| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Area transformation under matrices |
| Difficulty | Standard +0.3 This question tests standard FP1 content: computing a 2×2 determinant (routine) and applying the area scale factor property (area scales by |det M|). Part (b) requires finding the original triangle's area using the vertex formula, then solving |det M| × area(T) = 216. While it involves multiple steps, each is straightforward application of known formulas with no novel insight required, making it slightly easier than average. |
| Spec | 4.03h Determinant 2x2: calculation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{\det M = (8)(2) - (-1)(-4)\} \Rightarrow \det M = 12\) | B1 | 12 |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(T = \frac{216}{12}\{= 18\}\) | M1 | Area \(T = \frac{216}{\text{their "det M"}}\) |
| \(h = \pm(1-k)\) | M1 | Uses \((k-1)\) or \((1-k)\) in their solution. |
| \(\frac{1}{2}8(k-1) = 18\) or \(\frac{1}{2}8(1-k) = 18\) or \((k-1) = \frac{18}{4}\) or \((1-k) = \frac{18}{4}\) | ddM1 | dependent on the two previous M marks \(\frac{1}{2}8(k-1)\) or \(\frac{1}{2}8(1-k) = \frac{216}{\text{their "det M"}}\) or \((k-1)\) or \((1-k) = \frac{216}{4(\text{their "det M"})}\) or \(h = \frac{216}{4(\text{their "det M"})}\), \(k = 1 \pm \frac{216}{4(\text{their "det M"})}\) |
| \(\frac{1}{2}8h = 18; \Rightarrow h = 2, k = 1 \pm \frac{9}{2}\) | ||
| \(\Rightarrow k = 5.5\) or \(k = -3.5\) | A1 | At least one of either \(k = 5.5\) or \(k = -3.5\) |
| A1 | Both \(k = 5.5\) and \(k = -3.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T' = \begin{pmatrix} 8 & -1 \\ -4 & 2 \end{pmatrix}\begin{pmatrix} 4 & 6 & 12 \\ 1 & k & 1 \end{pmatrix}\) | ||
| \(T' = \begin{pmatrix} 31 & 48-k & 95 \\ -14 & -24+2k & -46 \end{pmatrix}\) or 18 seen | M1 | At least 5 out of 6 elements are correct or 18 seen.. |
| \(\frac{1}{2}\begin{vmatrix} 31 & 48-k & 95 \\ -14 & -24+2k & -46 \end{vmatrix} = 216\) or \(\frac{1}{2}\begin{vmatrix} 4 & 6 & 12 \\ 1 & k & 1 \end{vmatrix} = 18\) | M1 | \(\frac{1}{2} |
| \(\frac{1}{2} | -744 + 62k + 672 - 14k - 2208 + 46k | = 216\) or \(\frac{1}{2} |
| \(\frac{1}{2} | 4k - 6 + 6 - 12k + 12 - 4 | = 18\) |
| \(\frac{1}{2} | 96 - 96k | = 216\) or \(\frac{1}{2} |
| So, \(1 - k = 4.5\) or \(k - 1 = 4.5\) | ||
| \(\Rightarrow k = -3.5\) or \(k = 5.5\) | A1 | At least one of either \(k = -3.5\) or \(k = 5.5\) |
| A1 | Both \(k = -3.5\) and \(k = 5.5\) |
**Part (a)**
| $\{\det M = (8)(2) - (-1)(-4)\} \Rightarrow \det M = 12$ | B1 | 12 |
**Part (b)**
| Area $T = \frac{216}{12}\{= 18\}$ | M1 | Area $T = \frac{216}{\text{their "det M"}}$ |
| $h = \pm(1-k)$ | M1 | Uses $(k-1)$ or $(1-k)$ in their solution. |
| $\frac{1}{2}8(k-1) = 18$ or $\frac{1}{2}8(1-k) = 18$ or $(k-1) = \frac{18}{4}$ or $(1-k) = \frac{18}{4}$ | ddM1 | dependent on the two previous M marks $\frac{1}{2}8(k-1)$ or $\frac{1}{2}8(1-k) = \frac{216}{\text{their "det M"}}$ or $(k-1)$ or $(1-k) = \frac{216}{4(\text{their "det M"})}$ or $h = \frac{216}{4(\text{their "det M"})}$, $k = 1 \pm \frac{216}{4(\text{their "det M"})}$ |
| $\frac{1}{2}8h = 18; \Rightarrow h = 2, k = 1 \pm \frac{9}{2}$ | | |
| $\Rightarrow k = 5.5$ or $k = -3.5$ | A1 | At least one of either $k = 5.5$ or $k = -3.5$ |
| | A1 | Both $k = 5.5$ and $k = -3.5$ |
**ALT (b)**
| $T' = \begin{pmatrix} 8 & -1 \\ -4 & 2 \end{pmatrix}\begin{pmatrix} 4 & 6 & 12 \\ 1 & k & 1 \end{pmatrix}$ | | |
| $T' = \begin{pmatrix} 31 & 48-k & 95 \\ -14 & -24+2k & -46 \end{pmatrix}$ or 18 seen | M1 | At least 5 out of 6 elements are correct or 18 seen.. |
| $\frac{1}{2}\begin{vmatrix} 31 & 48-k & 95 \\ -14 & -24+2k & -46 \end{vmatrix} = 216$ or $\frac{1}{2}\begin{vmatrix} 4 & 6 & 12 \\ 1 & k & 1 \end{vmatrix} = 18$ | M1 | $\frac{1}{2}|\text{their } T'| = 216$ or $\frac{1}{2}\begin{vmatrix} 4 & 6 & 12 \\ 1 & k & 1 \end{vmatrix} = 18$ |
| $\frac{1}{2}|-744 + 62k + 672 - 14k - 2208 + 46k| = 216$ or $\frac{1}{2}|2280 - 190k - 1330 + 1426| = 216$ | ddM1 | Dependent on the two previous M marks. Full method of evaluating a determinant. |
| $\frac{1}{2}|4k - 6 + 6 - 12k + 12 - 4| = 18$ | | |
| $\frac{1}{2}|96 - 96k| = 216$ or $\frac{1}{2}|8 - 8k| = 18$ | | |
| So, $1 - k = 4.5$ or $k - 1 = 4.5$ | | |
| $\Rightarrow k = -3.5$ or $k = 5.5$ | A1 | At least one of either $k = -3.5$ or $k = 5.5$ |
| | A1 | Both $k = -3.5$ and $k = 5.5$ |
**Total: 6 marks**
---6.
$$\mathbf { M } = \left( \begin{array} { r r }
8 & - 1 \\
- 4 & 2
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\operatorname { det } \mathbf { M }$
The triangle $T$ has vertices at the points $( 4,1 ) , ( 6 , k )$ and $( 12,1 )$, where $k$ is a constant.\\
The triangle $T$ is transformed onto the triangle $T ^ { \prime }$ by the transformation represented by the matrix $\mathbf { M }$.
Given that the area of triangle $T ^ { \prime }$ is 216 square units,
\item find the possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2018 Q6 [6]}}