| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Find image coordinates under transformation |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths linear transformations question requiring standard techniques: matrix-vector multiplication (part a), solving a simple system for invariant points (part b), and finding the image of a line by transforming two points (part c). All parts are routine applications with no novel insight required, making it slightly easier than average even for Further Maths. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{bmatrix}3 & -1\\-2 & 6\end{bmatrix}\begin{bmatrix}4\\-5\end{bmatrix}=\begin{bmatrix}3\times4+(-1)\times(-5)\\-2\times4+6\times(-5)\end{bmatrix}=\begin{bmatrix}17\\-38\end{bmatrix}\) | M1 | Multiplies position vector of \(A\) by \(\mathbf{M}\) to achieve at least one correct calculation or coordinate |
| Image of \(A\) is \((17,-38)\) | A1 | Do not accept a position vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{bmatrix}3 & -1\\-2 & 6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}\) | M1 | Considers general point \((x,y)\); multiplies \(\mathbf{M}\begin{bmatrix}x\\y\end{bmatrix}\) to form at least one correct equation; or shows origin is invariant e.g. \(\mathbf{M}\begin{bmatrix}0\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\) |
| \(3x-y=x\) and \(-2x+6y=y\), giving \(y=2x\) and \(2x=5y\), so \(x=0\) and \(y=0\) | M1 | Forms two different correct equations in \(x\) and \(y\) |
| So \((0,0)\) is the only invariant point | R1 | Completes reasoned argument and concludes origin is only invariant point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{M}\begin{bmatrix}x\\x+1\end{bmatrix}=\begin{bmatrix}X\\mX+c\end{bmatrix}\) | M1 | Writes product \(\mathbf{M}\begin{bmatrix}x\\x+1\end{bmatrix}\) and equates to \(\begin{bmatrix}X\\mX+c\end{bmatrix}\); PI by one correct equation; or multiplies \(\mathbf{M}\) by a specific point on \(y=x+1\) |
| \(3x-1(x+1)=X\) and \(-2x+6(x+1)=mX+c\) | A1 | Obtains two correct equations; PI or obtains a correct specific point on \(y=2x+8\) |
| \(2x-1=X\) and \(4x+6=mX+c\), so \(4x+6\equiv m(2x-1)+c\) | M1 | Forms correct equation in either \(m\) or \(c\); FT their two equations |
| \(4=2m\) and \(6=-m+c\), so \(m=2\), \(c=8\) | M1 | Solves two equations in \(m\) and \(c\); accept one incorrect equation if clearly from correct method |
| The new line is \(y=2x+8\) | A1 |
## Question 14(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{bmatrix}3 & -1\\-2 & 6\end{bmatrix}\begin{bmatrix}4\\-5\end{bmatrix}=\begin{bmatrix}3\times4+(-1)\times(-5)\\-2\times4+6\times(-5)\end{bmatrix}=\begin{bmatrix}17\\-38\end{bmatrix}$ | M1 | Multiplies position vector of $A$ by $\mathbf{M}$ to achieve at least one correct calculation or coordinate |
| Image of $A$ is $(17,-38)$ | A1 | Do not accept a position vector |
---
## Question 14(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{bmatrix}3 & -1\\-2 & 6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$ | M1 | Considers general point $(x,y)$; multiplies $\mathbf{M}\begin{bmatrix}x\\y\end{bmatrix}$ to form at least one correct equation; or shows origin is invariant e.g. $\mathbf{M}\begin{bmatrix}0\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ |
| $3x-y=x$ and $-2x+6y=y$, giving $y=2x$ and $2x=5y$, so $x=0$ and $y=0$ | M1 | Forms two different correct equations in $x$ and $y$ |
| So $(0,0)$ is the only invariant point | R1 | Completes reasoned argument and concludes origin is only invariant point |
---
## Question 14(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{M}\begin{bmatrix}x\\x+1\end{bmatrix}=\begin{bmatrix}X\\mX+c\end{bmatrix}$ | M1 | Writes product $\mathbf{M}\begin{bmatrix}x\\x+1\end{bmatrix}$ and equates to $\begin{bmatrix}X\\mX+c\end{bmatrix}$; PI by one correct equation; or multiplies $\mathbf{M}$ by a specific point on $y=x+1$ |
| $3x-1(x+1)=X$ and $-2x+6(x+1)=mX+c$ | A1 | Obtains two correct equations; PI or obtains a correct specific point on $y=2x+8$ |
| $2x-1=X$ and $4x+6=mX+c$, so $4x+6\equiv m(2x-1)+c$ | M1 | Forms correct equation in either $m$ or $c$; FT their two equations |
| $4=2m$ and $6=-m+c$, so $m=2$, $c=8$ | M1 | Solves two equations in $m$ and $c$; accept one incorrect equation if clearly from correct method |
| The new line is $y=2x+8$ | A1 | |
---14 The matrix $\mathbf { M }$ represents the transformation T , and is given by
$$\mathbf { M } = \left[ \begin{array} { c c }
3 & - 1 \\
- 2 & 6
\end{array} \right]$$
14
\begin{enumerate}[label=(\alph*)]
\item The point $A$ has coordinates ( $4 , - 5$ )\\
Find the coordinates of the image of $A$ under T\\
14
\item Show that the only invariant point under T is the origin.\\
14
\item The line $L _ { 1 }$ has equation $y = x + 1$
The transformation $T$ maps the line $L _ { 1 }$ onto the line $L _ { 2 }$\\
Find the equation of $L _ { 2 }$ in the form $y = m x + c$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2024 Q14 [10]}}