Secondary 4 Elementary Mathematics Vectors Matrices Quiz

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Secondary 4 Elementary Mathematics Quiz - Vectors Matrices

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ______ / 40

Duration: 50 minutes
Total Marks: 40

Instructions:

• Answer all questions.
• Write your answers in the spaces provided.
• Show all working clearly for questions worth 2 marks or more.
• Omission of essential working will result in loss of marks.
• Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified.
• The use of an approved scientific calculator is expected, where appropriate.

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Section A: Vectors (Questions 1–10) [20 marks]

1. [2 marks]

Given that $\vec{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$, find the vector $2\vec{a} - 3\vec{b}$.

Answer: $\begin{pmatrix} \quad \\ \quad \end{pmatrix}$

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2. [2 marks]

The position vectors of points $A$ and $B$ relative to origin $O$ are $\vec{OA} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} -2 \\ 7 \end{pmatrix}$ respectively. Find the vector $\vec{AB}$.

Answer: $\begin{pmatrix} \quad \\ \quad \end{pmatrix}$

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3. [2 marks]

Point $P$ has position vector $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and point $Q$ has position vector $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$. Find the magnitude of $\vec{PQ}$, correct to 3 significant figures.

Answer: ___________________________

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4. [2 marks]

In the diagram, $OABC$ is a parallelogram with $\vec{OA} = \vec{a}$ and $\vec{OC} = \vec{c}$. $M$ is the midpoint of $AB$. Express $\vec{OM}$ in terms of $\vec{a}$ and $\vec{c}$.

Image pending generation: diagram for Q4.

Answer: $\vec{OM} =$ ___________________________

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5. [2 marks]

Vectors $\vec{p} = \begin{pmatrix} k \\ 6 \end{pmatrix}$ and $\vec{q} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ are parallel. Find the value of $k$.

Answer: $k =$ ___________________________

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6. [3 marks]

In triangle $OPQ$, $\vec{OP} = \vec{p}$ and $\vec{OQ} = \vec{q}$. Point $R$ lies on $PQ$ such that $PR : RQ = 2 : 1$. Express $\vec{OR}$ in terms of $\vec{p}$ and $\vec{q}$.

Answer: $\vec{OR} =$ ___________________________

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7. [3 marks]

The position vectors of points $A$, $B$, and $C$ are $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $\begin{pmatrix} 4 \\ 6 \end{pmatrix}$, and $\begin{pmatrix} 7 \\ 10 \end{pmatrix}$ respectively. Show that $A$, $B$, and $C$ are collinear.

Answer: ___________________________

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8. [2 marks]

Given $\vec{u} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$, find the unit vector in the direction of $\vec{u}$.

Answer: $\begin{pmatrix} \quad \\ \quad \end{pmatrix}$

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9. [2 marks]

In the diagram, $OXYZ$ is a trapezium with $OX \parallel ZY$. $\vec{OX} = \vec{x}$ and $\vec{OZ} = \vec{z}$. Given that $ZY = \frac{1}{2} OX$, express $\vec{OY}$ in terms of $\vec{x}$ and $\vec{z}$.

Image pending generation: diagram for Q9.

Answer: $\vec{OY} =$ ___________________________

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10. [2 marks]

A vector $\vec{v}$ has magnitude 13 and makes an angle of $60^\circ$ with the positive $x$-axis. Write $\vec{v}$ in component form $\begin{pmatrix} x \\ y \end{pmatrix}$, giving exact values.

Answer: $\begin{pmatrix} \quad \\ \quad \end{pmatrix}$

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Section B: Matrices (Questions 11–16) [12 marks]

11. [2 marks]

Let $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ -2 & 5 \end{pmatrix}$. Evaluate $3A - 2B$.

Answer: $\begin{pmatrix} \quad & \quad \\ \quad & \quad \end{pmatrix}$

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12. [2 marks]

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

Answer: $\begin{pmatrix} \quad & \quad \\ \quad & \quad \end{pmatrix}$

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13. [2 marks]

Find the inverse of the matrix $M = \begin{pmatrix} 4 & 3 \\ 2 & 2 \end{pmatrix}$.

Answer: $M^{-1} =$ ___________________________

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14. [2 marks]

The matrix $N = \begin{pmatrix} k & 2 \\ 6 & 3 \end{pmatrix}$ is singular. Find the value of $k$.

Answer: $k =$ ___________________________

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15. [2 marks]

Solve the matrix equation $\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}$ for $x$ and $y$.

Answer: $x =$ __________, $y =$ __________

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16. [2 marks]

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

Answer: ___________________________

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Section C: Combined Vectors and Matrices Applications (Questions 17–20) [8 marks]

17. [2 marks]

The vertices of triangle $ABC$ are $A(1, 2)$, $B(4, 6)$, and $C(5, 3)$. The triangle is transformed by the matrix $\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$. Find the coordinates of the image of $A$.

Answer: ___________________________

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18. [2 marks]

A vector $\vec{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ is transformed by the matrix $R = \begin{pmatrix} \cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ \end{pmatrix}$. Find the image vector $R\vec{v}$.

Answer: $\begin{pmatrix} \quad \\ \quad \end{pmatrix}$

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19. [2 marks]

Points $P$ and $Q$ have position vectors $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$ respectively. The transformation matrix $S = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ maps $P$ to $P'$ and $Q$ to $Q'$. Find the vector $\vec{P'Q'}$.

Answer: $\begin{pmatrix} \quad \\ \quad \end{pmatrix}$

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20. [2 marks]

The matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ represents a transformation. A triangle with area 5 square units is transformed by $A$. Find the area of the image triangle.

Answer: ___________________________ square units

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End of Quiz

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Secondary 4 Elementary Mathematics Quiz - Vectors Matrices (Answer Key)

Total Marks: 40

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Section A: Vectors (Questions 1–10) [20 marks]

1. [2 marks]

Answer: $\begin{pmatrix} 9 \\ -16 \end{pmatrix}$

Working:

$$2\vec{a} - 3\vec{b} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} - 3\begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} - \begin{pmatrix} -3 \\ 12 \end{pmatrix} = \begin{pmatrix} 6 - (-3) \\ -4 - 12 \end{pmatrix} = \begin{pmatrix} 9 \\ -16 \end{pmatrix}$$

Marking: 1 mark for correct scalar multiplication, 1 mark for correct subtraction and final answer.

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2. [2 marks]

Answer: $\begin{pmatrix} -7 \\ 10 \end{pmatrix}$

Working:

$$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} -2 \\ 7 \end{pmatrix} - \begin{pmatrix} 5 \\ -3 \end{pmatrix} = \begin{pmatrix} -2 - 5 \\ 7 - (-3) \end{pmatrix} = \begin{pmatrix} -7 \\ 10 \end{pmatrix}$$

Marking: 1 mark for correct formula $\vec{AB} = \vec{OB} - \vec{OA}$, 1 mark for correct calculation.

Common mistake: Writing $\vec{OA} - \vec{OB}$ instead.

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3. [2 marks]

Answer: $7.21$ (or $\sqrt{52} = 2\sqrt{13}$)

Working:

$$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} -6 \\ 4 \end{pmatrix}$$ $$|\vec{PQ}| = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.21$$

Marking: 1 mark for finding $\vec{PQ}$, 1 mark for correct magnitude (exact or 3 s.f.).

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4. [2 marks]

Answer: $\vec{OM} = \vec{a} + \frac{1}{2}\vec{c}$

Working: In parallelogram $OABC$, $\vec{AB} = \vec{OC} = \vec{c}$. $M$ is midpoint of $AB$, so $\vec{AM} = \frac{1}{2}\vec{AB} = \frac{1}{2}\vec{c}$.

$$\vec{OM} = \vec{OA} + \vec{AM} = \vec{a} + \frac{1}{2}\vec{c}$$

Marking: 1 mark for recognising $\vec{AB} = \vec{c}$, 1 mark for correct expression for $\vec{OM}$.

Alternative method: $\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2}(\vec{a} + \vec{a} + \vec{c}) = \vec{a} + \frac{1}{2}\vec{c}$.

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5. [2 marks]

Answer: $k = -4$

Working: Parallel vectors are scalar multiples: $\vec{p} = \lambda \vec{q}$ for some $\lambda$.

$$\begin{pmatrix} k \\ 6 \end{pmatrix} = \lambda \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 2\lambda \\ -3\lambda \end{pmatrix}$$

From the second component: $6 = -3\lambda \Rightarrow \lambda = -2$. Then $k = 2\lambda = 2(-2) = -4$.

Marking: 1 mark for setting up scalar multiple or using ratio $\frac{k}{2} = \frac{6}{-3}$, 1 mark for $k = -4$.

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6. [3 marks]

Answer: $\vec{OR} = \frac{1}{3}\vec{p} + \frac{2}{3}\vec{q}$

Working: $PR : RQ = 2 : 1$, so $R$ divides $PQ$ internally in ratio $2:1$. Using section formula: $\vec{OR} = \frac{1\vec{p} + 2\vec{q}}{1+2} = \frac{\vec{p} + 2\vec{q}}{3} = \frac{1}{3}\vec{p} + \frac{2}{3}\vec{q}$.

Alternative working: $\vec{PQ} = \vec{q} - \vec{p}$. $\vec{PR} = \frac{2}{3}\vec{PQ} = \frac{2}{3}(\vec{q} - \vec{p})$. $\vec{OR} = \vec{OP} + \vec{PR} = \vec{p} + \frac{2}{3}(\vec{q} - \vec{p}) = \vec{p} + \frac{2}{3}\vec{q} - \frac{2}{3}\vec{p} = \frac{1}{3}\vec{p} + \frac{2}{3}\vec{q}$.

Marking: 1 mark for correct ratio interpretation, 1 mark for correct section formula or vector addition method, 1 mark for simplified final answer.

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7. [3 marks]

Answer: (See working)

Working:

$$\vec{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$ $$\vec{BC} = \begin{pmatrix} 7 \\ 10 \end{pmatrix} - \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$

Since $\vec{AB} = \vec{BC}$, the vectors are parallel and share point $B$. Therefore $A$, $B$, and $C$ are collinear.

Alternative: Show $\vec{AC} = 2\vec{AB}$ or gradients are equal.

Marking: 1 mark for finding $\vec{AB}$, 1 mark for finding $\vec{BC}$ (or $\vec{AC}$), 1 mark for correct conclusion with reasoning.

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8. [2 marks]

Answer: $\begin{pmatrix} \frac{2}{\sqrt{29}} \\ -\frac{5}{\sqrt{29}} \end{pmatrix}$ or $\begin{pmatrix} \frac{2\sqrt{29}}{29} \\ -\frac{5\sqrt{29}}{29} \end{pmatrix}$

Working:

$$|\vec{u}| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}$$

Unit vector $= \frac{\vec{u}}{|\vec{u}|} = \frac{1}{\sqrt{29}}\begin{pmatrix} 2 \\ -5 \end{pmatrix} = \begin{pmatrix} \frac{2}{\sqrt{29}} \\ -\frac{5}{\sqrt{29}} \end{pmatrix}$

Marking: 1 mark for correct magnitude $\sqrt{29}$, 1 mark for correct unit vector.

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9. [2 marks]

Answer: $\vec{OY} = \frac{1}{2}\vec{x} + \vec{z}$

Working: In trapezium $OXYZ$, $\vec{ZY} = \frac{1}{2}\vec{OX} = \frac{1}{2}\vec{x}$ (since $ZY \parallel OX$ and same direction).

$$\vec{OY} = \vec{OZ} + \vec{ZY} = \vec{z} + \frac{1}{2}\vec{x} = \frac{1}{2}\vec{x} + \vec{z}$$

Marking: 1 mark for $\vec{ZY} = \frac{1}{2}\vec{x}$, 1 mark for correct $\vec{OY}$.

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10. [2 marks]

Answer: $\begin{pmatrix} \frac{13}{2} \\ \frac{13\sqrt{3}}{2} \end{pmatrix}$ or $\begin{pmatrix} 6.5 \\ \frac{13\sqrt{3}}{2} \end{pmatrix}$

Working: $x = 13 \cos 60^\circ = 13 \times \frac{1}{2} = \frac{13}{2}$ $y = 13 \sin 60^\circ = 13 \times \frac{\sqrt{3}}{2} = \frac{13\sqrt{3}}{2}$

Marking: 1 mark for correct $x$-component, 1 mark for correct $y$-component (exact values required).

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Section B: Matrices (Questions 11–16) [12 marks]

11. [2 marks]

Answer: $\begin{pmatrix} 4 & -3 \\ 13 & 2 \end{pmatrix}$

Working:

$$3A = 3\begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 6 & -3 \\ 9 & 12 \end{pmatrix}$$ $$2B = 2\begin{pmatrix} 1 & 0 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ -4 & 10 \end{pmatrix}$$ $$3A - 2B = \begin{pmatrix} 6-2 & -3-0 \\ 9-(-4) & 12-10 \end{pmatrix} = \begin{pmatrix} 4 & -3 \\ 13 & 2 \end{pmatrix}$$

Marking: 1 mark for correct scalar multiplication, 1 mark for correct subtraction.

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12. [2 marks]

Answer: $\begin{pmatrix} 7 & 3 \\ 9 & 11 \end{pmatrix}$

Working:

$$PQ = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 3(1)+2(2) & 3(-1)+2(3) \\ 1(1)+4(2) & 1(-1)+4(3) \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 9 & 11 \end{pmatrix}$$

Marking: 1 mark for correct row-by-column multiplication method, 1 mark for all four entries correct.

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13. [2 marks]

Answer: $M^{-1} = \begin{pmatrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{pmatrix}$ or $\frac{1}{2}\begin{pmatrix} 2 & -3 \\ -2 & 4 \end{pmatrix}$

Working: For $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ 2 & 2 \end{pmatrix}$, $\det(M) = ad - bc = 4(2) - 3(2) = 8 - 6 = 2$.

$$M^{-1} = \frac{1}{\det(M)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 2 & -3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{pmatrix}$$

Marking: 1 mark for correct determinant, 1 mark for correct inverse matrix.

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14. [2 marks]

Answer: $k = 4$

Working: Matrix is singular $\Rightarrow \det(N) = 0$.

$$\det(N) = k(3) - 2(6) = 3k - 12 = 0 \Rightarrow 3k = 12 \Rightarrow k = 4$$

Marking: 1 mark for setting determinant to 0, 1 mark for $k = 4$.

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15. [2 marks]

Answer: $x = 2$, $y = 1$

Working: Let $A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$, $\vec{x} = \begin{pmatrix} x \\ y \end{pmatrix}$, $\vec{b} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}$. $\det(A) = 2(2) - 1(3) = 4 - 3 = 1$. $A^{-1} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}$.

$$\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\vec{b} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} 5 \\ 8 \end{pmatrix} = \begin{pmatrix} 2(5) - 1(8) \\ -3(5) + 2(8) \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

Alternative: Solve simultaneous equations $2x + y = 5$ and $3x + 2y = 8$.

Marking: 1 mark for correct inverse or elimination method, 1 mark for correct $x$ and $y$.

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16. [2 marks]

Answer: Rotation of $90^\circ$ anticlockwise about the origin.

Working: The matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ represents rotation by $\theta$ anticlockwise about the origin. Here $\cos\theta = 0$, $\sin\theta = 1 \Rightarrow \theta = 90^\circ$.

Marking: 1 mark for identifying as rotation, 1 mark for correct angle ($90^\circ$) and direction (anticlockwise) and centre (origin).

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Section C: Combined Vectors and Matrices Applications (Questions 17–20) [8 marks]

17. [2 marks]

Answer: $(2, 6)$

Working: Image of $A(1,2)$ under $\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$:

$$\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2(1) + 0(2) \\ 0(1) + 3(2) \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$$

Marking: 1 mark for correct matrix multiplication setup, 1 mark for correct coordinates.

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18. [2 marks]

Answer: $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$

Working: $R = \begin{pmatrix} \cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

$$R\vec{v} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 0(3) + (-1)(4) \\ 1(3) + 0(4) \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}$$

Marking: 1 mark for correct rotation matrix, 1 mark for correct image vector.

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19. [2 marks]

Answer: $\begin{pmatrix} 9 \\ 4 \end{pmatrix}$

Working: $\vec{PQ} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$. Under linear transformation $S$, $\vec{P'Q'} = S(\vec{PQ})$.

$$\vec{P'Q'} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1(3) + 2(4) \\ 0(3) + 1(4) \end{pmatrix} = \begin{pmatrix} 11 \\ 4 \end{pmatrix}$$

Wait, let me recalculate: $P' = S\vec{p} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}$ $Q' = S\vec{q} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 11 \\ 3 \end{pmatrix}$ $\vec{P'Q'} = \begin{pmatrix} 11 \\ 3 \end{pmatrix} - \begin{pmatrix} 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 11 \\ 4 \end{pmatrix}$

Correction: The answer is $\begin{pmatrix} 11 \\ 4 \end{pmatrix}$.

Marking: 1 mark for correct method (transform both points or use $\vec{P'Q'} = S\vec{PQ}$), 1 mark for correct final vector.

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20. [2 marks]

Answer: $5$ square units

Working: Area scale factor = $|\det(A)|$. $\det(A) = 2(1) - 1(1) = 2 - 1 = 1$. Area of image = $|\det(A)| \times \text{original area} = 1 \times 5 = 5$ square units.

Marking: 1 mark for finding determinant = 1, 1 mark for correct area (5 square units).

Note: Since $\det(A) = 1$, the transformation preserves area.

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End of Answer Key