| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Find image coordinates under transformation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on matrix transformations requiring routine application of matrix multiplication to find image coordinates, recognition of a standard scaling transformation, finding a composite inverse transformation, and using the determinant to find area scaling. All parts are standard FP1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 4 \\ 0 & 2 & 1 \end{pmatrix}\] | M1 | Any valid method – may be implied |
| \[= \begin{pmatrix} 0 & -4 & 2 \\ 0 & 0 & 12 \end{pmatrix}\] | A1 | Correct position vectors found (need not be identified) |
| \(A' = (0, 0)\), \(B' = (-4, 0)\), \(C' = (2, 12)\) | A1ft | co-ordinates, ft their position vectors. \(A'\), \(B'\), \(C'\) identifiable. Coordinates only, M1A0A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{M}\) represents a two-way stretch, factor 4 parallel to the \(x\) axis, factor 2 parallel to the \(y\) axis | B1 | Stretch. (enlargement B0) |
| Directions indicated | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix}\] | M1 | Attempt at \(\mathbf{MT}\) in correct sequence |
| \[= \begin{pmatrix} 4 & -8 \\ 6 & 0 \end{pmatrix}\] | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 4 & -8 \\ 6 & 0 \end{pmatrix}^{-1} = \frac{1}{48}\begin{pmatrix} 0 & 8 \\ -6 & 4 \end{pmatrix}\] represents the single transformation | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{1}{6}\begin{pmatrix} 0 & 2 \\ -3 & 1 \end{pmatrix} \cdot \frac{1}{8}\begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix} = \frac{1}{48}\begin{pmatrix} 0 & 8 \\ -6 & 4 \end{pmatrix}\] | B1, M1, A1 | for \(\mathbf{T}^{-1}\) and \(\mathbf{M}^{-1}\) correct; for attempt at \(\mathbf{T}^{-1}\mathbf{M}^{-1}\); cao |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 0 & -16 & 8 \\ 0 & 0 & 24 \end{pmatrix}\] whence \(\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} 0 & -16 & 8 \\ 0 & 0 & 24 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 4 \\ 0 & 2 & 1 \end{pmatrix}\) | M1 | Finding \(A''\), \(B''\) and \(C''\) coordinates or position vectors |
| \[\Rightarrow \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \frac{1}{48}\begin{pmatrix} 0 & 8 \\ -6 & 4 \end{pmatrix}\] | A1, A1 | For correct position vectors; Inverse matrix correctly found |
| Answer | Marks |
|---|---|
| Area scale factor \(= 48\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Area of triangle \(A''B''C'' = 48 \times\) area of triangle \(ABC = 192\) (square units) | M1 | Using their "48" and their area of triangle \(ABC\), correct triangle. Or other valid method. |
| A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Finding \(A''B''C''\): \((0,0)\), \((-16, 0)\), \((8, 24)\) and using them | B1 | \(A''B''C''\) may be in (iii) |
| Finding the area of \(A''B''C''\) | M1 | Any valid method attempted |
| Area of triangle \(= 192\) (square units) | A1 | cao (possibly after rounding to 3 sf) |
# Question 9:
## Part (i):
$$\begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 4 \\ 0 & 2 & 1 \end{pmatrix}$$ | M1 | Any valid method – may be implied
$$= \begin{pmatrix} 0 & -4 & 2 \\ 0 & 0 & 12 \end{pmatrix}$$ | A1 | Correct position vectors found (need not be identified)
$A' = (0, 0)$, $B' = (-4, 0)$, $C' = (2, 12)$ | A1ft | co-ordinates, ft their position vectors. $A'$, $B'$, $C'$ identifiable. Coordinates only, M1A0A1
**[3]**
## Part (ii):
$\mathbf{M}$ represents a two-way stretch, factor 4 parallel to the $x$ axis, factor 2 parallel to the $y$ axis | B1 | Stretch. (enlargement B0)
Directions indicated | B1 B1
**[3]**
## Part (iii):
$$\begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix}$$ | M1 | Attempt at $\mathbf{MT}$ in correct sequence
$$= \begin{pmatrix} 4 & -8 \\ 6 & 0 \end{pmatrix}$$ | A1 | cao
Represents the composite transformation T followed by M
$$\begin{pmatrix} 4 & -8 \\ 6 & 0 \end{pmatrix}^{-1} = \frac{1}{48}\begin{pmatrix} 0 & 8 \\ -6 & 4 \end{pmatrix}$$ represents the single transformation | A1 | cao
**[3]**
**OR:**
$$\frac{1}{6}\begin{pmatrix} 0 & 2 \\ -3 & 1 \end{pmatrix} \cdot \frac{1}{8}\begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix} = \frac{1}{48}\begin{pmatrix} 0 & 8 \\ -6 & 4 \end{pmatrix}$$ | B1, M1, A1 | for $\mathbf{T}^{-1}$ and $\mathbf{M}^{-1}$ correct; for attempt at $\mathbf{T}^{-1}\mathbf{M}^{-1}$; cao
**[3]**
**OR:**
$$\begin{pmatrix} 0 & -16 & 8 \\ 0 & 0 & 24 \end{pmatrix}$$ whence $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} 0 & -16 & 8 \\ 0 & 0 & 24 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 4 \\ 0 & 2 & 1 \end{pmatrix}$ | M1 | Finding $A''$, $B''$ and $C''$ coordinates or position vectors
$$\Rightarrow \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \frac{1}{48}\begin{pmatrix} 0 & 8 \\ -6 & 4 \end{pmatrix}$$ | A1, A1 | For correct position vectors; Inverse matrix correctly found
**[3]**
## Part (iv):
Area scale factor $= 48$ | B1 |
Area of triangle $ABC = 4$ square units
Area of triangle $A''B''C'' = 48 \times$ area of triangle $ABC = 192$ (square units) | M1 | Using their "48" and their area of triangle $ABC$, correct triangle. Or other valid method.
| A1 | cao
**[3]**
**OR:**
Finding $A''B''C''$: $(0,0)$, $(-16, 0)$, $(8, 24)$ and using them | B1 | $A''B''C''$ may be in (iii)
Finding the area of $A''B''C''$ | M1 | Any valid method attempted
Area of triangle $= 192$ (square units) | A1 | cao (possibly after rounding to 3 sf)9 The triangle ABC has vertices at $\mathrm { A } ( 0,0 ) , \mathrm { B } ( 0,2 )$ and $\mathrm { C } ( 4,1 )$. The matrix $\left( \begin{array} { r r } 1 & - 2 \\ 3 & 0 \end{array} \right)$ represents a transformation T .\\
(i) The transformation $T$ maps triangle $A B C$ onto triangle $A ^ { \prime } B ^ { \prime } C ^ { \prime }$. Find the coordinates of $A ^ { \prime } , B ^ { \prime }$ and $C ^ { \prime }$.
Triangle $A ^ { \prime } B ^ { \prime } C ^ { \prime }$ is now mapped onto triangle $A ^ { \prime \prime } B ^ { \prime \prime } C ^ { \prime \prime }$ using the matrix $\mathbf { M } = \left( \begin{array} { l l } 4 & 0 \\ 0 & 2 \end{array} \right)$.\\
(ii) Describe fully the transformation represented by $\mathbf { M }$.\\
(iii) Triangle $\mathrm { A } ^ { \prime \prime } \mathrm { B } ^ { \prime \prime } \mathrm { C } ^ { \prime \prime }$ is now mapped back onto ABC by a single transformation. Find the matrix representing this transformation.\\
(iv) Calculate the area of $A ^ { \prime \prime } B ^ { \prime \prime } C ^ { \prime \prime }$.
\hfill \mbox{\textit{OCR MEI FP1 2015 Q9 [12]}}