| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Find image coordinates under transformation |
| Difficulty | Standard +0.3 This is a multi-part question on linear transformations requiring matrix multiplication and understanding of singular matrices. Parts (i)-(iv) involve routine application of transformation matrices to find images of lines and points. Part (v) requires calculating determinant and relating singularity to geometric properties. Part (vi) involves matrix composition. While comprehensive, each individual part uses standard techniques without requiring novel insight, making it slightly easier than average for Further Maths. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03l Singular/non-singular matrices4.03m det(AB) = det(A)*det(B) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{Q}\) represents a rotation | B1 | |
| 90 degrees clockwise about the origin | B1 | Angle, direction and centre |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\2\end{pmatrix} = \begin{pmatrix}-2\\2\end{pmatrix}\) | M1 | |
| \(P = (-2,\ 2)\) | A1 | Allow both marks for \(P(-2,2)\) www |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-y\\y\end{pmatrix}\) | M1 | Or use of a minimum of two points |
| \(l\) is the line \(y = -x\) | A1 | Allow both marks for \(y=-x\) www |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-y\\y\end{pmatrix} = \begin{pmatrix}-6\\6\end{pmatrix}\) | M1 | Use of a general point or two different points leading to \(\begin{pmatrix}-6\\6\end{pmatrix}\) |
| \(n\) is the line \(y = 6\) | B1 | \(y=6\); if seen alone M1B1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\det \mathbf{M} = 0 \Rightarrow \mathbf{M}\) is singular (or 'no inverse') | B1 | www |
| The transformation is many to one. | E1 | Accept area collapses to 0, or other equivalent statements |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{R} = \mathbf{QM} = \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix} = \begin{pmatrix}0 & 1\\0 & 1\end{pmatrix}\) | M1 | Attempt to multiply in correct order; or argue by rotation of the line \(y=-x\) |
| \(\begin{pmatrix}0 & 1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}y\\y\end{pmatrix}\) | ||
| \(q\) is the line \(y = x\) | A1 | \(y=x\) SC B1 following M0 |
| [2] |
# Question 9:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{Q}$ represents a rotation | B1 | |
| 90 degrees clockwise about the origin | B1 | Angle, direction and centre |
| **[2]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\2\end{pmatrix} = \begin{pmatrix}-2\\2\end{pmatrix}$ | M1 | |
| $P = (-2,\ 2)$ | A1 | Allow both marks for $P(-2,2)$ www |
| **[2]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-y\\y\end{pmatrix}$ | M1 | Or use of a minimum of two points |
| $l$ is the line $y = -x$ | A1 | Allow both marks for $y=-x$ www |
| **[2]** | | |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-y\\y\end{pmatrix} = \begin{pmatrix}-6\\6\end{pmatrix}$ | M1 | Use of a general point or two different points leading to $\begin{pmatrix}-6\\6\end{pmatrix}$ |
| $n$ is the line $y = 6$ | B1 | $y=6$; if seen alone M1B1 |
| **[2]** | | |
## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det \mathbf{M} = 0 \Rightarrow \mathbf{M}$ is singular (or 'no inverse') | B1 | www |
| The transformation is many to one. | E1 | Accept area collapses to 0, or other equivalent statements |
| **[2]** | | |
## Part (vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{R} = \mathbf{QM} = \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}\begin{pmatrix}0 & -1\\0 & 1\end{pmatrix} = \begin{pmatrix}0 & 1\\0 & 1\end{pmatrix}$ | M1 | Attempt to multiply in correct order; or argue by rotation of the line $y=-x$ |
| $\begin{pmatrix}0 & 1\\0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}y\\y\end{pmatrix}$ | | |
| $q$ is the line $y = x$ | A1 | $y=x$ SC B1 following M0 |
| **[2]** | | |9 (i) Describe fully the transformation Q , represented by the matrix $\mathbf { Q }$, where $\mathbf { Q } = \left( \begin{array} { r l } 0 & 1 \\ - 1 & 0 \end{array} \right)$.
The transformation M is represented by the matrix $\mathbf { M }$, where $\mathbf { M } = \left( \begin{array} { r r } 0 & - 1 \\ 0 & 1 \end{array} \right)$.\\
(ii) M maps all points on the line $y = 2$ onto a single point, P. Find the coordinates of P.\\
(iii) M maps all points on the plane onto a single line, $l$. Find the equation of $l$.\\
(iv) M maps all points on the line $n$ onto the point ( - 6 , 6). Find the equation of $n$.\\
(v) Show that $\mathbf { M }$ is singular. Relate this to the transformation it represents.\\
(vi) R is the composite transformation M followed by Q . R maps all points on the plane onto the line $q$. Find the equation of $q$.
\hfill \mbox{\textit{OCR MEI FP1 2013 Q9 [12]}}