Secondary 3 Additional Mathematics Vectors Matrices Quiz

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Secondary 3 Additional Mathematics Quiz - Vectors Matrices

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

Duration: 45 minutes
Total Marks: 40

Instructions:

• Answer all questions.
• Write your answers in the spaces provided.
• Show all working clearly.
• 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 in the question.

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Section A (20 marks)

Answer all questions. Each question carries 2 marks.

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

Answer: $\begin{pmatrix} \phantom{-} \end{pmatrix}$ [2]

2. The position vectors of points $A$ and $B$ relative to origin $O$ are $\overrightarrow{OA} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$. Find the vector $\overrightarrow{AB}$.

Answer: $\begin{pmatrix} \phantom{-} \end{pmatrix}$ [2]

3. Find the magnitude of the vector $\mathbf{v} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$.

Answer: _______________________ [2]

4. A vector $\mathbf{u}$ has magnitude 13 and makes an angle of $60^\circ$ with the positive $x$-axis. Express $\mathbf{u}$ in the form $\begin{pmatrix} x \\ y \end{pmatrix}$.

Answer: $\begin{pmatrix} \phantom{-} \end{pmatrix}$ [2]

5. Given that $\mathbf{p} = \begin{pmatrix} 4 \\ k \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}$ are parallel, find the value of $k$.

Answer: $k =$ _______________________ [2]

6. The matrix $M = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$. Find $M^2$.

Answer: $\begin{pmatrix} \phantom{-} & \phantom{-} \\ \phantom{-} & \phantom{-} \end{pmatrix}$ [2]

7. Given matrices $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & -1 \\ 2 & 3 \end{pmatrix}$, find $3A - 2B$.

Answer: $\begin{pmatrix} \phantom{-} & \phantom{-} \\ \phantom{-} & \phantom{-} \end{pmatrix}$ [2]

8. Find the determinant of the matrix $\begin{pmatrix} 5 & -2 \\ 3 & 4 \end{pmatrix}$.

Answer: _______________________ [2]

9. The matrix $P = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$. Find the inverse matrix $P^{-1}$.

Answer: $\begin{pmatrix} \phantom{-} & \phantom{-} \\ \phantom{-} & \phantom{-} \end{pmatrix}$ [2]

10. Solve the matrix equation $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 11 \end{pmatrix}$ for $x$ and $y$.

Answer: $x =$ __________, $y =$ __________ [2]

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Section B (12 marks)

Answer all questions. Each question carries 3 marks.

11. Vectors $\mathbf{a}$ and $\mathbf{b}$ are such that $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
(a) Find the unit vector in the direction of $\mathbf{a} + 2\mathbf{b}$.
(b) Hence, find a vector of magnitude 10 in the same direction as $\mathbf{a} + 2\mathbf{b}$.

Answer:
(a) $\begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$
(b) $\begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$ [3]

12. In the diagram, $OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. The point $M$ lies on $AB$ such that $AM : MB = 2 : 1$. The point $N$ lies on $BC$ such that $BN : NC = 1 : 2$.
<image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: Parallelogram OABC with vertices labelled O, A, B, C in order. Point M on AB with AM:MB = 2:1. Point N on BC with BN:NC = 1:2. Vectors a = OA, c = OC shown. labels: O, A, B, C, M, N, a, c values: AM:MB = 2:1, BN:NC = 1:2 must_show: Parallelogram with labelled vertices, points M and N on sides AB and BC respectively, vector labels a and c from O </image_placeholder>

Express the following vectors in terms of $\mathbf{a}$ and $\mathbf{c}$:
(a) $\overrightarrow{OM}$
(b) $\overrightarrow{ON}$
(c) $\overrightarrow{MN}$

Answer:
(a) $\overrightarrow{OM} =$ _______________________
(b) $\overrightarrow{ON} =$ _______________________
(c) $\overrightarrow{MN} =$ _______________________ [3]

13. The matrix $A = \begin{pmatrix} 2 & k \\ 1 & 3 \end{pmatrix}$ is singular. Find the value of $k$.

Answer: $k =$ _______________________ [3]

14. Given that $M = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $M^2 - kM - 2I = \mathbf{0}$, where $I$ is the identity matrix and $\mathbf{0}$ is the zero matrix, find the value of $k$.

Answer: $k =$ _______________________ [3]

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Section C (8 marks)

Answer all questions. Each question carries 4 marks.

15. The points $A$, $B$, and $C$ have position vectors $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $\begin{pmatrix} 4 \\ 6 \end{pmatrix}$, and $\begin{pmatrix} 7 \\ 10 \end{pmatrix}$ respectively, relative to the origin $O$.
(a) Find $\overrightarrow{AB}$ and $\overrightarrow{BC}$.
(b) Show that $A$, $B$, and $C$ are collinear.
(c) Find the ratio $AB : BC$.

Answer:
(a) $\overrightarrow{AB} = \begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$
(b) _________________________________________________________________
(c) $AB : BC =$ _______________________ [4]

16. A transformation $T$ is represented by the matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.
(a) Describe fully the geometric transformation $T$.
(b) The triangle $PQR$ has vertices $P(1, 2)$, $Q(3, 1)$, and $R(2, 4)$. Find the coordinates of the image of triangle $PQR$ under transformation $T$.
(c) Find the matrix representing the transformation $T^2$ and describe it geometrically.

Answer:
(a) _________________________________________________________________
(b) $P'($$,$$)$, $Q'($$,$$)$, $R'($$,$$)$
(c) Matrix: $\begin{pmatrix} \phantom{-} & \phantom{-} \\ \phantom{-} & \phantom{-} \end{pmatrix}$, Description: _________________________________________________________________ [4]

17. The matrix $A = \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}$.
(a) Find $A^{-1}$.
(b) Use your answer to (a) to solve the simultaneous equations:
$3x + y = 7$
$2x + y = 5$

Answer:
(a) $A^{-1} = \begin{pmatrix} \phantom{-} & \phantom{-} \\ \phantom{-} & \phantom{-} \end{pmatrix}$
(b) $x =$ __________, $y =$ __________ [4]

18. In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. The point $P$ lies on $OA$ such that $OP : PA = 1 : 2$. The point $Q$ lies on $OB$ such that $OQ : QB = 2 : 1$. The lines $AQ$ and $BP$ intersect at $R$.
<image_placeholder> id: Q18-fig1 type: diagram linked_question: Q18 description: Triangle OAB with vertices O, A, B. Point P on OA with OP:PA = 1:2. Point Q on OB with OQ:QB = 2:1. Lines AQ and BP intersect at R. labels: O, A, B, P, Q, R, a, b values: OP:PA = 1:2, OQ:QB = 2:1 must_show: Triangle with labelled vertices, points P and Q on sides OA and OB, intersection point R of AQ and BP, vector labels a and b </image_placeholder>

(a) Express $\overrightarrow{AP}$ and $\overrightarrow{BQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
(b) Express $\overrightarrow{AQ}$ and $\overrightarrow{BP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
(c) Given that $\overrightarrow{AR} = \lambda \overrightarrow{AQ}$ and $\overrightarrow{BR} = \mu \overrightarrow{BP}$, find the values of $\lambda$ and $\mu$.

Answer:
(a) $\overrightarrow{AP} =$ _______________________, $\overrightarrow{BQ} =$ _______________________
(b) $\overrightarrow{AQ} =$ _______________________, $\overrightarrow{BP} =$ _______________________
(c) $\lambda =$ __________, $\mu =$ __________ [4]

19. The matrix $M = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$.
(a) Find the eigenvalues of $M$.
(b) For each eigenvalue, find a corresponding eigenvector.

Answer:
(a) Eigenvalues: _______________________
(b) Eigenvector for first eigenvalue: $\begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$, Eigenvector for second eigenvalue: $\begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$ [4]

20. A sequence of vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \dots$ is defined by $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{v}_{n+1} = M \mathbf{v}_n$ for $n \ge 1$, where $M = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$.
(a) Find $\mathbf{v}_2$, $\mathbf{v}_3$, and $\mathbf{v}_4$.
(b) Describe the pattern you observe in the components of $\mathbf{v}_n$.
(c) Hence, or otherwise, find $\mathbf{v}_5$.

Answer:
(a) $\mathbf{v}_2 = \begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$, $\mathbf{v}_3 = \begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$, $\mathbf{v}_4 = \begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$
(b) _________________________________________________________________
(c) $\mathbf{v}_5 = \begin{pmatrix} \phantom{-} \\ \phantom{-} \end{pmatrix}$ [4]

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End of Quiz

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Secondary 3 Additional Mathematics Quiz - Vectors Matrices (Answer Key)

Total Marks: 40

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Section A (20 marks)

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

2. $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} -5 \\ -4 \end{pmatrix}$ Answer: $\begin{pmatrix} -5 \\ -4 \end{pmatrix}$ [2]

3. $|\mathbf{v}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ Answer: 10 [2]

4. $\mathbf{u} = \begin{pmatrix} 13\cos 60^\circ \\ 13\sin 60^\circ \end{pmatrix} = \begin{pmatrix} 13 \times \frac{1}{2} \\ 13 \times \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} 6.5 \\ \frac{13\sqrt{3}}{2} \end{pmatrix} \approx \begin{pmatrix} 6.5 \\ 11.3 \end{pmatrix}$ Answer: $\begin{pmatrix} 6.5 \\ \frac{13\sqrt{3}}{2} \end{pmatrix}$ or $\begin{pmatrix} 6.5.5 \\ 11.3 \end{pmatrix}$ [2]

5. Parallel vectors: $\mathbf{p} = k\mathbf{q}$ for some scalar $k$.
$\begin{pmatrix} 4 \\ k \end{pmatrix} = t \begin{pmatrix} -2 \\ 6 \end{pmatrix} \Rightarrow 4 = -2t \Rightarrow t = -2$.
Then $k = 6t = 6(-2) = -12$. Answer: $k = -12$ [2]

6. $M^2 = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 4-3 & -2-4 \\ 6+12 & -3+16 \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 18 & 13 \end{pmatrix}$ Answer: $\begin{pmatrix} 1 & -6 \\ 18 & 13 \end{pmatrix}$ [2]

7. $3A - 2B = 3\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - 2\begin{pmatrix} 0 & -1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 9 & 12 \end{pmatrix} - \begin{pmatrix} 0 & -2 \\ 4 & 6 \end{pmatrix} = \begin{pmatrix} 3 & 8 \\ 5 & 6 \end{pmatrix}$ Answer: $\begin{pmatrix} 3 & 8 \\ 5 & 6 \end{pmatrix}$ [2]

8. $\det = (5)(4) - (-2)(3) = 20 + 6 = 26$ Answer: 26 [2]

9. For $P = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$, $\det P = 4 - 3 = 1$.
$P^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$ Answer: $\begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$ [2]

10. $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 11 \end{pmatrix}$
$\det = 4 - 6 = -2$. Inverse: $-\frac{1}{2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$
$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 5 \\ 11 \end{pmatrix} = \begin{pmatrix} -10 + 11 \\ 7.5 - 5.5 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$
Alternatively: $x + 2y = 5$, $3x + 4y = 11 \Rightarrow x = 1, y = 2$. Answer: $x = 1$, $y = 2$ [2]

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Section B (12 marks)

11. $\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \end{pmatrix}$
Magnitude: $|\mathbf{a} + 2\mathbf{b}| = \sqrt{8^2 + 7^2} = \sqrt{64 + 49} = \sqrt{113}$
(a) Unit vector: $\frac{1}{\sqrt{113}}\begin{pmatrix} 8 \\ 7 \end{pmatrix} = \begin{pmatrix} \frac{8}{\sqrt{113}} \\ \frac{7}{\sqrt{113}} \end{pmatrix}$
(b) Vector of magnitude 10: $10 \times \begin{pmatrix} \frac{8}{\sqrt{113}} \\ \frac{7}{\sqrt{113}} \end{pmatrix} = \begin{pmatrix} \frac{80}{\sqrt{113}} \\ \frac{70}{\sqrt{113}} \end{pmatrix} \approx \begin{pmatrix} 7.52 \\ 6.58 \end{pmatrix}$

Answer:
(a) $\begin{pmatrix} \frac{8}{\sqrt{113}} \\ \frac{7}{\sqrt{113}} \end{pmatrix}$
(b) $\begin{pmatrix} \frac{80}{\sqrt{113}} \\ \frac{70}{\sqrt{113}} \end{pmatrix}$ [3]

12. In parallelogram $OABC$, $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$, $\overrightarrow{AB} = \mathbf{c}$, $\overrightarrow{BC} = \mathbf{a}$.
(a) $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \frac{2}{3}\overrightarrow{AB} = \mathbf{a} + \frac{2}{3}\mathbf{c}$
(b) $\overrightarrow{ON} = \overrightarrow{OC} + \overrightarrow{CN} = \mathbf{c} + \frac{2}{3}\overrightarrow{CB} = \mathbf{c} + \frac{2}{3}(-\mathbf{a}) = \mathbf{c} - \frac{2}{3}\mathbf{a}$
(Alternatively: $\overrightarrow{ON} = \overrightarrow{OB} + \overrightarrow{BN} = \mathbf{a} + \mathbf{c} + \frac{1}{3}(-\mathbf{a}) = \frac{2}{3}\mathbf{a} + \mathbf{c}$ — wait, check: $BN:NC = 1:2$ so $BN = \frac{1}{3}BC = \frac{1}{3}\mathbf{a}$, so $\overrightarrow{ON} = \mathbf{a} + \mathbf{c} + \frac{1}{3}\mathbf{a} = \frac{4}{3}\mathbf{a} + \mathbf{c}$? No: $BC$ goes from B to C, so $\overrightarrow{BC} = \mathbf{c} - (\mathbf{a}+\mathbf{c}) = -\mathbf{a}$. Wait, let's redo carefully.)

In parallelogram $OABC$: $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OC} = \mathbf{c}$. Then $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{c}$.
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \mathbf{c} - (\mathbf{a} + \mathbf{c}) = -\mathbf{a}$.

$M$ on $AB$: $AM:MB = 2:1 \Rightarrow \overrightarrow{AM} = \frac{2}{3}\overrightarrow{AB} = \frac{2}{3}\mathbf{c}$.
$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \frac{2}{3}\mathbf{c}$. ✓

$N$ on $BC$: $BN:NC = 1:2 \Rightarrow \overrightarrow{BN} = \frac{1}{3}\overrightarrow{BC} = -\frac{1}{3}\mathbf{a}$.
$\overrightarrow{ON} = \overrightarrow{OB} + \overrightarrow{BN} = (\mathbf{a} + \mathbf{c}) - \frac{1}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} + \mathbf{c}$. ✓

(c) $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = (\frac{2}{3}\mathbf{a} + \mathbf{c}) - (\mathbf{a} + \frac{2}{3}\mathbf{c}) = -\frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{c} = \frac{1}{3}(\mathbf{c} - \mathbf{a})$.

Answer:
(a) $\overrightarrow{OM} = \mathbf{a} + \frac{2}{3}\mathbf{c}$
(b) $\overrightarrow{ON} = \frac{2}{3}\mathbf{a} + \mathbf{c}$
(c) $\overrightarrow{MN} = \frac{1}{3}(\mathbf{c} - \mathbf{a})$ [3]

13. Matrix $A = \begin{pmatrix} 2 & k \\ 1 & 3 \end{pmatrix}$ is singular $\Rightarrow \det A = 0$.
$\det A = (2)(3) - (k)(1) = 6 - k = 0 \Rightarrow k = 6$. Answer: $k = 6$ [3]

14. $M = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $M^2 = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}$ (from Q6 pattern).
$M^2 - kM - 2I = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix} - k\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - 2\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
$\begin{pmatrix} 7 - k - 2 & 10 - 2k \\ 15 - 3k & 22 - 4k - 2 \end{pmatrix} = \begin{pmatrix} 5 - k & 10 - 2k \\ 15 - 3k & 20 - 4k \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
From $5 - k = 0 \Rightarrow k = 5$. Check: $10 - 2(5) = 0$, $15 - 3(5) = 0$, $20 - 4(5) = 0$. Consistent. Answer: $k = 5$ [3]

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Section C (8 marks)

15. (a) $\overrightarrow{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
$\overrightarrow{BC} = \begin{pmatrix} 7 \\ 10 \end{pmatrix} - \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
(b) $\overrightarrow{AB} = \overrightarrow{BC}$, so vectors are parallel and share point $B$. Hence $A$, $B$, $C$ are collinear.
(c) $|\overrightarrow{AB}| = \sqrt{3^2 + 4^2} = 5$, $|\overrightarrow{BC}| = 5$. Ratio $AB:BC = 5:5 = 1:1$.

Answer:
(a) $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
(b) $\overrightarrow{AB} = \overrightarrow{BC}$, so they are parallel and share point $B$, hence $A$, $B$, $C$ are collinear.
(c) $AB : BC = 1 : 1$ [4]

16. (a) Matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ represents a rotation of $90^\circ$ anticlockwise about the origin.
(b) $P(1,2) \to P'(-2, 1)$, $Q(3,1) \to Q'(-1, 3)$, $R(2,4) \to R'(-4, 2)$.
(c) $T^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I$. This represents a rotation of $180^\circ$ about the origin (or a half-turn, or point reflection in the origin).

Answer:
(a) Rotation of $90^\circ$ anticlockwise about the origin.
(b) $P'(-2, 1)$, $Q'(-1, 3)$, $R'(-4, 2)$
(c) Matrix: $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$, Description: Rotation of $180^\circ$ about the origin (half-turn). [4]

17. (a) $A = \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}$, $\det A = 3 - 2 = 1$.
$A^{-1} = \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix}$.
(b) $\begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 5 \end{pmatrix}$
$\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1} \begin{pmatrix} 7 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} 7 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 - 5 \\ -14 + 15 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

Answer:
(a) $A^{-1} = \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix}$
(b) $x = 2$, $y = 1$ [4]

18. (a) $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA}$. $P$ divides $OA$ in ratio $1:2$, so $\overrightarrow{OP} = \frac{1}{3}\mathbf{a}$.
$\overrightarrow{AP} = \frac{1}{3}\mathbf{a} - \mathbf{a} = -\frac{2}{3}\mathbf{a}$.
$\overrightarrow{BQ} = \overrightarrow{OQ} - \overrightarrow{OB}$. $Q$ divides $OB$ in ratio $2:1$, so $\overrightarrow{OQ} = \frac{2}{3}\mathbf{b}$.
$\overrightarrow{BQ} = \frac{2}{3}\mathbf{b} - \mathbf{b} = -\frac{1}{3}\mathbf{b}$.

(b) $\overrightarrow{AQ} = \overrightarrow{OQ} - \overrightarrow{OA} = \frac{2}{3}\mathbf{b} - \mathbf{a}$.
$\overrightarrow{BP} = \overrightarrow{OP} - \overrightarrow{OB} = \frac{1}{3}\mathbf{a} - \mathbf{b}$.

(c) $\overrightarrow{AR} = \lambda \overrightarrow{AQ} = \lambda(\frac{2}{3}\mathbf{b} - \mathbf{a})$.
$\overrightarrow{OR} = \overrightarrow{OA} + \overrightarrow{AR} = \mathbf{a} + \lambda(\frac{2}{3}\mathbf{b} - \mathbf{a}) = (1-\lambda)\mathbf{a} + \frac{2\lambda}{3}\mathbf{b}$.

Also $\overrightarrow{BR} = \mu \overrightarrow{BP} = \mu(\frac{1}{3}\mathbf{a} - \mathbf{b})$.
$\overrightarrow{OR} = \overrightarrow{OB} + \overrightarrow{BR} = \mathbf{b} + \mu(\frac{1}{3}\mathbf{a} - \mathbf{b}) = \frac{\mu}{3}\mathbf{a} + (1-\mu)\mathbf{b}$.

Equate coefficients:
$1 - \lambda = \frac{\mu}{3}$ ...(1)
$\frac{2\lambda}{3} = 1 - \mu$ ...(2)

From (1): $\mu = 3(1-\lambda)$. Sub into (2):
$\frac{2\lambda}{3} = 1 - 3(1-\lambda) = 1 - 3 + 3\lambda = 3\lambda - 2$
$2\lambda = 9\lambda - 6 \Rightarrow 7\lambda = 6 \Rightarrow \lambda = \frac{6}{7}$.
$\mu = 3(1 - \frac{6}{7}) = 3(\frac{1}{7}) = \frac{3}{7}$.

Answer:
(a) $\overrightarrow{AP} = -\frac{2}{3}\mathbf{a}$, $\overrightarrow{BQ} = -\frac{1}{3}\mathbf{b}$
(b) $\overrightarrow{AQ} = \frac{2}{3}\mathbf{b} - \mathbf{a}$, $\overrightarrow{BP} = \frac{1}{3}\mathbf{a} - \mathbf{b}$
(c) $\lambda = \frac{6}{7}$, $\mu = \frac{3}{7}$ [4]

19. (a) Characteristic equation: $\det(M - \lambda I) = 0$.
$\det\begin{pmatrix} 2-\lambda & 1 \\ -1 & 3-\lambda \end{pmatrix} = (2-\lambda)(3-\lambda) + 1 = \lambda^2 - 5\lambda + 6 + 1 = \lambda^2 - 5\lambda + 7 = 0$.
$\lambda = \frac{5 \pm \sqrt{25 - 28}}{2} = \frac{5 \pm \sqrt{-3}}{2} = \frac{5 \pm i\sqrt{3}}{2}$.
Eigenvalues are complex: $\lambda_1 = \frac{5 + i\sqrt{3}}{2}$, $\lambda_2 = \frac{5 - i\sqrt{3}}{2}$.

(b) For $\lambda_1 = \frac{5 + i\sqrt{3}}{2}$:
$(M - \lambda_1 I)\mathbf{v} = \mathbf{0}$
$\begin{pmatrix} 2 - \frac{5 + i\sqrt{3}}{2} & 1 \\ -1 & 3 - \frac{5 + i\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} \frac{-1 - i\sqrt{3}}{2} & 1 \\ -1 & \frac{1 - i\sqrt{3}}{2} \end{pmatrix}$
First row: $\frac{-1 - i\sqrt{3}}{2}x + y = 0 \Rightarrow y = \frac{1 + i\sqrt{3}}{2}x$.
Eigenvector: $\begin{pmatrix} 2 \\ 1 + i\sqrt{3} \end{pmatrix}$ (or any scalar multiple).

For \lambda_2 = \frac{5 - i\sqrt{ - i\sqrt{3}}{2}:
Eigenvector: $\begin{pmatrix} 2 \\ 1 - i\sqrt{3} \end{pmatrix}$.

Answer:
(a) Eigenvalues: $\frac{5 + i\sqrt{3}}{2}$, $\frac{5 - i\sqrt{3}}{2}$
(b) Eigenvector for $\lambda_1$: $\begin{pmatrix} 2 \\ 1 + i\sqrt{3} \end{pmatrix}$, Eigenvector for $\lambda_2$: $\begin{pmatrix} 2 \\ 1 - i\sqrt{3} \end{pmatrix}$ [4]

20. $M = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$, $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$
(a) $\mathbf{v}_2 = M\mathbf{v}_1 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
$\mathbf{v}_3 = M\mathbf{v}_2 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
$\mathbf{v}_4 = M\mathbf{v}_3 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$

(b) The components follow the Fibonacci sequence: $\mathbf{v}_n = \begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix}$ where $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, \dots$
(c) $\mathbf{v}_5 = M\mathbf{v}_4 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$.

Answer:
(a) $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, $\mathbf{v}_3 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, $\mathbf{v}_4 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$
(b) The components are consecutive Fibonacci numbers: $\mathbf{v}_n = \begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix}$ where $F_1=1, F_2=1, F_{n+1}=F_n+F_{n-1}$.
(c) $\mathbf{v}_5 = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ [4]

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End of Answer Key