| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Area transformation under matrices |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard matrix operations (determinant, inverse) and the area scale factor property. Part (a)-(c) are routine calculations, while (d) requires applying the inverse matrix to three points—mechanical but slightly more work. The concepts are fundamental FP1 material with no novel problem-solving required. |
| Spec | 4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det \mathbf{A} = 2(3) - (-1)(-2) = 6 - 2 = 4\) | B1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{A}^{-1} = \frac{1}{4}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}\) | M1 | \(\frac{1}{\det\mathbf{A}}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}\) |
| A1 | \(\frac{1}{4}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Area}(R) = \frac{72}{4} = 18 \text{ (units)}^2\) | M1 | \(\frac{72}{\text{their det}\,\mathbf{A}}\) or \(72(\text{their det}\,\mathbf{A})\) |
| A1\(\sqrt{}\) | 18 or ft answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{AR} = \mathbf{S} \Rightarrow \mathbf{A}^{-1}\mathbf{AR} = \mathbf{A}^{-1}\mathbf{S} \Rightarrow \mathbf{R} = \mathbf{A}^{-1}\mathbf{S}\) | ||
| \(\mathbf{R} = \frac{1}{4}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}\begin{pmatrix}0 & 8 & 12\\4 & 16 & 4\end{pmatrix}\) | M1 | At least one attempt to apply \(\mathbf{A}^{-1}\) by any of the three vertices in \(\mathbf{S}\) |
| \(= \frac{1}{4}\begin{pmatrix}8 & 56 & 44\\8 & 40 & 20\end{pmatrix}\) | ||
| \(= \begin{pmatrix}2 & 14 & 11\\2 & 10 & 5\end{pmatrix}\) | A1\(\sqrt{}\) | At least one correct column |
| A1 | At least two correct columns | |
| Vertices are \((2,2)\), \((14,10)\) and \((11,5)\) | A1 | All three coordinates correct |
## Question 8:
### Part (a):
| $\det \mathbf{A} = 2(3) - (-1)(-2) = 6 - 2 = 4$ | B1 | 4 |
|---|---|---|
(1 mark)
### Part (b):
| $\mathbf{A}^{-1} = \frac{1}{4}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}$ | M1 | $\frac{1}{\det\mathbf{A}}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}$ |
|---|---|---|
| | A1 | $\frac{1}{4}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}$ |
(2 marks)
### Part (c):
| $\text{Area}(R) = \frac{72}{4} = 18 \text{ (units)}^2$ | M1 | $\frac{72}{\text{their det}\,\mathbf{A}}$ or $72(\text{their det}\,\mathbf{A})$ |
|---|---|---|
| | A1$\sqrt{}$ | 18 or ft answer |
(2 marks)
### Part (d):
| $\mathbf{AR} = \mathbf{S} \Rightarrow \mathbf{A}^{-1}\mathbf{AR} = \mathbf{A}^{-1}\mathbf{S} \Rightarrow \mathbf{R} = \mathbf{A}^{-1}\mathbf{S}$ | | |
|---|---|---|
| $\mathbf{R} = \frac{1}{4}\begin{pmatrix}3 & 2\\1 & 2\end{pmatrix}\begin{pmatrix}0 & 8 & 12\\4 & 16 & 4\end{pmatrix}$ | M1 | At least one attempt to apply $\mathbf{A}^{-1}$ by any of the three vertices in $\mathbf{S}$ |
| $= \frac{1}{4}\begin{pmatrix}8 & 56 & 44\\8 & 40 & 20\end{pmatrix}$ | | |
| $= \begin{pmatrix}2 & 14 & 11\\2 & 10 & 5\end{pmatrix}$ | A1$\sqrt{}$ | At least one correct column |
| | A1 | At least two correct columns |
| Vertices are $(2,2)$, $(14,10)$ and $(11,5)$ | A1 | All three coordinates correct |
(4 marks)8.
$$\mathbf { A } = \left( \begin{array} { r r }
2 & - 2 \\
- 1 & 3
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find $\operatorname { det } \mathbf { A }$.
\item Find $\mathbf { A } ^ { - 1 }$.
The triangle $R$ is transformed to the triangle $S$ by the matrix $\mathbf { A }$. Given that the area of triangle $S$ is 72 square units,
\item find the area of triangle $R$.
The triangle $S$ has vertices at the points $( 0,4 ) , ( 8,16 )$ and $( 12,4 )$.
\item Find the coordinates of the vertices of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q8 [9]}}