Secondary 3 Additional Mathematics Vectors Matrices Quiz

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Secondary 3 Additional Mathematics Quiz - Vectors Matrices

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

1. Answer all questions.
2. Show all necessary working clearly. No marks will be given for correct answers without working.
3. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
4. The use of an approved scientific calculator is expected, where appropriate.

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Section A: Vector Algebra and Geometry (20 Marks)

1. Given that $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$, find the column vector representing $2\mathbf{a} - 3\mathbf{b}$.
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2. The position vectors of points $A$ and $B$ relative to the origin $O$ are $\vec{OA} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} 8 \\ -1 \end{pmatrix}$. Find the unit vector in the direction of $\vec{AB}$.
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3. Points $P, Q$ and $R$ have position vectors $\mathbf{p} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$, $\mathbf{q} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} 10 \\ 15 \end{pmatrix}$ respectively. Show that $P, Q$ and $R$ are collinear.
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4. In triangle $OAB$, $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. Point $M$ is the midpoint of $AB$. Express $\vec{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
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5. Given vectors $\mathbf{u} = \begin{pmatrix} 4 \\ k \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Find the value of $k$ if $\mathbf{u}$ is perpendicular to $\mathbf{v}$.
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6. A vector $\mathbf{w}$ has magnitude 10 and makes an angle of $120^\circ$ with the positive x-axis. Express $\mathbf{w}$ in the form $\begin{pmatrix} x \\ y \end{pmatrix}$, leaving your answer in exact surd form.
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7. Relative to an origin $O$, the position vectors of points $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ respectively. Point $C$ lies on $AB$ such that $AC : CB = 2 : 1$. Find $\vec{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
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Section B: Matrix Operations and Transformations (15 Marks)

8. Given matrices $A = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ -2 & 0 \end{pmatrix}$. Calculate the matrix $A - 2B$.
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9. Using the matrices from Question 8, calculate the product $AB$.
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10. Find the inverse of the matrix $M = \begin{pmatrix} 5 & 2 \\ 3 & 1 \end{pmatrix}$.
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11. Solve the following simultaneous equations using the matrix method:

$$\begin{cases} 4x + 3y = 10 \\ 2x - y = 5 \end{cases}$$

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12. The transformation $T$ is represented by the matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Describe the geometric transformation $T$ fully.
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Section C: Advanced Applications and Proofs (15 Marks)

13. Given that $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, verify that $A^2 - 5A - 2I = \mathbf{0}$, where $I$ is the identity matrix and $\mathbf{0}$ is the zero matrix.
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14. Points $A(1, 2)$, $B(4, 6)$ and $C(7, 10)$ are given. (a) Find the vectors $\vec{AB}$ and $\vec{BC}$.
(b) Hence, determine if $A, B$ and $C$ form a triangle or are collinear. Justify your answer.
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15. A rectangle $OABC$ has vertices $O(0,0)$, $A(4,0)$, $B(4,3)$ and $C(0,3)$. The rectangle is transformed by the matrix $M = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. (a) Find the coordinates of the image $A'B'C'O'$.
(b) Calculate the ratio of the area of $O'A'B'C'$ to the area of $OABC$.
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16. Given that $\mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$. Find the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$.
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17. If $\begin{pmatrix} x & 2 \\ 3 & y \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \end{pmatrix}$, find the values of $x$ and $y$.
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18. The matrix $P = \begin{pmatrix} k & 1 \\ 2 & k \end{pmatrix}$ is singular. Find the possible values of $k$.
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19. In a parallelogram $OABC$, $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$. $M$ is the midpoint of $OA$ and $N$ is the midpoint of $BC$. Express $\vec{MN}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.
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20. Given vectors $\mathbf{p} = 2\mathbf{i} - \mathbf{j}$ and $\mathbf{q} = \mathbf{i} + 3\mathbf{j}$. Find the angle between $\mathbf{p}$ and $\mathbf{q}$, correct to 1 decimal place.
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Secondary 3 Additional Mathematics Quiz - Vectors Matrices (Answer Key)

1. $2\mathbf{a} - 3\mathbf{b} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} - 3\begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} - \begin{pmatrix} -3 \\ 12 \end{pmatrix} = \begin{pmatrix} 6 - (-3) \\ -4 - 12 \end{pmatrix} = \begin{pmatrix} 9 \\ -16 \end{pmatrix}$ Answer: $\begin{pmatrix} 9 \\ -16 \end{pmatrix}$ [2]

2. $\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -6 \end{pmatrix}$ Magnitude $|\vec{AB}| = \sqrt{6^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$. Unit vector = $\frac{1}{6\sqrt{2}} \begin{pmatrix} 6 \\ -6 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix}$ or $\begin{pmatrix} \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} \end{pmatrix}$. Answer: $\begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix}$ [3]

3. $\vec{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ $\vec{QR} = \mathbf{r} - \mathbf{q} = \begin{pmatrix} 10 \\ 15 \end{pmatrix} - \begin{pmatrix} 4 \\ 7 \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$ Since $\vec{QR} = 2 \begin{pmatrix} 3 \\ 4 \end{pmatrix} = 2\vec{PQ}$, the vectors are parallel. Since they share a common point $Q$, $P, Q, R$ are collinear. [3]

4. Using the midpoint formula for vectors: $\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2}(\mathbf{a} + \mathbf{b})$ Answer: $\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$ [2]

5. If perpendicular, dot product is zero: $\mathbf{u} \cdot \mathbf{v} = (4)(2) + (k)(-1) = 0$ $8 - k = 0 \implies k = 8$ Answer: $k = 8$ [2]

6. $x = 10 \cos(120^\circ) = 10 \left(-\frac{1}{2}\right) = -5$ $y = 10 \sin(120^\circ) = 10 \left(\frac{\sqrt{3}}{2}\right) = 5\sqrt{3}$ Answer: $\begin{pmatrix} -5 \\ 5\sqrt{3} \end{pmatrix}$ [3]

7. Using the section formula: $\vec{OC} = \frac{1\mathbf{a} + 2\mathbf{b}}{1+2} = \frac{\mathbf{a} + 2\mathbf{b}}{3}$ Answer: $\frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}$ [3]

8. $2B = \begin{pmatrix} 2 & 8 \\ -4 & 0 \end{pmatrix}$ $A - 2B = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} - \begin{pmatrix} 2 & 8 \\ -4 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -9 \\ 4 & 3 \end{pmatrix}$ Answer: $\begin{pmatrix} 0 & -9 \\ 4 & 3 \end{pmatrix}$ [2]

9. $AB = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ -2 & 0 \end{pmatrix}$ Row 1, Col 1: $(2)(1) + (-1)(-2) = 2 + 2 = 4$ Row 1, Col 2: $(2)(4) + (-1)(0) = 8 + 0 = 8$ Row 2, Col 1: $(0)(1) + (3)(-2) = 0 - 6 = -6$ Row 2, Col 2: $(0)(4) + (3)(0) = 0 + 0 = 0$ Answer: $\begin{pmatrix} 4 & 8 \\ -6 & 0 \end{pmatrix}$ [3]

10. Determinant of $M = (5)(1) - (2)(3) = 5 - 6 = -1$. $M^{-1} = \frac{1}{-1} \begin{pmatrix} 1 & -2 \\ -3 & 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 3 & -5 \end{pmatrix}$ Answer: $\begin{pmatrix} -1 & 2 \\ 3 & -5 \end{pmatrix}$ [3]

11. Matrix form: $\begin{pmatrix} 4 & 3 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ 5 \end{pmatrix}$. Determinant $D = (4)(-1) - (3)(2) = -4 - 6 = -10$. Inverse: $\frac{1}{-10} \begin{pmatrix} -1 & -3 \\ -2 & 4 \end{pmatrix}$. $\begin{pmatrix} x \\ y \end{pmatrix} = -\frac{1}{10} \begin{pmatrix} -1 & -3 \\ -2 & 4 \end{pmatrix} \begin{pmatrix} 10 \\ 5 \end{pmatrix}$ $= -\frac{1}{10} \begin{pmatrix} -10 - 15 \\ -20 + 20 \end{pmatrix} = -\frac{1}{10} \begin{pmatrix} -25 \\ 0 \end{pmatrix} = \begin{pmatrix} 2.5 \\ 0 \end{pmatrix}$ Answer: $x = 2.5, y = 0$ [4]

12. The matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ maps $(1,0) \to (0,1)$ and $(0,1) \to (-1,0)$. This is a rotation of $90^\circ$ anti-clockwise about the origin. Answer: Rotation $90^\circ$ anti-clockwise about the origin. [3]

13. $A^2 = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1+6 & 2+8 \\ 3+12 & 6+16 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}$ $5A = \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix}, \quad 2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ $A^2 - 5A - 2I = \begin{pmatrix} 7-5-2 & 10-10-0 \\ 15-15-0 & 22-20-2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{0}$ Verified. [4]

14. (a) $\vec{AB} = \begin{pmatrix} 4-1 \\ 6-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, $\vec{BC} = \begin{pmatrix} 7-4 \\ 10-6 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$. [2] (b) Since $\vec{AB} = \vec{BC}$, the vectors are parallel and share point $B$. Thus, $A, B, C$ are collinear. They do not form a triangle. [2]

15. (a) $M \begin{pmatrix} 4 \\ 0 \end{pmatrix} = \begin{pmatrix} 8 \\ 0 \end{pmatrix} \implies A'(8,0)$. $M \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ 6 \end{pmatrix} \implies B'(8,6)$. $M \begin{pmatrix} 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 6 \end{pmatrix} \implies C'(0,6)$. $O'(0,0)$. [2] (b) Area $OABC = 4 \times 3 = 12$. Area $O'A'B'C' = 8 \times 6 = 48$. Ratio $48:12 = 4:1$. (Alternatively, determinant of $M$ is 4, so area scale factor is 4). [2]

16. Scalar projection of $\mathbf{a}$ on $\mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$. $\mathbf{a} \cdot \mathbf{b} = (3)(1) + (4)(2) = 3 + 8 = 11$. $|\mathbf{b}| = \sqrt{1^2 + 2^2} = \sqrt{5}$. Answer: $\frac{11}{\sqrt{5}}$ or $\frac{11\sqrt{5}}{5}$ [3]

17. $\begin{pmatrix} x+2 \\ 3+y \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \end{pmatrix}$. $x + 2 = 5 \implies x = 3$. $3 + y = 7 \implies y = 4$. Answer: $x=3, y=4$ [2]

18. Singular means determinant is 0. $\det(P) = k^2 - 2 = 0$. $k^2 = 2 \implies k = \pm\sqrt{2}$. Answer: $k = \sqrt{2}, -\sqrt{2}$ [2]

19. $\vec{OM} = \frac{1}{2}\mathbf{a}$. $\vec{ON} = \vec{OC} + \vec{CN} = \mathbf{c} + \frac{1}{2}\mathbf{a}$ (since $\vec{CB} = \mathbf{a}$ and $N$ is midpoint). $\vec{MN} = \vec{ON} - \vec{OM} = (\mathbf{c} + \frac{1}{2}\mathbf{a}) - \frac{1}{2}\mathbf{a} = \mathbf{c}$. Answer: $\mathbf{c}$ [3]

20. $\mathbf{p} \cdot \mathbf{q} = (2)(1) + (-1)(3) = 2 - 3 = -1$. $|\mathbf{p}| = \sqrt{2^2 + (-1)^2} = \sqrt{5}$. $|\mathbf{q}| = \sqrt{1^2 + 3^2} = \sqrt{10}$. $\cos \theta = \frac{-1}{\sqrt{5}\sqrt{10}} = \frac{-1}{\sqrt{50}} = \frac{-1}{5\sqrt{2}}$. $\theta = \cos^{-1}\left(\frac{-1}{5\sqrt{2}}\right) \approx 98.1^\circ$. Answer: $98.1^\circ$ [3]