Question 4 - A-Level Maths

CAIE Further Paper 1 2020 November — Question 4 13 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2020
Session	November
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Find invariant lines through origin
Difficulty	Standard +0.3 This is a standard Further Maths question on matrix transformations requiring recognition of geometric transformations (reflection and rotation), basic matrix multiplication, and finding invariant lines via eigenvalues. All techniques are routine for Further Maths students, though part (e) requires systematic application of the characteristic equation. Slightly easier than average Further Maths material due to straightforward matrix forms.
Spec	4.03d Linear transformations 2D: reflection, rotation, enlargement, shear 4.03e Successive transformations: matrix products 4.03g Invariant points and lines 4.03h Determinant 2x2: calculation 4.03n Inverse 2x2 matrix

4 The matrices $\mathbf { A }$ and $\mathbf { B }$ are given by $$\mathbf { A } = \left( \begin{array} { l l } 0 & 1 \\ 1 & 0 \end{array} \right) \text { and } \mathbf { B } = \left( \begin{array} { c c } \frac { 1 } { 2 } & - \frac { 1 } { 2 } \sqrt { 3 } \\ \frac { 1 } { 2 } \sqrt { 3 } & \frac { 1 } { 2 } \end{array} \right)$$

Give full details of the geometrical transformation in the $x - y$ plane represented by $\mathbf { A }$.
Give full details of the geometrical transformation in the $x - y$ plane represented by $\mathbf { B }$.
The triangle $D E F$ in the $x - y$ plane is transformed by $\mathbf { A B }$ onto triangle $P Q R$.
Show that the triangles $D E F$ and $P Q R$ have the same area.
Find the matrix which transforms triangle $P Q R$ onto triangle $D E F$.
Find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { A B }$.

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks	Guidance
Reflection in the line $y = x$.	B1	Only mention reflection (and no other transformation).
Total: 1

Part (b):

Answer	Marks	Guidance
Rotation	B1	Writes 'rotation' or 'rotate' (and no other transformation).
$\frac{1}{3}\pi$ anticlockwise about the origin.	B1	Or $60°$
Total: 2

Part (c):

Answer	Marks	Guidance
$\mathbf{AB} = \begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}$ or $\det \mathbf{AB} = \det \mathbf{A} \det \mathbf{B}$	M1	Finds $\mathbf{AB}$ or uses product of determinants. Full marks may be obtained by arguing that both reflection and rotation preserve [absolute] value of area and so also does their combination.
$\det \mathbf{AB} = -1$	B1
Area of \(PQR = \	-1\	\) Area of $DEF$
Total: 3

Part (d):

Answer	Marks	Guidance
$(\mathbf{AB})^{-1} = -\begin{pmatrix} -\frac{1}{2}\sqrt{3} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\sqrt{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}$	M1 A1	or using $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$
Total: 2

Part (e):

Answer	Marks	Guidance
$\begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{1}{2}x\sqrt{3}+\frac{1}{2}y \\ \frac{1}{2}x - \frac{1}{2}y\sqrt{3} \end{pmatrix}$	B1	Transforms $\begin{pmatrix} x \\ y \end{pmatrix}$ to $\begin{pmatrix} X \\ Y \end{pmatrix}$. Accept $t\begin{pmatrix} 1 \\ m \end{pmatrix}$ instead of $\begin{pmatrix} x \\ y \end{pmatrix}$.
$\frac{1}{2} - \frac{1}{2}m\sqrt{3} = \frac{1}{2}m\sqrt{3} + \frac{1}{2}m^2$	M1 A1	Uses $Y = mX$. Allow working with $y = mx + c$ as long as $c = 0$ is stated explicitly.
$1 - m\sqrt{3} = m\sqrt{3} + m^2 \Rightarrow m^2 + 2m\sqrt{3} - 1 = 0 \Rightarrow m = \pm 2 - \sqrt{3}$	A1
$y = (2-\sqrt{3})x$ and $y + (2+\sqrt{3})x = 0$	A1	OE
Total: 5

## Question 4:

### Part (a):
| Reflection in the line $y = x$. | B1 | Only mention reflection (and no other transformation). |
| **Total: 1** | | |

### Part (b):
| Rotation | B1 | Writes 'rotation' or 'rotate' (and no other transformation). |
| $\frac{1}{3}\pi$ **anticlockwise** about the origin. | B1 | Or $60°$ |
| **Total: 2** | | |

### Part (c):
| $\mathbf{AB} = \begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}$ or $\det \mathbf{AB} = \det \mathbf{A} \det \mathbf{B}$ | M1 | Finds $\mathbf{AB}$ or uses product of determinants. Full marks may be obtained by arguing that both reflection and rotation preserve [absolute] value of area and so also does their combination. |
| $\det \mathbf{AB} = -1$ | B1 | |
| Area of $PQR = \|-1\|$ Area of $DEF$ | A1 | |
| **Total: 3** | | |

### Part (d):
| $(\mathbf{AB})^{-1} = -\begin{pmatrix} -\frac{1}{2}\sqrt{3} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\sqrt{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}$ | M1 A1 | or using $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$ |
| **Total: 2** | | |

### Part (e):
| $\begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{1}{2}x\sqrt{3}+\frac{1}{2}y \\ \frac{1}{2}x - \frac{1}{2}y\sqrt{3} \end{pmatrix}$ | B1 | Transforms $\begin{pmatrix} x \\ y \end{pmatrix}$ to $\begin{pmatrix} X \\ Y \end{pmatrix}$. Accept $t\begin{pmatrix} 1 \\ m \end{pmatrix}$ instead of $\begin{pmatrix} x \\ y \end{pmatrix}$. |
| $\frac{1}{2} - \frac{1}{2}m\sqrt{3} = \frac{1}{2}m\sqrt{3} + \frac{1}{2}m^2$ | M1 A1 | Uses $Y = mX$. Allow working with $y = mx + c$ as long as $c = 0$ is stated explicitly. |
| $1 - m\sqrt{3} = m\sqrt{3} + m^2 \Rightarrow m^2 + 2m\sqrt{3} - 1 = 0 \Rightarrow m = \pm 2 - \sqrt{3}$ | A1 | |
| $y = (2-\sqrt{3})x$ and $y + (2+\sqrt{3})x = 0$ | A1 | OE |
| **Total: 5** | | |

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4 The matrices $\mathbf { A }$ and $\mathbf { B }$ are given by

$$\mathbf { A } = \left( \begin{array} { l l } 
0 & 1 \\
1 & 0
\end{array} \right) \text { and } \mathbf { B } = \left( \begin{array} { c c } 
\frac { 1 } { 2 } & - \frac { 1 } { 2 } \sqrt { 3 } \\
\frac { 1 } { 2 } \sqrt { 3 } & \frac { 1 } { 2 }
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Give full details of the geometrical transformation in the $x - y$ plane represented by $\mathbf { A }$.
\item Give full details of the geometrical transformation in the $x - y$ plane represented by $\mathbf { B }$.\\

The triangle $D E F$ in the $x - y$ plane is transformed by $\mathbf { A B }$ onto triangle $P Q R$.
\item Show that the triangles $D E F$ and $P Q R$ have the same area.
\item Find the matrix which transforms triangle $P Q R$ onto triangle $D E F$.
\item Find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { A B }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q4 [13]}}

This paper (7 questions)

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Q1 8 Q2 5 Q3 7 Q4 13 Q5 14 Q6 13 Q7 15

Answer	Marks	Guidance
Reflection in the line \(y = x\).	B1	Only mention reflection (and no other transformation).
Total: 1

Answer	Marks	Guidance
Rotation	B1	Writes 'rotation' or 'rotate' (and no other transformation).
\(\frac{1}{3}\pi\) anticlockwise about the origin.	B1	Or \(60°\)
Total: 2

Answer	Marks	Guidance
\(\mathbf{AB} = \begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}\) or \(\det \mathbf{AB} = \det \mathbf{A} \det \mathbf{B}\)	M1	Finds \(\mathbf{AB}\) or uses product of determinants. Full marks may be obtained by arguing that both reflection and rotation preserve [absolute] value of area and so also does their combination.
\(\det \mathbf{AB} = -1\)	B1
Area of \(PQR = \	-1\	\) Area of \(DEF\)
Total: 3

Answer	Marks	Guidance
\((\mathbf{AB})^{-1} = -\begin{pmatrix} -\frac{1}{2}\sqrt{3} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\sqrt{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}\)	M1 A1	or using \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\)
Total: 2

Answer	Marks	Guidance
\(\begin{pmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{1}{2}x\sqrt{3}+\frac{1}{2}y \\ \frac{1}{2}x - \frac{1}{2}y\sqrt{3} \end{pmatrix}\)	B1	Transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} X \\ Y \end{pmatrix}\). Accept \(t\begin{pmatrix} 1 \\ m \end{pmatrix}\) instead of \(\begin{pmatrix} x \\ y \end{pmatrix}\).
\(\frac{1}{2} - \frac{1}{2}m\sqrt{3} = \frac{1}{2}m\sqrt{3} + \frac{1}{2}m^2\)	M1 A1	Uses \(Y = mX\). Allow working with \(y = mx + c\) as long as \(c = 0\) is stated explicitly.
\(1 - m\sqrt{3} = m\sqrt{3} + m^2 \Rightarrow m^2 + 2m\sqrt{3} - 1 = 0 \Rightarrow m = \pm 2 - \sqrt{3}\)	A1
\(y = (2-\sqrt{3})x\) and \(y + (2+\sqrt{3})x = 0\)	A1	OE
Total: 5