| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | December |
| Marks | 8 |
| Topic | Linear transformations |
| Type | Write down transformation matrix |
| Difficulty | Moderate -0.8 This is a straightforward Further Pure question testing basic matrix transformations and inverse properties. Parts (a)-(d) require simple recall of standard transformations and matrix multiplication, while part (e) is a routine verification of the reverse order law for matrix inverses. Despite being Further Maths content, the question involves no problem-solving or novel insight—just mechanical application of well-drilled techniques. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03n Inverse 2x2 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(A = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\) | B1 | |
| (b) Stretch scale factor 1/3 parallel to x-axis | M1 | Must be complete description (except no need to specify 2-D) |
| \(A^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}\) | A1 | |
| (c) Reflection in the line \(y = -x\) | B1 | |
| (d) \(BA = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = \ldots\) | M1 | For understanding that the matrix representing successive transformations is the product in the correct order. ie BA, not AB |
| \(\ldots = \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix}\) | A1 | |
| (e) \((BA)^{-1} = \frac{1}{3}\begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix}\) | M1 | For carrying out the procedure for inverting the matrix found in (d) (or BA worked out from scratch) |
| \(A^{-1}B^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -\frac{1}{3} \\ -1 & 0 \end{pmatrix}\) | A1 | |
| \(= \frac{1}{3}\begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix} = (BA)^{-1}\) AG | A1 | Calculation and statement of equality. Needs to be in the same form, not just an assertion |
**(a)** $A = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$ | B1 |
**(b)** Stretch scale factor 1/3 parallel to x-axis | M1 | Must be complete description (except no need to specify 2-D)
$A^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}$ | A1 |
**(c)** Reflection in the line $y = -x$ | B1 |
**(d)** $BA = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = \ldots$ | M1 | For understanding that the matrix representing successive transformations is the product in the correct order. ie BA, not AB
$\ldots = \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix}$ | A1 |
**(e)** $(BA)^{-1} = \frac{1}{3}\begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix}$ | M1 | For carrying out the procedure for inverting the matrix found in (d) (or BA worked out from scratch)
$A^{-1}B^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -\frac{1}{3} \\ -1 & 0 \end{pmatrix}$ | A1 |
$= \frac{1}{3}\begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix} = (BA)^{-1}$ AG | A1 | Calculation and statement of equality. Needs to be in the same form, not just an assertion
---\textbf{In this question you must show detailed reasoning.}
S is the 2-D transformation which is a stretch of scale factor 3 parallel to the $x$-axis. A is the matrix which represents S.
\begin{enumerate}[label=(\alph*)]
\item Write down A. [1]
\item By considering the transformation represented by $\mathbf{A}^{-1}$, determine the matrix $\mathbf{A}^{-1}$. [2]
\end{enumerate}
Matrix B is given by $\mathbf{B} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$. T is the transformation represented by B.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Describe T. [1]
\item Determine the matrix which represents the transformation S followed by T. [2]
\item Demonstrate, by direct calculation, that $(\mathbf{BA})^{-1} = \mathbf{A}^{-1}\mathbf{B}^{-1}$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q2 [8]}}