatrix}
D. (2−7)\begin{pmatrix} 2 \\ -7 \end{pmatrix}

 

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

A. 7\sqrt{7}
B. 55
C. 77
D. 13\sqrt{13}

 

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

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

 

4. Vectors u=(2k)\mathbf{u} = \begin{pmatrix} 2 \\ k \end{pmatrix} and v=(615)\mathbf{v} = \begin{pmatrix} 6 \\ 15 \end{pmatrix} are parallel. Find the value of kk.

A. 33
B. 55
C. 77
D. 1010

 

5. A unit vector in the direction of a=(512)\mathbf{a} = \begin{pmatrix} 5 \\ 12 \end{pmatrix} is:

A. (5121)\begin{pmatrix} \frac{5}{12} \\ 1 \end{pmatrix}
B. (5131213)\begin{pmatrix} \frac{5}{13} \\ \frac{12}{13} \end{pmatrix}
C. (1213513)\begin{pmatrix} \frac{12}{13} \\ \frac{5}{13} \end{pmatrix}
D. (512)\begin{pmatrix} 5 \\ 12 \end{pmatrix}

 

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

6. Given m=(−37)\mathbf{m} = \begin{pmatrix} -3 \\ 7 \end{pmatrix} and n=(4−1)\mathbf{n} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}, find:

(a) 3m−2n3\mathbf{m} - 2\mathbf{n}
(b) ∣3m−2n∣|3\mathbf{m} - 2\mathbf{n}|, giving your answer correct to 2 decimal places.

 

 

7. Points PP, QQ, and RR have position vectors p⃗=i+3j\vec{p} = \mathbf{i} + 3\mathbf{j}, q⃗=5i−j\vec{q} = 5\mathbf{i} - \mathbf{j}, and r⃗=−3i+7j\vec{r} = -3\mathbf{i} + 7\mathbf{j} respectively.

(a) Find PQ→\overrightarrow{PQ} and QR→\overrightarrow{QR} as column vectors.
(b) Hence determine whether PP, QQ, and RR are collinear. Justify your answer.

 

 

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

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

 

 

9. Given a=(12)\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} and b=(−34)\mathbf{b} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}, find the scalar kk such that a+kb\mathbf{a} + k\mathbf{b} is parallel to (5−6)\begin{pmatrix} 5 \\ -6 \end{pmatrix}.

 

 

10. The points A(2,1)A(2, 1) and B(8,7)B(8, 7) are given. Point CC lies on the line ABAB such that AC→=23AB→\overrightarrow{AC} = \frac{2}{3}\overrightarrow{AB}.

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

 

 

Section B: Matrices (Questions 11–16)

11. Given A=(2−134)A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} and B=(50−23)B = \begin{pmatrix} 5 & 0 \\ -2 & 3 \end{pmatrix}, find:

(a) A+BA + B
(b) 3A−2B3A - 2B

 

 

12. Given P=(13−24)P = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} and Q=(2−150)Q = \begin{pmatrix} 2 & -1 \\ 5 & 0 \end{pmatrix}, find the product PQPQ.

 

 

13. Find the inverse of the matrix M=(4312)M = \begin{pmatrix} 4 & 3 \\ 1 & 2 \end{pmatrix}, if it exists.

 

 

14. Solve the simultaneous equations using a matrix method:

3x+2y=123x + 2y = 12 5x−y=75x - y = 7

Write the equations in the form Ax=bA\mathbf{x} = \mathbf{b}, find A−1A^{-1}, and hence solve for xx and yy.

 

 

15. A transformation is represented by the matrix T=(0−110)T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.

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

 

 

16. Given A=(21−13)A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} and B=(102−1)B = \begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}, find:

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

 

 

Section C: Application Problems (Questions 17–20)

17. A boat travels with a velocity vector vb=(86)\mathbf{v}_b = \begin{pmatrix} 8 \\ 6 \end{pmatrix} km/h in still water. The river current has a velocity vector vc=(−21)\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.

 

 

18. In a factory, two products XX and YY require different amounts of materials M1M_1 and M2M_2. The requirements are summarised in matrix form:

R=(3524)R = \begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix}

where the rows represent materials M1M_1 and M2M_2 (in kg), and the columns represent products XX and YY respectively.

An order is placed for 20 units of product XX and 15 units of product YY, represented by the matrix O=(2015)O = \begin{pmatrix} 20 \\ 15 \end{pmatrix}.

(a) Calculate the total amount of each material required using matrix multiplication.
(b) If material M1M_1 costs $4 per kg and material M2M_2 costs $6 per kg, find the total cost of materials for this order.

 

 

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

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

 

 

20. A transformation TT maps the point (x,y)(x, y) to (x′,y′)(x', y') where:

(x′y′)=(2113)(xy)\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+1y = 2x + 1 under this transformation. Express your answer in the form y′=mx′+cy' = mx' + c.
(b) Find the area scale factor of this transformation.
(c) Determine whether the point (5,8)(5, 8) lies on the image of the line y=2x+1y = 2x + 1. Justify your answer.

 

 

END OF QUIZ

Answers

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Secondary 3 Additional Mathematics Quiz - Vectors Matrices

Answer Key

Section A: Vectors (Questions 1–10)

1. A. (23)\begin{pmatrix} 2 \\ 3 \end{pmatrix}
[2 marks]
a+b=(3+(−1)−2+5)=(23)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 + (-1) \\ -2 + 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

2. B. 55
[2 marks]
∣p∣=42+(−3)2=16+9=25=5|\mathbf{p}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

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

4. B. 55
[2 marks]
For parallel vectors: 62=15k⇒3=15k⇒k=5\frac{6}{2} = \frac{15}{k} \Rightarrow 3 = \frac{15}{k} \Rightarrow k = 5

5. B. (5131213)\begin{pmatrix} \frac{5}{13} \\ \frac{12}{13} \end{pmatrix}
[2 marks]
∣a∣=25+144=169=13|\mathbf{a}| = \sqrt{25 + 144} = \sqrt{169} = 13
Unit vector =113(512)=(5131213)= \frac{1}{13}\begin{pmatrix} 5 \\ 12 \end{pmatrix} = \begin{pmatrix} \frac{5}{13} \\ \frac{12}{13} \end{pmatrix}

6. (a) 3m−2n=(−921)−(8−2)=(−1723)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) ∣3m−2n∣=(−17)2+232=289+529=818≈28.60|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.

7. (a) PQ→=q⃗−p⃗=(5−1)−(13)=(4−4)\overrightarrow{PQ} = \vec{q} - \vec{p} = \begin{pmatrix} 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}
QR→=r⃗−q⃗=(−37)−(5−1)=(−88)\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) QR→=−2×PQ→\overrightarrow{QR} = -2 \times \overrightarrow{PQ} since (−88)=−2(4−4)\begin{pmatrix} -8 \\ 8 \end{pmatrix} = -2\begin{pmatrix} 4 \\ -4 \end{pmatrix}.
Since QR→\overrightarrow{QR} is a scalar multiple of PQ→\overrightarrow{PQ}, the vectors are parallel and share point QQ, so PP, QQ, and RR are collinear.
[2 marks]
Marking note: Award 1 mark for showing the scalar multiple relationship; 1 mark for the collinearity conclusion with justification.

8. (a) Unit vector =125+144(5−12)=113(5−12)=(513−1213)= \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) v=13×(513−1213)=(5−12)\mathbf{v} = 13 \times \begin{pmatrix} \frac{5}{13} \\ -\frac{12}{13} \end{pmatrix} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}
[1 mark]

9. a+kb=(12)+k(−34)=(1−3k2+4k)\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 (5−6)\begin{pmatrix} 5 \\ -6 \end{pmatrix}:
1−3k5=2+4k−6\frac{1-3k}{5} = \frac{2+4k}{-6}
−6(1−3k)=5(2+4k)-6(1-3k) = 5(2+4k)
−6+18k=10+20k-6 + 18k = 10 + 20k
−16=2k-16 = 2k
k=−8k = -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.

10. (a) AB→=(8−27−1)=(66)\overrightarrow{AB} = \begin{pmatrix} 8-2 \\ 7-1 \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}
[1 mark]
(b) AC→=23(66)=(44)\overrightarrow{AC} = \frac{2}{3}\begin{pmatrix} 6 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}
C=A+AC→=(2+4,1+4)=(6,5)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.

Section B: Matrices (Questions 11–16)

11. (a) A+B=(2+5−1+03+(−2)4+3)=(7−117)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=(6−3912)−(100−46)=(−4−3136)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]

12. PQ=(1×2+3×51×(−1)+3×0−2×2+4×5−2×(−1)+4×0)=(2+15−1+0−4+202+0)=(17−1162)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.

13. det⁡(M)=(4)(2)−(3)(1)=8−3=5≠0\det(M) = (4)(2) - (3)(1) = 8 - 3 = 5 \neq 0, so the inverse exists.
M−1=15(2−3−14)=(25−35−1545)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.

14. A=(325−1)A = \begin{pmatrix} 3 & 2 \\ 5 & -1 \end{pmatrix}, x=(xy)\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, b=(127)\mathbf{b} = \begin{pmatrix} 12 \\ 7 \end{pmatrix}
det⁡(A)=(3)(−1)−(2)(5)=−3−10=−13\det(A) = (3)(-1) - (2)(5) = -3 - 10 = -13
A−1=1−13(−1−2−53)=(113213513−313)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}
(xy)=A−1b=(113×12+213×7513×12+(−313)×7)=(12+141360−2113)=(23)\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=2x = 2, y=3y = 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.

15. (a) (0−110)(35)=(0×3+(−1)×51×3+0×5)=(−53)\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)(-5, 3)
[2 marks]
(b) This is a rotation of 90°90° anticlockwise about the origin.
[1 mark]
Marking note: Accept equivalent descriptions such as "rotation about O through 90° in the anticlockwise direction."

16. (a) det⁡(A)=(2)(3)−(1)(−1)=6+1=7\det(A) = (2)(3) - (1)(-1) = 6 + 1 = 7
[1 mark]
(b) det⁡(B)=(1)(−1)−(0)(2)=−1\det(B) = (1)(-1) - (0)(2) = -1
[1 mark]
(c) AB=(2×1+1×22×0+1×(−1)−1×1+3×2−1×0+3×(−1))=(4−15−3)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(AB) = (4)(-3) - (-1)(5) = -12 + 5 = -7
det⁡(A)det⁡(B)=7×(−1)=−7\det(A)\det(B) = 7 \times (-1) = -7
Hence det⁡(AB)=det⁡(A)det⁡(B)\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.

Section C: Application Problems (Questions 17–20)

17. (a) vr=vb+vc=(86)+(−21)=(67)\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) ∣vr∣=62+72=36+49=85≈9.22|\mathbf{v}_r| = \sqrt{6^2 + 7^2} = \sqrt{36 + 49} = \sqrt{85} \approx 9.22 km/h
[1 mark]
(c) tan⁡θ=76⇒θ=tan⁡−1(76)≈49.4°\tan\theta = \frac{7}{6} \Rightarrow \theta = \tan^{-1}\left(\frac{7}{6}\right) \approx 49.4°
Bearing =90°−49.4°=40.6°≈041°= 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°.

18. (a) (3524)(2015)=(3×20+5×152×20+4×15)=(60+7540+60)=(135100)\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 M1M_1: 135 kg; Material M2M_2: 100 kg
[2 marks]
(b) Total cost = 135 \times 4 + 100 \times 6 = 540 + 600 = \1140$
[2 marks]

19. (a) AB→=(4−16−2)=(34)\overrightarrow{AB} = \begin{pmatrix} 4-1 \\ 6-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}
AC→=(7−14−2)=(62)\overrightarrow{AC} = \begin{pmatrix} 7-1 \\ 4-2 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}
[2 marks]
(b) AB→⋅AC→=3×6+4×2=18+8=26\overrightarrow{AB} \cdot \overrightarrow{AC} = 3\times6 + 4\times2 = 18 + 8 = 26
∣AB→∣=9+16=5|\overrightarrow{AB}| = \sqrt{9+16} = 5
∣AC→∣=36+4=40=210|\overrightarrow{AC}| = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}
cos⁡∠BAC=265×210=261010=13510≈0.8222\cos\angle BAC = \frac{26}{5 \times 2\sqrt{10}} = \frac{26}{10\sqrt{10}} = \frac{13}{5\sqrt{10}} \approx 0.8222
∠BAC=cos⁡−1(0.8222)≈34.7°≈35°\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.

20. (a) Let (x,y)(x, y) be a point on y=2x+1y = 2x + 1. Then:
(x′y′)=(2x+yx+3y)\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2x+y \\ x+3y \end{pmatrix}
Substitute y=2x+1y = 2x + 1:
x′=2x+(2x+1)=4x+1x' = 2x + (2x+1) = 4x + 1
y′=x+3(2x+1)=x+6x+3=7x+3y' = x + 3(2x+1) = x + 6x + 3 = 7x + 3
From x′=4x+1x' = 4x + 1: x=x′−14x = \frac{x'-1}{4}
Substitute into y′y': y′=7(x′−14)+3=7x′−74+3=7x′−7+124=7x′+54y' = 7\left(\frac{x'-1}{4}\right) + 3 = \frac{7x'-7}{4} + 3 = \frac{7x'-7+12}{4} = \frac{7x'+5}{4}
y′=74x′+54y' = \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= |\det(T)| = |(2)(3)-(1)(1)| = |6-1| = 5
[1 mark]
(c) Check: 8=74(5)+54=35+54=404=108 = \frac{7}{4}(5) + \frac{5}{4} = \frac{35+5}{4} = \frac{40}{4} = 10
Since 8≠108 \neq 10, the point (5,8)(5, 8) does not lie on the image.
[2 marks]
Marking note: Award 1 mark for substitution; 1 mark for conclusion.

Total: 60 marks