| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| 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 rotation and stretch matrices, finding invariant lines (eigenvalue method), and using determinants for area scaling. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average for Further Maths content. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03e Successive transformations: matrix products4.03g Invariant points and lines4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| [One way] stretch, rotation | B1 | |
| Stretch followed by rotation | B1 | Correct order |
| Stretch parallel to the \(x\)-axis, scale factor 14 | B1 | |
| Rotation, \(\frac{1}{3}\pi\) [anticlockwise] about the origin | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}\frac{1}{2} & -\frac{1}{2}\sqrt{3}\\ \frac{1}{2}\sqrt{3} & \frac{1}{2}\end{pmatrix}^{-1} = \begin{pmatrix}\frac{1}{2} & \frac{1}{2}\sqrt{3}\\ -\frac{1}{2}\sqrt{3} & \frac{1}{2}\end{pmatrix}\), \(\begin{pmatrix}14 & 0\\ 0 & 1\end{pmatrix}^{-1} = \frac{1}{14}\begin{pmatrix}1 & 0\\ 0 & 14\end{pmatrix}\) | B1 | Both inverses correct |
| \(\mathbf{M}^{-1} = \begin{pmatrix}\frac{1}{14} & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}\frac{1}{2} & \frac{1}{2}\sqrt{3}\\ -\frac{1}{2}\sqrt{3} & \frac{1}{2}\end{pmatrix}\) | B1 | Correct order |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{M} = \begin{pmatrix}7 & -\frac{1}{2}\sqrt{3}\\ 7\sqrt{3} & \frac{1}{2}\end{pmatrix}\) | B1 | |
| \(\begin{pmatrix}7 & -\frac{1}{2}\sqrt{3}\\ 7\sqrt{3} & \frac{1}{2}\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}7x - \frac{1}{2}\sqrt{3}y\\ 7\sqrt{3}x + \frac{1}{2}y\end{pmatrix}\) | B1 | Transforms \(\begin{pmatrix}x\\y\end{pmatrix}\) to \(\begin{pmatrix}X\\Y\end{pmatrix}\) |
| \(7\sqrt{3}x + \frac{1}{2}mx = m\left(7x - \frac{1}{2}\sqrt{3}mx\right)\) | M1 | Uses \(y = mx\) and \(Y = mX\) |
| \(7\sqrt{3} + \frac{1}{2}m = 7m - \frac{1}{2}\sqrt{3}m^2 \Rightarrow \frac{1}{2}\sqrt{3}m^2 - \frac{13}{2}m + 7\sqrt{3} = 0\) | A1 | |
| \(y = 2\sqrt{3}x\) and \(y = \frac{7}{3}\sqrt{3}x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(28 = 14 \times | ABC | \) |
| \(2 \text{ cm}^2\) | A1 | Allow with units missing |
## Question 4(a):
| [One way] stretch, rotation | B1 | |
| Stretch followed by rotation | B1 | Correct order |
| Stretch parallel to the $x$-axis, scale factor 14 | B1 | |
| Rotation, $\frac{1}{3}\pi$ [anticlockwise] about the origin | B1 | |
---
## Question 4(b):
| $\begin{pmatrix}\frac{1}{2} & -\frac{1}{2}\sqrt{3}\\ \frac{1}{2}\sqrt{3} & \frac{1}{2}\end{pmatrix}^{-1} = \begin{pmatrix}\frac{1}{2} & \frac{1}{2}\sqrt{3}\\ -\frac{1}{2}\sqrt{3} & \frac{1}{2}\end{pmatrix}$, $\begin{pmatrix}14 & 0\\ 0 & 1\end{pmatrix}^{-1} = \frac{1}{14}\begin{pmatrix}1 & 0\\ 0 & 14\end{pmatrix}$ | B1 | Both inverses correct |
| $\mathbf{M}^{-1} = \begin{pmatrix}\frac{1}{14} & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}\frac{1}{2} & \frac{1}{2}\sqrt{3}\\ -\frac{1}{2}\sqrt{3} & \frac{1}{2}\end{pmatrix}$ | B1 | Correct order |
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## Question 4(c):
| $\mathbf{M} = \begin{pmatrix}7 & -\frac{1}{2}\sqrt{3}\\ 7\sqrt{3} & \frac{1}{2}\end{pmatrix}$ | B1 | |
| $\begin{pmatrix}7 & -\frac{1}{2}\sqrt{3}\\ 7\sqrt{3} & \frac{1}{2}\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}7x - \frac{1}{2}\sqrt{3}y\\ 7\sqrt{3}x + \frac{1}{2}y\end{pmatrix}$ | B1 | Transforms $\begin{pmatrix}x\\y\end{pmatrix}$ to $\begin{pmatrix}X\\Y\end{pmatrix}$ |
| $7\sqrt{3}x + \frac{1}{2}mx = m\left(7x - \frac{1}{2}\sqrt{3}mx\right)$ | M1 | Uses $y = mx$ and $Y = mX$ |
| $7\sqrt{3} + \frac{1}{2}m = 7m - \frac{1}{2}\sqrt{3}m^2 \Rightarrow \frac{1}{2}\sqrt{3}m^2 - \frac{13}{2}m + 7\sqrt{3} = 0$ | A1 | |
| $y = 2\sqrt{3}x$ and $y = \frac{7}{3}\sqrt{3}x$ | A1 | |
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## Question 4(d):
| $28 = 14 \times |ABC|$ | M1 | Uses $|DEF| = |\det\mathbf{M}||ABC|$ |
| $2 \text{ cm}^2$ | A1 | Allow with units missing |
---4 The matrix $\mathbf { M }$ is given by $\mathbf { M } = \left( \begin{array} { c c } \frac { 1 } { 2 } & - \frac { 1 } { 2 } \sqrt { 3 } \\ \frac { 1 } { 2 } \sqrt { 3 } & \frac { 1 } { 2 } \end{array} \right) \left( \begin{array} { c c } 14 & 0 \\ 0 & 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item The matrix $\mathbf { M }$ represents a sequence of two geometrical transformations in the $x - y$ plane.
Give full details of each transformation, and make clear the order in which they are applied.
\item Write $\mathbf { M } ^ { - 1 }$ as the product of two matrices, neither of which is $\mathbf { I }$.
\item Find the equations of the invariant lines, through the origin, of the transformation represented by $\mathbf { M }$.
\item The triangle $A B C$ in the $x - y$ plane is transformed by $\mathbf { M }$ onto triangle $D E F$.
Given that the area of triangle $D E F$ is $28 \mathrm {~cm} ^ { 2 }$, find the area of triangle $A B C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q4 [13]}}