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O Level Elementary Mathematics Vectors Matrices Quiz

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Questions

O-Level Elementary Mathematics Quiz - Vectors Matrices

Name: __________________________
Class: __________________________
Date: __________________________
Score: ______ / 45

Duration: 45 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly. No marks will be given for correct answers without working.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless otherwise specified.
An approved calculator is expected to be used where appropriate.

Section A: Vector Notation and Geometry (Questions 1–5)

[12 Marks]

1. In the diagram below, $OABC$ is a parallelogram. $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$ . $M$ is the midpoint of $AB$ .

Express the following vectors in terms of $\mathbf{a}$ and $\mathbf{c}$ , in their simplest form.

(a) $\vec{OB}$
______________________________________________________________________ [1]

(b) $\vec{OM}$
______________________________________________________________________ [2]

2. The position vectors of points $A$ and $B$ relative to an origin $O$ are $\vec{OA} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ .

(a) Find the vector $\vec{AB}$ .
______________________________________________________________________ [1]

(b) Calculate the magnitude of $\vec{AB}$ , denoted as $|\vec{AB}|$ .
______________________________________________________________________ [2]

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

______________________________________________________________________ [3]

4. Points $A$ , $B$ , and $C$ have position vectors $\mathbf{a}$ , $\mathbf{b}$ , and $\mathbf{c}$ respectively. Given that $\vec{AB} = 2\mathbf{i} + 3\mathbf{j}$ and $\vec{BC} = 4\mathbf{i} + 6\mathbf{j}$ , explain why $A$ , $B$ , and $C$ are collinear.

______________________________________________________________________ [1]

5. Let $\mathbf{u} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ . Find the unit vector in the direction of $\mathbf{u} + \mathbf{v}$ .

______________________________________________________________________ [3]

Section B: Matrix Operations (Questions 6–10)

[13 Marks]

6. Let $A = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ -2 & 0 \end{pmatrix}$ .

Calculate the following:

(a) $A + B$
______________________________________________________________________ [1]

(b) $2A - B$
______________________________________________________________________ [2]

______________________________________________________________________ [3]

7. Given matrix $M = \begin{pmatrix} 3 & 1 \\ 2 & k \end{pmatrix}$ . If the determinant of $M$ is 10, find the value of $k$ .

______________________________________________________________________ [2]

8. Find the inverse of the matrix $P = \begin{pmatrix} 4 & 1 \\ 2 & 1 \end{pmatrix}$ .

______________________________________________________________________ [2]

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

\begin{cases} 3x + 2y = 12 \\ x - y = -1 \end{cases}

______________________________________________________________________ [3]

10. Given that $X = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $Y = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$ , verify whether $XY = YX$ . Show your working.

______________________________________________________________________ [2]

Section C: Transformations and Applications (Questions 11–15)

[10 Marks]

11. Triangle $T$ has vertices at $A(1, 2)$ , $B(3, 2)$ , and $C(1, 5)$ .

(a) Find the image of triangle $T$ under the transformation represented by the matrix $\mathbf{M} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ . State the coordinates of the new vertices $A'$ , $B'$ , and $C'$ .

$A'$ : (______, )
$B'$ : (, )
$C'$ : (, ______) [2]

(b) Describe the geometric transformation represented by matrix $\mathbf{M}$ .

______________________________________________________________________ [1]

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

(a) Find the image of the point $(4, 3)$ under this transformation.

______________________________________________________________________ [1]

(b) Describe the transformation fully.

______________________________________________________________________ [2]

13. The matrix $\mathbf{R} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ represents an enlargement with scale factor 3 centered at the origin.

(a) State the value of $k$ .
______________________________________________________________________ [1]

(b) If a shape has an area of $5 \text{ cm}^2$ , calculate the area of its image under this transformation.

______________________________________________________________________ [2]

14. Consider the matrix equation $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$ .

Without finding the inverse matrix explicitly, verify if $x=3$ and $y=1$ is the solution. Show your working.

______________________________________________________________________ [2]

15. Points $P(2, 1)$ and $Q(5, 3)$ are mapped to $P'(4, 2)$ and $Q'(10, 6)$ by a single transformation matrix $\mathbf{T}$ .

(a) Determine the matrix $\mathbf{T}$ .

______________________________________________________________________ [3]

(b) Calculate the determinant of $\mathbf{T}$ and explain what this value tells you about the area of any shape transformed by $\mathbf{T}$ .

______________________________________________________________________ [2]

Section D: Advanced Vector and Matrix Problems (Questions 16–20)

[10 Marks]

16. In a parallelogram $OABC$ , $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$ . The diagonals $OB$ and $AC$ intersect at $M$ .

Using vector methods, show that $\vec{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{c})$ .

______________________________________________________________________ [3]

17. A shop sells two types of fruit baskets. Basket A contains 3 apples and 2 oranges. Basket B contains 2 apples and 4 oranges. The price of an apple is $\$ x $and an orange is$ $y$.

(a) Write down a matrix equation to represent the total cost of Basket A ( $C_A$ ) and Basket B ( $C_B$ ).

______________________________________________________________________ [2]

(b) If Basket A costs $\$ 7.00 $and Basket B costs$ $10.00$, use the matrix method to find the price of one apple and one orange.

______________________________________________________________________ [3]

18. Given vectors $\mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$ .

(a) Calculate $\mathbf{a} \cdot \mathbf{b}$ (the scalar product).
______________________________________________________________________ [1]

(b) Hence, or otherwise, find the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ correct to 1 decimal place.

______________________________________________________________________ [2]

19. The matrix $S = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ represents a reflection.

(a) Describe the line of reflection.
______________________________________________________________________ [1]

(b) Find the image of the line $y = 2x + 1$ under this transformation.

______________________________________________________________________ [2]

20. Let $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ .

(a) Calculate $A^2$ .
______________________________________________________________________ [1]

(b) Calculate $A^3$ .
______________________________________________________________________ [1]

(c) Deduce the general form of $A^n$ for any positive integer $n$ .
______________________________________________________________________ [1]

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Answers

O-Level Elementary Mathematics Quiz - Vectors Matrices (Answer Key)

1. (a) $\vec{OB} = \vec{OA} + \vec{AB}$ . Since $OABC$ is a parallelogram, $\vec{AB} = \vec{OC} = \mathbf{c}$ . $\vec{OB} = \mathbf{a} + \mathbf{c}$ [1]

(b) $\vec{OM} = \vec{OA} + \vec{AM}$ . Since $M$ is midpoint of $AB$ , $\vec{AM} = \frac{1}{2}\vec{AB} = \frac{1}{2}\mathbf{c}$ . $\vec{OM} = \mathbf{a} + \frac{1}{2}\mathbf{c}$ [2]

(c) $\vec{CM} = \vec{CO} + \vec{OM} = -\mathbf{c} + (\mathbf{a} + \frac{1}{2}\mathbf{c}) = \mathbf{a} - \frac{1}{2}\mathbf{c}$ [2]

2. (a) $\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ [1]

(b) $|\vec{AB}| = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \approx 7.21$ [2]

3. $3\mathbf{p} - 2\mathbf{r} = \mathbf{q} \implies 2\mathbf{r} = 3\mathbf{p} - \mathbf{q}$ $3\mathbf{p} = \begin{pmatrix} 6 \\ 15 \end{pmatrix}$ $3\mathbf{p} - \mathbf{q} = \begin{pmatrix} 6 \\ 15 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 7 \\ 12 \end{pmatrix}$ $\mathbf{r} = \frac{1}{2} \begin{pmatrix} 7 \\ 12 \end{pmatrix} = \begin{pmatrix} 3.5 \\ 6 \end{pmatrix}$ [3]

4. $\vec{BC} = 2(2\mathbf{i} + 3\mathbf{j}) = 2\vec{AB}$ . Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ and they share a common point $B$ , the vectors are parallel and the points are collinear. [1]

5. $\mathbf{u} + \mathbf{v} = \begin{pmatrix} 4 \\ -1 \end{pmatrix} + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$ . Magnitude $|\mathbf{u}+\mathbf{v}| = \sqrt{6^2 + 2^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$ . Unit vector = $\frac{1}{2\sqrt{10}} \begin{pmatrix} 6 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{\sqrt{10}} \\ \frac{1}{\sqrt{10}} \end{pmatrix}$ or $\begin{pmatrix} \frac{3\sqrt{10}}{10} \\ \frac{\sqrt{10}}{10} \end{pmatrix}$ . [3]

6. (a) $A+B = \begin{pmatrix} 2+1 & -1+4 \\ 0-2 & 3+0 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ -2 & 3 \end{pmatrix}$ [1]

(b) $2A = \begin{pmatrix} 4 & -2 \\ 0 & 6 \end{pmatrix}$ . $2A - B = \begin{pmatrix} 4-1 & -2-4 \\ 0-(-2) & 6-0 \end{pmatrix} = \begin{pmatrix} 3 & -6 \\ 2 & 6 \end{pmatrix}$ [2]

(c) $AB = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} (2)(1)+(-1)(-2) & (2)(4)+(-1)(0) \\ (0)(1)+(3)(-2) & (0)(4)+(3)(0) \end{pmatrix} = \begin{pmatrix} 4 & 8 \\ -6 & 0 \end{pmatrix}$ [3]

7. $\det(M) = (3)(k) - (1)(2) = 3k - 2$ . $3k - 2 = 10 \implies 3k = 12 \implies k = 4$ . [2]

8. $\det(P) = (4)(1) - (1)(2) = 2$ . $P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} 0.5 & -0.5 \\ -1 & 2 \end{pmatrix}$ [2]

9. Matrix form: $\begin{pmatrix} 3 & 2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 12 \\ -1 \end{pmatrix}$ . $\det = (3)(-1) - (2)(1) = -5$ . Inverse: $\frac{1}{-5} \begin{pmatrix} -1 & -2 \\ -1 & 3 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 1 & -3 \end{pmatrix}$ . $\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 1 & -3 \end{pmatrix} \begin{pmatrix} 12 \\ -1 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 12-2 \\ 12+3 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 10 \\ 15 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ . $x=2, y=3$ . [3]

10. $XY = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix}$ . $YX = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 10 & 14 \end{pmatrix}$ . $XY \neq YX$ . [2]

11. (a) $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} \implies A'(1, -2)$ . $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \implies B'(3, -2)$ . $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \end{pmatrix} \implies C'(1, -5)$ . [2]

(b) Reflection in the x-axis. [1]

12. (a) $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$ . Image is $(-3, 4)$ . [1]

(b) Rotation $90^\circ$ anti-clockwise about the origin. [2]

13. (a) Scale factor 3 implies $k=3$ . [1]

(b) Area scale factor = $k^2 = 3^2 = 9$ . New Area = $5 \times 9 = 45 \text{ cm}^2$ . [2]

14. LHS: $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} (2)(3)+(1)(1) \\ (1)(3)+(1)(1) \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$ . RHS: $\begin{pmatrix} 7 \\ 4 \end{pmatrix}$ . LHS = RHS, so it is the solution. [2]

15. (a) Let $\mathbf{T} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . $\mathbf{T} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} \implies 2a+b=4, 2c+d=2$ . $\mathbf{T} \begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 10 \\ 6 \end{pmatrix} \implies 5a+3b=10, 5c+3d=6$ . Solving for a,b: $b=4-2a \implies 5a+3(4-2a)=10 \implies 5a+12-6a=10 \implies -a=-2 \implies a=2$ . $b=4-4=0$ . Solving for c,d: $d=2-2c \implies 5c+3(2-2c)=6 \implies 5c+6-6c=6 \implies -c=0 \implies c=0$ . $d=2$ . $\mathbf{T} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ . [3]

(b) $\det(\mathbf{T}) = 4$ . The area of the image is 4 times the area of the object. [2]

16. $\vec{OB} = \mathbf{a} + \mathbf{c}$ . Since diagonals of a parallelogram bisect each other, $M$ is the midpoint of $OB$ . $\vec{OM} = \frac{1}{2}\vec{OB} = \frac{1}{2}(\mathbf{a} + \mathbf{c})$ . [3]

17. (a) $\begin{pmatrix} 3 & 2 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} C_A \\ C_B \end{pmatrix}$ [2]

(b) $\begin{pmatrix} 3 & 2 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 10 \end{pmatrix}$ . Det = $12 - 4 = 8$ . Inverse = $\frac{1}{8} \begin{pmatrix} 4 & -2 \\ -2 & 3 \end{pmatrix}$ . $\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 4 & -2 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} 7 \\ 10 \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 28-20 \\ -14+30 \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 8 \\ 16 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ . Apple = $\$ 1.00 $, Orange =$ $2.00$. [3]

18. (a) $\mathbf{a} \cdot \mathbf{b} = (3)(-1) + (4)(2) = -3 + 8 = 5$ . [1]

(b) $|\mathbf{a}| = \sqrt{3^2+4^2} = 5$ . $|\mathbf{b}| = \sqrt{(-1)^2+2^2} = \sqrt{5}$ . $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}$ . $\theta = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \approx 63.4^\circ$ . [2]

19. (a) Reflection in the line $y = x$ . [1]

(b) The transformation swaps $x$ and $y$ . So $x = y'$ and $y = x'$ . Substitute into $y = 2x + 1$ : $x' = 2y' + 1 \implies 2y' = x' - 1 \implies y' = \frac{1}{2}x' - \frac{1}{2}$ . Equation: $y = \frac{1}{2}x - \frac{1}{2}$ . [2]

20. (a) $A^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}$ . [1]

(b) $A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix}$ . [1]