Question 6 - A-Level Maths

AQA FP1 2013 January — Question 6 12 marks

Exam Board	AQA
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	January
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear transformations
Type	Extract enlargement and rotation parameters
Difficulty	Moderate -0.3 This is a routine Further Maths question testing standard matrix operations (multiplication, powers), basic geometric transformations (reflection, rotation), and straightforward application of transformation composition. All parts follow textbook procedures with no novel problem-solving required, though being Further Maths places it slightly above average A-level difficulty overall.
Spec	1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form 4.03b Matrix operations: addition, multiplication, scalar 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear 4.03e Successive transformations: matrix products

The matrix $\mathbf { X }$ is defined by $\left[ \begin{array} { l l } 1 & 2 \\ 3 & 0 \end{array} \right]$.
1. Given that $\mathbf { X } ^ { 2 } = \left[ \begin{array} { c c } m & 2 \\ 3 & 6 \end{array} \right]$, find the value of $m$.
2. Show that $\mathbf { X } ^ { 3 } - 7 \mathbf { X } = n \mathbf { I }$, where $n$ is an integer and $\mathbf { I }$ is the $2 \times 2$ identity matrix.
It is given that $\mathbf { A } = \left[ \begin{array} { r r } 1 & 0 \\ 0 & - 1 \end{array} \right]$.
1. Describe the geometrical transformation represented by $\mathbf { A }$.
2. The matrix $\mathbf { B }$ represents an anticlockwise rotation through $45 ^ { \circ }$ about the origin. Show that $\mathbf { B } = k \left[ \begin{array} { r r } 1 & - 1 \\ 1 & 1 \end{array} \right]$, where $k$ is a surd.
3. Find the image of the point $P ( - 1,2 )$ under an anticlockwise rotation through $45 ^ { \circ }$ about the origin, followed by the transformation represented by $\mathbf { A }$. $7 \quad$ The variables $y$ and $x$ are related by an equation of the form $$y = a x ^ { n }$$ where $a$ and $n$ are constants. Let $Y = \log _ { 10 } y$ and $X = \log _ { 10 } x$.

Show mark scheme Show mark scheme source

Question 6:

Part (a)(i):

Answer	Marks	Guidance
$\mathbf{X}^2 = \begin{pmatrix} 7 & 2 \\ 3 & 6 \end{pmatrix}$; $(m=)7$	B1	$(m=)7$ or 7 as top left element of $\mathbf{X}^2$

Part (a)(ii):

Answer	Marks	Guidance
$\mathbf{X}^3 = \begin{pmatrix} 13 & 14 \\ 21 & 6 \end{pmatrix}$	M1	At least 2 elements correct
$7\mathbf{X} = \begin{pmatrix} 7 & 14 \\ 21 & 0 \end{pmatrix}$	B1	PI
$\mathbf{X}^3 - 7\mathbf{X} = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}$	A1F	Ft on c's $m$ value
$= 6\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 6\mathbf{I}$	A1	CSO. Accept either form but at least one must be shown explicitly

Part (b)(i):

Answer	Marks	Guidance
Reflection in the $x$-axis	B1	OE

Part (b)(ii):

Answer	Marks	Guidance
$\mathbf{B} = \begin{pmatrix} \cos 45° & -\sin 45° \\ \sin 45° & \cos 45° \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$	M1	Either OE. For M mark, accept dec. equiv. (at least 3sf) for $\frac{1}{\sqrt{2}}$
$= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$	A1	NMS SC1 for $k = \frac{1}{\sqrt{2}}$ or better

Part (b)(iii):

Answer	Marks	Guidance
$\mathbf{AB}\begin{pmatrix} -1 \\ 2 \end{pmatrix} = k\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix}$	M1	Attempt to find $\mathbf{AB}\begin{pmatrix} -1 \\ 2 \end{pmatrix}$
$= k\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} -3 \\ 1 \end{pmatrix}$	A1	Either $\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ or $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}$
$= k\begin{pmatrix} -3 \\ -1 \end{pmatrix}$	m1	Completing the matrix mult. to reach a $2\times1$ matrix
Image of $P$ is the point $\left(-\dfrac{3}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}\right)$	A1	CSO. SC Wrong order, works with $\mathbf{BA}\begin{pmatrix}-1\\2\end{pmatrix}$, mark out of max M1A0 m1A0

## Question 6:

### Part (a)(i):
$\mathbf{X}^2 = \begin{pmatrix} 7 & 2 \\ 3 & 6 \end{pmatrix}$; $(m=)7$ | B1 | $(m=)7$ or 7 as top left element of $\mathbf{X}^2$

### Part (a)(ii):
$\mathbf{X}^3 = \begin{pmatrix} 13 & 14 \\ 21 & 6 \end{pmatrix}$ | M1 | At least 2 elements correct

$7\mathbf{X} = \begin{pmatrix} 7 & 14 \\ 21 & 0 \end{pmatrix}$ | B1 | PI

$\mathbf{X}^3 - 7\mathbf{X} = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}$ | A1F | Ft on c's $m$ value

$= 6\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 6\mathbf{I}$ | A1 | CSO. Accept either form but at least one must be shown explicitly

### Part (b)(i):
Reflection in the $x$-axis | B1 | OE

### Part (b)(ii):
$\mathbf{B} = \begin{pmatrix} \cos 45° & -\sin 45° \\ \sin 45° & \cos 45° \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$ | M1 | Either OE. For M mark, accept dec. equiv. (at least 3sf) for $\frac{1}{\sqrt{2}}$

$= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ | A1 | NMS SC1 for $k = \frac{1}{\sqrt{2}}$ or better

### Part (b)(iii):
$\mathbf{AB}\begin{pmatrix} -1 \\ 2 \end{pmatrix} = k\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix}$ | M1 | Attempt to find $\mathbf{AB}\begin{pmatrix} -1 \\ 2 \end{pmatrix}$

$= k\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} -3 \\ 1 \end{pmatrix}$ | A1 | Either $\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ or $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}$

$= k\begin{pmatrix} -3 \\ -1 \end{pmatrix}$ | m1 | Completing the matrix mult. to reach a $2\times1$ matrix

Image of $P$ is the point $\left(-\dfrac{3}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}\right)$ | A1 | CSO. SC Wrong order, works with $\mathbf{BA}\begin{pmatrix}-1\\2\end{pmatrix}$, mark out of max M1A0 m1A0

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Show LaTeX source

6
\begin{enumerate}[label=(\alph*)]
\item The matrix $\mathbf { X }$ is defined by $\left[ \begin{array} { l l } 1 & 2 \\ 3 & 0 \end{array} \right]$.
\begin{enumerate}[label=(\roman*)]
\item Given that $\mathbf { X } ^ { 2 } = \left[ \begin{array} { c c } m & 2 \\ 3 & 6 \end{array} \right]$, find the value of $m$.
\item Show that $\mathbf { X } ^ { 3 } - 7 \mathbf { X } = n \mathbf { I }$, where $n$ is an integer and $\mathbf { I }$ is the $2 \times 2$ identity matrix.
\end{enumerate}\item It is given that $\mathbf { A } = \left[ \begin{array} { r r } 1 & 0 \\ 0 & - 1 \end{array} \right]$.
\begin{enumerate}[label=(\roman*)]
\item Describe the geometrical transformation represented by $\mathbf { A }$.
\item The matrix $\mathbf { B }$ represents an anticlockwise rotation through $45 ^ { \circ }$ about the origin. Show that $\mathbf { B } = k \left[ \begin{array} { r r } 1 & - 1 \\ 1 & 1 \end{array} \right]$, where $k$ is a surd.
\item Find the image of the point $P ( - 1,2 )$ under an anticlockwise rotation through $45 ^ { \circ }$ about the origin, followed by the transformation represented by $\mathbf { A }$.\\
$7 \quad$ The variables $y$ and $x$ are related by an equation of the form

$$y = a x ^ { n }$$

where $a$ and $n$ are constants.

Let $Y = \log _ { 10 } y$ and $X = \log _ { 10 } x$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2013 Q6 [12]}}

This paper (9 questions)

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Q1 5 Q2 9 Q3 8 Q4 4 Q5 9 Q6 12 Q7 7 Q8 8 Q9 13

Answer	Marks	Guidance
\(\mathbf{AB}\begin{pmatrix} -1 \\ 2 \end{pmatrix} = k\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix}\)	M1	Attempt to find \(\mathbf{AB}\begin{pmatrix} -1 \\ 2 \end{pmatrix}\)
\(= k\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} -3 \\ 1 \end{pmatrix}\)	A1	Either \(\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}\) or \(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}\)
\(= k\begin{pmatrix} -3 \\ -1 \end{pmatrix}\)	m1	Completing the matrix mult. to reach a \(2\times1\) matrix
Image of \(P\) is the point \(\left(-\dfrac{3}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}\right)\)	A1	CSO. SC Wrong order, works with \(\mathbf{BA}\begin{pmatrix}-1\\2\end{pmatrix}\), mark out of max M1A0 m1A0

Answer	Marks	Guidance
\(\mathbf{X}^3 = \begin{pmatrix} 13 & 14 \\ 21 & 6 \end{pmatrix}\)	M1	At least 2 elements correct
\(7\mathbf{X} = \begin{pmatrix} 7 & 14 \\ 21 & 0 \end{pmatrix}\)	B1	PI
\(\mathbf{X}^3 - 7\mathbf{X} = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\)	A1F	Ft on c's \(m\) value
\(= 6\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 6\mathbf{I}\)	A1	CSO. Accept either form but at least one must be shown explicitly

Answer	Marks	Guidance
\(\mathbf{B} = \begin{pmatrix} \cos 45° & -\sin 45° \\ \sin 45° & \cos 45° \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\)	M1	Either OE. For M mark, accept dec. equiv. (at least 3sf) for \(\frac{1}{\sqrt{2}}\)
\(= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\)	A1	NMS SC1 for \(k = \frac{1}{\sqrt{2}}\) or better