Question 9 - A-Level Maths

OCR MEI FP1 2012 January — Question 9 12 marks

Exam Board	OCR MEI
Module	FP1 (Further Pure Mathematics 1)
Year	2012
Session	January
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear transformations
Type	Matrix powers and repeated transformations
Difficulty	Standard +0.3 This is a straightforward Further Maths question on matrix transformations requiring recognition that R is a 90° rotation, standard matrix inverse/composition, and basic geometric reasoning. All parts follow predictable patterns with no novel problem-solving required, making it slightly easier than average even for FP1 level.
Spec	4.03d Linear transformations 2D: reflection, rotation, enlargement, shear 4.03o Inverse 3x3 matrix 4.03q Inverse transformations

9 The matrix \(\mathbf { R }\) is \(\left( \begin{array} { r r } 0 & - 1 \\ 1 & 0 \end{array} \right)\).

Explain in terms of transformations why \(\mathbf { R } ^ { 4 } = \mathbf { I }\).
Describe the transformation represented by \(\mathbf { R } ^ { - 1 }\) and write down the matrix \(\mathbf { R } ^ { - 1 }\).
\(\mathbf { S }\) is the matrix representing rotation through \(60 ^ { \circ }\) anticlockwise about the origin. Find \(\mathbf { S }\).
Write down the smallest positive integers \(m\) and \(n\) such that \(\mathbf { S } ^ { m } = \mathbf { R } ^ { n }\), explaining your answer in terms of transformations.
Find \(\mathbf { R S }\) and explain in terms of transformations why \(\mathbf { R S } = \mathbf { S R }\). \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE}

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Question 9:

Part (i):

Answer	Marks	Guidance
\(\mathbf{R}\) represents a rotation through \(90°\)	B1
\(\mathbf{R}^4\) represents 4 successive rotations through \(90°\), making \(360°\), which is a full turn, which is equivalent to the identity	B1, E1	4 successive rotations; Interpretation of \(\mathbf{R}^4\) and \(\mathbf{I}\) required

Part (ii):

Answer	Marks	Guidance
\(\mathbf{R}^{-1}\) represents a rotation of \(90°\) clockwise about the origin	B1	Rotation, angle, centre and sense
\(\mathbf{R}^{-1} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\)	B1

Part (iii):

Answer	Marks	Guidance
\(\mathbf{S} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\)	B2	One mark for each correct column (allow 3sf)

Part (iv):

Answer	Marks	Guidance
\(m = 3\); \(n = 2\)	B1, E1	\(m=3\) and \(n=2\); \(\mathbf{S}^3 = \mathbf{R}^2\) because both represent a rotation through \(180°\)

Part (v):

Answer	Marks	Guidance
\(\mathbf{RS} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix}\)	M1 A1ft	ft their \(\mathbf{S}\); \(-1\) each error
\(\mathbf{RS} = \mathbf{SR}\) because \(\mathbf{RS}\) represents a \(60°\) rotation anticlockwise about the origin followed by a \(90°\) rotation anticlockwise about the origin, making a total rotation of \(150°\) anticlockwise about the origin. \(\mathbf{SR}\) represents these two rotations in the opposite order, but the net effect is still a rotation of \(150°\) anticlockwise about the origin.	E1	Convincing explanation, correct, no ft

## Question 9:

### Part (i):
$\mathbf{R}$ represents a rotation through $90°$ | B1 | |

$\mathbf{R}^4$ represents 4 successive rotations through $90°$, making $360°$, which is a full turn, which is equivalent to the identity | B1, E1 | 4 successive rotations; Interpretation of $\mathbf{R}^4$ and $\mathbf{I}$ required | **[3]**

### Part (ii):
$\mathbf{R}^{-1}$ represents a rotation of $90°$ clockwise about the origin | B1 | Rotation, angle, centre and sense |

$\mathbf{R}^{-1} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ | B1 | | **[2]**

### Part (iii):
$\mathbf{S} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$ | B2 | One mark for each correct column (allow 3sf) | **[2]**

### Part (iv):
$m = 3$; $n = 2$ | B1, E1 | $m=3$ and $n=2$; $\mathbf{S}^3 = \mathbf{R}^2$ because both represent a rotation through $180°$ | **[2]**

### Part (v):
$\mathbf{RS} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix}$ | M1 A1ft | ft their $\mathbf{S}$; $-1$ each error |

$\mathbf{RS} = \mathbf{SR}$ because $\mathbf{RS}$ represents a $60°$ rotation anticlockwise about the origin followed by a $90°$ rotation anticlockwise about the origin, making a total rotation of $150°$ anticlockwise about the origin. $\mathbf{SR}$ represents these two rotations in the opposite order, but the net effect is still a rotation of $150°$ anticlockwise about the origin. | E1 | Convincing explanation, correct, no ft | **[3]**

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9 The matrix $\mathbf { R }$ is $\left( \begin{array} { r r } 0 & - 1 \\ 1 & 0 \end{array} \right)$.\\
(i) Explain in terms of transformations why $\mathbf { R } ^ { 4 } = \mathbf { I }$.\\
(ii) Describe the transformation represented by $\mathbf { R } ^ { - 1 }$ and write down the matrix $\mathbf { R } ^ { - 1 }$.\\
(iii) $\mathbf { S }$ is the matrix representing rotation through $60 ^ { \circ }$ anticlockwise about the origin. Find $\mathbf { S }$.\\
(iv) Write down the smallest positive integers $m$ and $n$ such that $\mathbf { S } ^ { m } = \mathbf { R } ^ { n }$, explaining your answer in terms of transformations.\\
(v) Find $\mathbf { R S }$ and explain in terms of transformations why $\mathbf { R S } = \mathbf { S R }$.

\section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE}

\hfill \mbox{\textit{OCR MEI FP1 2012 Q9 [12]}}

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