CAIE Further Paper 1 2023 June — Question 4 14 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear transformations
TypeDecompose matrix into transformation sequence
DifficultyChallenging +1.2 This is a standard Further Maths question on matrix transformations requiring identification of transformations (rotation and shear), matrix inversion, singularity conditions, and invariant points. While it involves multiple parts and some algebraic manipulation (particularly parts c and d), each component uses well-practiced techniques without requiring novel insight. The topic is core Further Maths content, making it moderately above average difficulty but not exceptional.
Spec4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03e Successive transformations: matrix products4.03g Invariant points and lines4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix

4 The matrix \(\mathbf { M }\) is given by \(\mathbf { M } = \left( \begin{array} { r r } \cos 2 \theta & - \sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array} \right) \left( \begin{array} { l l } 1 & k \\ 0 & 1 \end{array} \right)\), where \(0 < \theta < \pi\) and \(k\) is a non-zero constant. The matrix \(\mathbf { M }\) represents a sequence of two geometrical transformations, one of which is a shear.
  1. Describe fully the other transformation and state the order in which the transformations are applied.
  2. Write \(\mathbf { M } ^ { - 1 }\) as the product of two matrices, neither of which is \(\mathbf { I }\).
  3. Find, in terms of \(k\), the value of \(\tan \theta\) for which \(\mathbf { M - I }\) is singular.
  4. Given that \(k = 2 \sqrt { 3 }\) and \(\theta = \frac { 1 } { 3 } \pi\), show that the invariant points of the transformation represented by \(\mathbf { M }\) lie on the line \(3 y + \sqrt { 3 } x = 0\).

Question 4(a):
AnswerMarks Guidance
AnswerMark Guidance
Rotation [anticlockwise]B1
about the origin through angle \(2\theta\)B1
Shear [in the \(x\)-direction] followed by a rotation [anticlockwise about the origin through angle \(2\theta\)]B1
Question 4(b):
AnswerMarks Guidance
AnswerMark Guidance
\(\begin{pmatrix}1 & k\\0 & 1\end{pmatrix}^{-1} = \begin{pmatrix}1 & -k\\0 & 1\end{pmatrix}\), \(\begin{pmatrix}\cos2\theta & -\sin2\theta\\\sin2\theta & \cos2\theta\end{pmatrix}^{-1} = \begin{pmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{pmatrix}\)B1
\(\mathbf{M}^{-1} = \begin{pmatrix}1 & -k\\0 & 1\end{pmatrix}\begin{pmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{pmatrix}\)B1 Correct order
Question 4(c):
AnswerMarks Guidance
AnswerMark Guidance
\(\mathbf{M}-\mathbf{I} = \begin{pmatrix}\cos2\theta-1 & k\cos2\theta-\sin2\theta\\\sin2\theta & k\sin2\theta+\cos2\theta-1\end{pmatrix}\)B1
\((\cos2\theta-1)(k\sin2\theta+\cos2\theta-1)-k\sin2\theta\cos2\theta+\sin^22\theta[=0]\)M1 Evaluates \(\det(\mathbf{M}-\mathbf{I})\)
\(2-2\cos2\theta - k\sin2\theta = 0\)A1 Brackets removed correctly and \(=0\)
\(4\sin^2\theta = 2k\sin\theta\cos\theta\)M1 Uses \(1-\cos2\theta = 2\sin^2\theta\) and \(\sin2\theta = 2\sin\theta\cos\theta\) or all necessary double angle formulae
\(\tan\theta = \frac{1}{2}k\)A1
Question 4(d):
AnswerMarks Guidance
AnswerMark Guidance
\(\mathbf{M} = \begin{pmatrix}-\frac{1}{2} & -\frac{3}{2}\sqrt{3}\\\frac{1}{2}\sqrt{3} & \frac{5}{2}\end{pmatrix}\)B1
\(\begin{pmatrix}-\frac{1}{2} & -\frac{3}{2}\sqrt{3}\\\frac{1}{2}\sqrt{3} & \frac{5}{2}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}x-\frac{3}{2}\sqrt{3}y\\\frac{1}{2}\sqrt{3}x+\frac{5}{2}y\end{pmatrix}\)B1FT Transforms \(\begin{pmatrix}x\\y\end{pmatrix}\) to \(\begin{pmatrix}X\\Y\end{pmatrix}\)
\(-\frac{1}{2}x - \frac{3}{2}\sqrt{3}y = x \Rightarrow -\frac{3}{2}x - \frac{3}{2}\sqrt{3}y = 0 \Rightarrow x+\sqrt{3}y=0\) and \(\frac{1}{2}\sqrt{3}x+\frac{5}{2}y = y \Rightarrow \frac{1}{2}\sqrt{3}x+\frac{3}{2}y=0 \Rightarrow \sqrt{3}x+3y=0\)M1 Sets \(\begin{pmatrix}X\\Y\end{pmatrix} = \begin{pmatrix}x\\y\end{pmatrix}\)
\(\sqrt{3}x+3y=0\)A1 AG
## Question 4(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Rotation [anticlockwise] | B1 | |
| about the origin through angle $2\theta$ | B1 | |
| Shear [in the $x$-direction] followed by a rotation [anticlockwise about the origin through angle $2\theta$] | B1 | |

## Question 4(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}1 & k\\0 & 1\end{pmatrix}^{-1} = \begin{pmatrix}1 & -k\\0 & 1\end{pmatrix}$, $\begin{pmatrix}\cos2\theta & -\sin2\theta\\\sin2\theta & \cos2\theta\end{pmatrix}^{-1} = \begin{pmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{pmatrix}$ | B1 | |
| $\mathbf{M}^{-1} = \begin{pmatrix}1 & -k\\0 & 1\end{pmatrix}\begin{pmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{pmatrix}$ | B1 | Correct order |

## Question 4(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{M}-\mathbf{I} = \begin{pmatrix}\cos2\theta-1 & k\cos2\theta-\sin2\theta\\\sin2\theta & k\sin2\theta+\cos2\theta-1\end{pmatrix}$ | B1 | |
| $(\cos2\theta-1)(k\sin2\theta+\cos2\theta-1)-k\sin2\theta\cos2\theta+\sin^22\theta[=0]$ | M1 | Evaluates $\det(\mathbf{M}-\mathbf{I})$ |
| $2-2\cos2\theta - k\sin2\theta = 0$ | A1 | Brackets removed correctly and $=0$ |
| $4\sin^2\theta = 2k\sin\theta\cos\theta$ | M1 | Uses $1-\cos2\theta = 2\sin^2\theta$ and $\sin2\theta = 2\sin\theta\cos\theta$ or all necessary double angle formulae |
| $\tan\theta = \frac{1}{2}k$ | A1 | |

## Question 4(d):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{M} = \begin{pmatrix}-\frac{1}{2} & -\frac{3}{2}\sqrt{3}\\\frac{1}{2}\sqrt{3} & \frac{5}{2}\end{pmatrix}$ | B1 | |
| $\begin{pmatrix}-\frac{1}{2} & -\frac{3}{2}\sqrt{3}\\\frac{1}{2}\sqrt{3} & \frac{5}{2}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}x-\frac{3}{2}\sqrt{3}y\\\frac{1}{2}\sqrt{3}x+\frac{5}{2}y\end{pmatrix}$ | B1FT | Transforms $\begin{pmatrix}x\\y\end{pmatrix}$ to $\begin{pmatrix}X\\Y\end{pmatrix}$ |
| $-\frac{1}{2}x - \frac{3}{2}\sqrt{3}y = x \Rightarrow -\frac{3}{2}x - \frac{3}{2}\sqrt{3}y = 0 \Rightarrow x+\sqrt{3}y=0$ and $\frac{1}{2}\sqrt{3}x+\frac{5}{2}y = y \Rightarrow \frac{1}{2}\sqrt{3}x+\frac{3}{2}y=0 \Rightarrow \sqrt{3}x+3y=0$ | M1 | Sets $\begin{pmatrix}X\\Y\end{pmatrix} = \begin{pmatrix}x\\y\end{pmatrix}$ |
| $\sqrt{3}x+3y=0$ | A1 | AG |
4 The matrix $\mathbf { M }$ is given by $\mathbf { M } = \left( \begin{array} { r r } \cos 2 \theta & - \sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array} \right) \left( \begin{array} { l l } 1 & k \\ 0 & 1 \end{array} \right)$, where $0 < \theta < \pi$ and $k$ is a non-zero constant. The matrix $\mathbf { M }$ represents a sequence of two geometrical transformations, one of which is a shear.
\begin{enumerate}[label=(\alph*)]
\item Describe fully the other transformation and state the order in which the transformations are applied.
\item Write $\mathbf { M } ^ { - 1 }$ as the product of two matrices, neither of which is $\mathbf { I }$.
\item Find, in terms of $k$, the value of $\tan \theta$ for which $\mathbf { M - I }$ is singular.
\item Given that $k = 2 \sqrt { 3 }$ and $\theta = \frac { 1 } { 3 } \pi$, show that the invariant points of the transformation represented by $\mathbf { M }$ lie on the line $3 y + \sqrt { 3 } x = 0$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q4 [14]}}