Secondary 3 Additional Mathematics Vectors Matrices Quiz

Secondary 3 Additional Mathematics Quiz - Vectors Matrices

ANSWER KEY

Section A: Basic Concepts (Questions 1–5)

1. $2\mathbf{a} - 3\mathbf{b} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} - 3\begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} - \begin{pmatrix} -3 \\ 12 \end{pmatrix} = \begin{pmatrix} 9 \\ -16 \end{pmatrix}$
[2 marks] – Award 1 mark for correct scalar multiplication, 1 mark for correct final answer.

2. $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} -4 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} -6 \\ -4 \end{pmatrix}$
[2 marks] – Award 1 mark for correct subtraction setup, 1 mark for correct answer.

3. $P + 2Q = \begin{pmatrix} 2 & 1 \\ 0 & -3 \end{pmatrix} + 2\begin{pmatrix} -1 & 4 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & -3 \end{pmatrix} + \begin{pmatrix} -2 & 8 \\ 4 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 9 \\ 4 & -3 \end{pmatrix}$
[2 marks] – Award 1 mark for $2Q$, 1 mark for correct sum.

4. $|\mathbf{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
[2 marks] – Award 1 mark for correct formula, 1 mark for correct answer.

5. $\det(A) = (1)(4) - (2)(3) = 4 - 6 = -2$
[2 marks] – Award 1 mark for correct formula, 1 mark for correct answer.

Section B: Intermediate Applications (Questions 6–10)

6. $\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$; $\overrightarrow{QR} = \mathbf{r} - \mathbf{q} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$.
Since $\overrightarrow{PQ} = \overrightarrow{QR}$, the vectors are parallel and share point $Q$, so $P$, $Q$, $R$ are collinear.
[3 marks] – Award 1 mark for each vector, 1 mark for conclusion with reasoning.

7. For parallel vectors, $\mathbf{u} = \lambda \mathbf{v}$ for some scalar $\lambda$.
$\begin{pmatrix} 4 \\ 1 \end{pmatrix} = \lambda \begin{pmatrix} -2 \\ k \end{pmatrix} \implies 4 = -2\lambda \implies \lambda = -2$.
Then $1 = \lambda k = -2k \implies k = -\frac{1}{2}$.
[3 marks] – Award 1 mark for setting up proportionality, 1 mark for finding $\lambda$, 1 mark for $k$.

8. $\det(M) = (2)(3) - (1)(5) = 6 - 5 = 1$.
$M^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$
[3 marks] – Award 1 mark for determinant, 1 mark for correct adjugate matrix, 1 mark for final answer.

9. Unit vector $= \frac{\mathbf{w}}{|\mathbf{w}|} = \frac{1}{5} \begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} \\ -\frac{4}{5} \end{pmatrix}$
[3 marks] – Award 1 mark for formula, 1 mark for correct division, 1 mark for final answer.

10. $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$
So $y = 3$ and $x + 2(3) = 5 \implies x = -1$.
[3 marks] – Award 1 mark for matrix multiplication, 1 mark for $y$, 1 mark for $x$.

Section C: Advanced Problem Solving (Questions 11–15)

11. Using section formula for internal division in ratio $1:2$:
$\mathbf{c} = \frac{2\mathbf{a} + 1\mathbf{b}}{1+2} = \frac{2\begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 8 \\ 5 \end{pmatrix}}{3} = \frac{\begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} 8 \\ 5 \end{pmatrix}}{3} = \frac{\begin{pmatrix} 12 \\ 7 \end{pmatrix}}{3} = \begin{pmatrix} 4 \\ \frac{7}{3} \end{pmatrix}$
[3 marks] – Award 1 mark for correct formula, 1 mark for substitution, 1 mark for correct answer.

12. $AB = \begin{pmatrix} 3 & -1 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 3(1)+(-1)(-2) & 3(4)+(-1)(3) \\ 2(1)+0(-2) & 2(4)+0(3) \end{pmatrix} = \begin{pmatrix} 5 & 9 \\ 2 & 8 \end{pmatrix}$
[3 marks] – Award 1 mark for correct setup, 1 mark for two correct entries, 1 mark for all correct.

13. $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}$
$|\mathbf{a} + \mathbf{b}| = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$
[3 marks] – Award 1 mark for vector sum, 1 mark for magnitude formula, 1 mark for simplified answer.

14. $\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} 4p + 2q \\ p + 3q \end{pmatrix} = \begin{pmatrix} 10 \\ 8 \end{pmatrix}$
$4p + 2q = 10$ ... (1)
$p + 3q = 8$ ... (2)
From (2): $p = 8 - 3q$. Substitute into (1): $4(8-3q) + 2q = 10 \implies 32 - 12q + 2q = 10 \implies -10q = -22 \implies q = 2.2$.
Then $p = 8 - 3(2.2) = 8 - 6.6 = 1.4$.
[3 marks] – Award 1 mark for matrix multiplication, 1 mark for solving, 1 mark for both correct values.

15. The fourth vertex has position vector $\mathbf{p} + \mathbf{q} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$.
[3 marks] – Award 1 mark for recognising parallelogram property, 1 mark for addition, 1 mark for correct answer.

Section D: Challenge Questions (Questions 16–20)

16. $|\mathbf{u}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$
$|\mathbf{v}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$
$\mathbf{u}$ has the greater magnitude.
[2 marks] – Award 1 mark for both magnitudes, 1 mark for correct conclusion.

17. $T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ represents a rotation of $90^\circ$ anticlockwise about the origin.
[2 marks] – Award 1 mark for angle, 1 mark for direction.

18. $AB = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}$
$BA = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}$
Since $AB = BA$, the matrices commute.
[2 marks] – Award 1 mark for each product, or 2 marks for complete correct working.

19. $\overrightarrow{DE} = \mathbf{e} - \mathbf{d} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$; $\overrightarrow{EF} = \mathbf{f} - \mathbf{e} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$
$|\overrightarrow{DE}| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$; $|\overrightarrow{EF}| = 4\sqrt{2}$
$DE : EF = 1 : 1$
[2 marks] – Award 1 mark for vectors, 1 mark for ratio.

20. From the two equations:
$2a - b = 5$ ... (1)
$2c - d = 0$ ... (2)
$a + 3b = -1$ ... (3)
$c + 3d = 11$ ... (4)

From (1) and (3): $b = 2a - 5$. Substitute into (3): $a + 3(2a - 5) = -1 \implies a + 6a - 15 = -1 \implies 7a = 14 \implies a = 2$.
Then $b = 2(2) - 5 = -1$.

From (2): $d = 2c$. Substitute into (4): $c + 3(2c) = 11 \implies c + 6c = 11 \implies 7c = 11 \implies c = \frac{11}{7}$.
Then $d = 2\left(\frac{11}{7}\right) = \frac{22}{7}$.

[2 marks] – Award 1 mark for setting up equations, 1 mark for correct values.

Total: 50 marks