Secondary 3 Additional Mathematics Vectors Matrices Quiz

Secondary 3 Additional Mathematics Quiz - Vectors Matrices — ANSWER KEY

Total Marks: 50

Section A: Short Answer (10 marks)

1. a + 2b = (3i − 2j) + 2(−i + 4j)
= 3i − 2j − 2i + 8j
= i + 6j
[M1 for correct substitution; A1 for correct answer]
Answer: i + 6j or (\begin{pmatrix} 1 \ 6 \end{pmatrix})
[2 marks]

2. (\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5\mathbf{i} - 3\mathbf{j}) - (2\mathbf{i} + \mathbf{j}))
= 3i − 4j
[M1 for subtraction; A1 for correct answer]
Answer: 3i − 4j or (\begin{pmatrix} 3 \ -4 \end{pmatrix})
[2 marks]

3. p − q = (\begin{pmatrix} 4 \ -1 \end{pmatrix} - \begin{pmatrix} -2 \ 5 \end{pmatrix} = \begin{pmatrix} 6 \ -6 \end{pmatrix})
|p − q| = (\sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2})
[M1 for subtraction and magnitude formula; A1 for correct simplified answer]
Answer: (6\sqrt{2})
[2 marks]

4. |v| = (\sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10)
Unit vector = (\frac{1}{10}(6\mathbf{i} - 8\mathbf{j}) = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j})
[M1 for magnitude; A1 for correct unit vector]
Answer: (\frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}) or (\begin{pmatrix} 3/5 \ -4/5 \end{pmatrix})
[2 marks]

5. For parallel vectors, u = λv for some scalar λ.
(\begin{pmatrix} 3 \ k \end{pmatrix} = \lambda \begin{pmatrix} -6 \ 4 \end{pmatrix})
From first component: (3 = -6\lambda \Rightarrow \lambda = -\frac{1}{2})
From second component: (k = 4\lambda = 4\left(-\frac{1}{2}\right) = -2)
[M1 for setting up parallel condition; A1 for correct k]
Answer: (k = -2)
[2 marks]

Section B: Calculation and Proof (24 marks)

6. (a) (\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (6\mathbf{i} - \mathbf{j}) - (2\mathbf{i} + 3\mathbf{j}) = 4\mathbf{i} - 4\mathbf{j})
(\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = (10\mathbf{i} - 5\mathbf{j}) - (6\mathbf{i} - \mathbf{j}) = 4\mathbf{i} - 4\mathbf{j})
[A1 each; 2 marks total]

(b) Since (\overrightarrow{AB} = \overrightarrow{BC}), the vectors are equal, meaning AB and BC are parallel and share point B. Therefore, A, B, and C are collinear.
[M1 for noting vectors are equal/parallel; A1 for conclusion with reasoning]
[2 marks]

(c) AB = |(\overrightarrow{AB})| = (\sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2})
BC = |(\overrightarrow{BC})| = (\sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2})
Ratio AB : BC = (4\sqrt{2} : 4\sqrt{2} = 1 : 1)
[M1 for calculating magnitudes; A1 for correct ratio]
[2 marks]

7. (a) (A + B = \begin{pmatrix} 2+1 & -1+0 \ 3+(-2) & 4+5 \end{pmatrix} = \begin{pmatrix} 3 & -1 \ 1 & 9 \end{pmatrix})
[A1]
[1 mark]

(b) (3A = \begin{pmatrix} 3(2) & 3(-1) \ 3(3) & 3(4) \end{pmatrix} = \begin{pmatrix} 6 & -3 \ 9 & 12 \end{pmatrix})
[A1]
[1 mark]

(c) (AB = \begin{pmatrix} 2 & -1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 0 \ -2 & 5 \end{pmatrix})
= (\begin{pmatrix} 2(1) + (-1)(-2) & 2(0) + (-1)(5) \ 3(1) + 4(-2) & 3(0) + 4(5) \end{pmatrix})
= (\begin{pmatrix} 2 + 2 & 0 - 5 \ 3 - 8 & 0 + 20 \end{pmatrix} = \begin{pmatrix} 4 & -5 \ -5 & 20 \end{pmatrix})
[M1 for correct multiplication method; A1 for correct answer]
[2 marks]

(d) (B^T = \begin{pmatrix} 1 & -2 \ 0 & 5 \end{pmatrix})
[A1]
[1 mark]

8. (M = \begin{pmatrix} 4 & 1 \ 2 & 3 \end{pmatrix})
det(M) = (4)(3) − (1)(2) = 12 − 2 = 10
(M^{-1} = \frac{1}{10} \begin{pmatrix} 3 & -1 \ -2 & 4 \end{pmatrix} = \begin{pmatrix} \frac{3}{10} & -\frac{1}{10} \ -\frac{2}{10} & \frac{4}{10} \end{pmatrix} = \begin{pmatrix} 0.3 & -0.1 \ -0.2 & 0.4 \end{pmatrix})
[M1 for determinant; M1 for correct adjugate; A1 for correct inverse]
[3 marks]

9. For a singular matrix, det(P) = 0.
det(P) = (x)(x−1) − (2)(3) = x² − x − 6
Set x² − x − 6 = 0
(x − 3)(x + 2) = 0
x = 3 or x = −2
[M1 for setting det = 0; M1 for solving quadratic; A1 for both values]
Answer: (x = 3) or (x = -2)
[3 marks]

10. Write as matrix equation: (\begin{pmatrix} 2 & 1 \ 3 & -2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 7 \ 0 \end{pmatrix})
Let (A = \begin{pmatrix} 2 & 1 \ 3 & -2 \end{pmatrix})
det(A) = (2)(−2) − (1)(3) = −4 − 3 = −7
(A^{-1} = -\frac{1}{7} \begin{pmatrix} -2 & -1 \ -3 & 2 \end{pmatrix} = \begin{pmatrix} \frac{2}{7} & \frac{1}{7} \ \frac{3}{7} & -\frac{2}{7} \end{pmatrix})
(\begin{pmatrix} x \ y \end{pmatrix} = A^{-1} \begin{pmatrix} 7 \ 0 \end{pmatrix} = \begin{pmatrix} \frac{2}{7}(7) + \frac{1}{7}(0) \ \frac{3}{7}(7) + (-\frac{2}{7})(0) \end{pmatrix} = \begin{pmatrix} 2 \ 3 \end{pmatrix})
[M1 for matrix form; M1 for inverse; M1 for multiplication; A1 for correct x, y]
Answer: (x = 2, y = 3)
[4 marks]

Section C: Structured Response (16 marks)

11. Let (A = \begin{pmatrix} 5 & 2 \ 3 & 1 \end{pmatrix})
det(A) = (5)(1) − (2)(3) = 5 − 6 = −1
(A^{-1} = \frac{1}{-1} \begin{pmatrix} 1 & -2 \ -3 & 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \ 3 & -5 \end{pmatrix})
(\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -1 & 2 \ 3 & -5 \end{pmatrix} \begin{pmatrix} 8 \ 5 \end{pmatrix} = \begin{pmatrix} -8 + 10 \ 24 - 25 \end{pmatrix} = \begin{pmatrix} 2 \ -1 \end{pmatrix})
[M1 for determinant; M1 for inverse; A1 for correct x, y]
Answer: (x = 2, y = -1)
[3 marks]

12. (a) (\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = (4\mathbf{i} - \mathbf{j}) - (\mathbf{i} + 2\mathbf{j}) = 3\mathbf{i} - 3\mathbf{j})
[A1]
[1 mark]

(b) (\overrightarrow{PR} = \mathbf{r} - \mathbf{p} = (7\mathbf{i} + k\mathbf{j}) - (\mathbf{i} + 2\mathbf{j}) = 6\mathbf{i} + (k-2)\mathbf{j})
Given (\overrightarrow{PR} = 2\overrightarrow{PQ} = 2(3\mathbf{i} - 3\mathbf{j}) = 6\mathbf{i} - 6\mathbf{j})
Equating: (6\mathbf{i} + (k-2)\mathbf{j} = 6\mathbf{i} - 6\mathbf{j})
Therefore, (k - 2 = -6 \Rightarrow k = -4)
[M1 for finding PR; M1 for equating; A1 for correct k]
[3 marks]

(c) Midpoint of PR = (\frac{\mathbf{p} + \mathbf{r}}{2} = \frac{(\mathbf{i} + 2\mathbf{j}) + (7\mathbf{i} - 4\mathbf{j})}{2} = \frac{8\mathbf{i} - 2\mathbf{j}}{2} = 4\mathbf{i} - \mathbf{j})
[M1 for midpoint formula; A1 for correct answer]
[2 marks]

13. (a) a = i + 2j, b = 5i + 8j, c = 9i + 4j
[A1 for all three correct]
[1 mark]

(b) (\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5\mathbf{i} + 8\mathbf{j}) - (\mathbf{i} + 2\mathbf{j}) = 4\mathbf{i} + 6\mathbf{j})
(\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (9\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} + 2\mathbf{j}) = 8\mathbf{i} + 2\mathbf{j})
[A1 each; 2 marks total]

(c) (\overrightarrow{AB} \cdot \overrightarrow{AC} = (4)(8) + (6)(2) = 32 + 12 = 44)
[M1 for dot product formula; A1 for correct value]
[2 marks]

(d) For a right angle at A, (\overrightarrow{AB} \cdot \overrightarrow{AC} = 0).
Since (\overrightarrow{AB} \cdot \overrightarrow{AC} = 44 \neq 0), triangle ABC is NOT right-angled at A.
[M1 for condition; A1 for correct conclusion with justification]
[2 marks]

14. (a) det(Q) = (3)(4) − (−2)(1) = 12 + 2 = 14
[A1]
[1 mark]

(b) (Q^{-1} = \frac{1}{14} \begin{pmatrix} 4 & 2 \ -1 & 3 \end{pmatrix} = \begin{pmatrix} \frac{4}{14} & \frac{2}{14} \ -\frac{1}{14} & \frac{3}{14} \end{pmatrix} = \begin{pmatrix} \frac{2}{7} & \frac{1}{7} \ -\frac{1}{14} & \frac{3}{14} \end{pmatrix})
[M1 for correct adjugate; A1 for correct inverse]
[2 marks]

15. Vector projection of u onto v = (\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2} \mathbf{v})
u · v = (2)(3) + (5)(−1) = 6 − 5 = 1
|v|² = 3² + (−1)² = 9 + 1 = 10
Projection = (\frac{1}{10}(3\mathbf{i} - \mathbf{j}) = \frac{3}{10}\mathbf{i} - \frac{1}{10}\mathbf{j})
[M1 for dot product; M1 for magnitude squared; A1 for correct projection]
Answer: (\frac{3}{10}\mathbf{i} - \frac{1}{10}\mathbf{j}) or (\begin{pmatrix} 0.3 \ -0.1 \end{pmatrix})
[3 marks]

Section D: Application and Challenge (0 marks)

16. Reserved for future use.
[0 marks]

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[0 marks]

18. Reserved for future use.
[0 marks]

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[0 marks]

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[0 marks]

END OF ANSWER KEY