| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Augmented matrices for translations |
| Difficulty | Standard +0.8 This is a Further Maths question requiring homogeneous coordinates to represent translations as matrices, composition of transformations, and finding fixed points. While the individual concepts are standard, combining translation with reflection using 3×3 matrices and finding the inverse through T² requires solid understanding of transformation geometry beyond typical A-level, placing it moderately above average difficulty. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Translation matrix: \(\begin{pmatrix}1 & 0 & -1\\0 & 1 & 1\\0 & 0 & 1\end{pmatrix}\) | B1 | |
| Answer: Reflection matrix: \(\begin{pmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{pmatrix}\) | B1 | |
| Answer: \(T = \begin{pmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & -1\\0 & 1 & 1\\0 & 0 & 1\end{pmatrix}\) | M1 | |
| Answer: \(T = \begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix}\) | A1 | Multiplying the wrong way gives \(T = \begin{pmatrix}0 & 1 & -1\\1 & 0 & 1\\0 & 0 & 1\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(T\begin{pmatrix}x\\y\\1\end{pmatrix} = \begin{pmatrix}x'\\y'\\1\end{pmatrix}\) FT their \(T\) from (a) | M1 | |
| Answer: Giving \(y+1 = x'\) and \(x - 1 = y'\). Therefore the line of fixed points is \(y = x - 1\). | A1 | Must be identical equations if FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(T^2 = \begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\) | M1 A1 | FT (a). At least 4 correct entries. |
| Answer: \(T^{-1} = \begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix}\) | B1 | Accept \(T^{-1} = T\) |
**Part a)**
Answer: Translation matrix: $\begin{pmatrix}1 & 0 & -1\\0 & 1 & 1\\0 & 0 & 1\end{pmatrix}$ | B1 |
Answer: Reflection matrix: $\begin{pmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{pmatrix}$ | B1 |
Answer: $T = \begin{pmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & -1\\0 & 1 & 1\\0 & 0 & 1\end{pmatrix}$ | M1 |
Answer: $T = \begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix}$ | A1 | Multiplying the wrong way gives $T = \begin{pmatrix}0 & 1 & -1\\1 & 0 & 1\\0 & 0 & 1\end{pmatrix}$
**Part b)**
Answer: $T\begin{pmatrix}x\\y\\1\end{pmatrix} = \begin{pmatrix}x'\\y'\\1\end{pmatrix}$ FT their $T$ from (a) | M1 |
Answer: Giving $y+1 = x'$ and $x - 1 = y'$. Therefore the line of fixed points is $y = x - 1$. | A1 | Must be identical equations if FT
**Part c)**
Answer: $T^2 = \begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$ | M1 A1 | FT (a). At least 4 correct entries.
Answer: $T^{-1} = \begin{pmatrix}0 & 1 & 1\\1 & 0 & -1\\0 & 0 & 1\end{pmatrix}$ | B1 | Accept $T^{-1} = T$The transformation $T$ in the plane consists of a translation in which the point $(x, y)$ is transformed to the point $(x - 1, y + 1)$, followed by a reflection in the line $y = x$.
\begin{enumerate}[label=(\alph*)]
\item Determine the $3 \times 3$ matrix which represents $T$. [4]
\item Find the equation of the line of fixed points of $T$. [2]
\item Find $T^2$ and hence write down $T^{-1}$. [3]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2018 Q8 [9]}}