CAIE Further Paper 1 2021 November — Question 1 9 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2021
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear transformations
TypeFind invariant points
DifficultyModerate -0.3 This is a multi-part question covering standard linear transformation concepts (geometric interpretation, area scaling by determinant, matrix equation solving, and invariant points). All parts use routine techniques: (a) is immediate recognition, (b) requires det(A)=-2, (c) requires finding A^{-1}, and (d) solves (A-I)v=0. While it's Further Maths content, these are foundational exercises requiring no novel insight, making it slightly easier than an average A-level question overall.
Spec4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix

1
  1. Give full details of the geometrical transformation in the \(x - y\) plane represented by the matrix \(\left( \begin{array} { l l } 6 & 0 \\ 0 & 6 \end{array} \right)\). Let \(\mathbf { A } = \left( \begin{array} { l l } 3 & 4 \\ 2 & 2 \end{array} \right)\).
  2. The triangle \(D E F\) in the \(x - y\) plane is transformed by \(\mathbf { A }\) onto triangle \(P Q R\). Given that the area of triangle \(D E F\) is \(13 \mathrm {~cm} ^ { 2 }\), find the area of triangle \(P Q R\).
  3. Find the matrix \(\mathbf { B }\) such that \(\mathbf { A B } = \left( \begin{array} { l l } 6 & 0 \\ 0 & 6 \end{array} \right)\).
  4. Show that the origin is the only invariant point of the transformation in the \(x - y\) plane represented by \(\mathbf { A }\).

Question 1:
Part (a):
AnswerMarks Guidance
AnswerMarks Guidance
Enlargement, scale factor 6B1
Total: 1
Part (b):
AnswerMarks Guidance
AnswerMarks Guidance
\(\det \mathbf{A} = 6 - 8 = -2\)M1 Finds \(\det \mathbf{A}\)
Area \(= 2 \times 13 = 26 \text{ cm}^2\)A1
Total: 2
Part (c):
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{A}^{-1} = -\frac{1}{2}\begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix}\)M1 Finds \(\mathbf{A}^{-1}\)
\(\mathbf{B} = -3\begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} -6 & 12 \\ 6 & -9 \end{pmatrix}\)A1 AEF; could be solved by equations
Total: 2
Part (d):
AnswerMarks Guidance
AnswerMarks Guidance
\(\begin{pmatrix} 3 & 4 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x+4y \\ 2x+2y \end{pmatrix}\)B1 Finds \(\begin{pmatrix} X \\ Y \end{pmatrix}\)
\(3x+4y = x\) leading to \(\begin{cases} 2x+4y=0 \\ 2x+y=0 \end{cases}\)M1 Uses \(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} X \\ Y \end{pmatrix}\) to form simultaneous equations
\(y = -2x\) leading to \(-6x = 0\) leading to \(x = 0,\ y = 0\)M1 A1 Solves equations or states that \(\det\begin{pmatrix} 2 & 4 \\ 2 & 1 \end{pmatrix} \neq 0\), AG
Total: 4
## Question 1:

### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Enlargement, scale factor 6 | **B1** | |
| **Total: 1** | | |

### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det \mathbf{A} = 6 - 8 = -2$ | **M1** | Finds $\det \mathbf{A}$ |
| Area $= 2 \times 13 = 26 \text{ cm}^2$ | **A1** | |
| **Total: 2** | | |

### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}^{-1} = -\frac{1}{2}\begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix}$ | **M1** | Finds $\mathbf{A}^{-1}$ |
| $\mathbf{B} = -3\begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} -6 & 12 \\ 6 & -9 \end{pmatrix}$ | **A1** | AEF; could be solved by equations |
| **Total: 2** | | |

### Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 3 & 4 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x+4y \\ 2x+2y \end{pmatrix}$ | **B1** | Finds $\begin{pmatrix} X \\ Y \end{pmatrix}$ |
| $3x+4y = x$ leading to $\begin{cases} 2x+4y=0 \\ 2x+y=0 \end{cases}$ | **M1** | Uses $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} X \\ Y \end{pmatrix}$ to form simultaneous equations |
| $y = -2x$ leading to $-6x = 0$ leading to $x = 0,\ y = 0$ | **M1 A1** | Solves equations or states that $\det\begin{pmatrix} 2 & 4 \\ 2 & 1 \end{pmatrix} \neq 0$, AG |
| **Total: 4** | | |
1
\begin{enumerate}[label=(\alph*)]
\item Give full details of the geometrical transformation in the $x - y$ plane represented by the matrix $\left( \begin{array} { l l } 6 & 0 \\ 0 & 6 \end{array} \right)$.

Let $\mathbf { A } = \left( \begin{array} { l l } 3 & 4 \\ 2 & 2 \end{array} \right)$.
\item The triangle $D E F$ in the $x - y$ plane is transformed by $\mathbf { A }$ onto triangle $P Q R$.

Given that the area of triangle $D E F$ is $13 \mathrm {~cm} ^ { 2 }$, find the area of triangle $P Q R$.
\item Find the matrix $\mathbf { B }$ such that $\mathbf { A B } = \left( \begin{array} { l l } 6 & 0 \\ 0 & 6 \end{array} \right)$.
\item Show that the origin is the only invariant point of the transformation in the $x - y$ plane represented by $\mathbf { A }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q1 [9]}}