| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Matrix satisfying given equation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question involving matrix multiplication, solving simultaneous equations from matrix equality, and applying the determinant-area relationship. All techniques are standard FP1 material with no novel insight required, making it slightly easier than average even for Further Maths. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03i Determinant: area scale factor and orientation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{AB\} = \begin{pmatrix}-5p+12 & 4p-10\\-15+6p & 12-5p\end{pmatrix}\) | M1 | At least 2 elements correct |
| A1 | Correct matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB + 2A = kI\) gives \(\begin{pmatrix}-3p+12 & 4p-6\\-9+6p & 12-3p\end{pmatrix} = \begin{pmatrix}k & 0\\0 & k\end{pmatrix}\) | M1 | Forms an equation in \(p\) |
| \(p = \frac{3}{2}\) | A1 | |
| Substitutes \(p = \frac{3}{2}\) into \((-5p+12)+2p\) to find \(k\) | M1 | |
| \(k = \frac{15}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm\frac{270}{15}\ \{= \pm 18\}\) | B1 | Can be implied from calculations |
| \(\det M = (a)(2) - (-9)(1)\) | M1 | Applies \(ad - bc\) to \(M\). Require clear evidence of correct formula |
| \(2a + 9 = 18\) or \(2a + 9 = -18\) | M1 | Equates their \(\det A\) to either \(18\) or \(-18\) |
| \(a = 4.5\) or \(a = -13.5\) | A1 | At least one correct |
| Both \(a = 4.5\) and \(a = -13.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Consider vertices of triangle with area 15 units e.g. \((0,0), (15,0)\) and \((0,2)\) | B1 | |
| Pre-multiplies matrix by \(M\) and obtains single matrix | M1 | |
| Equates their determinant to \(270\) and attempts to solve | M1 | |
| \(a = 4.5\) or \(a = -13.5\) | A1 | At least one correct |
| Both \(a = 4.5\) and \(a = -13.5\) | A1 |
# Question 5:
## Part (i)(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{AB\} = \begin{pmatrix}-5p+12 & 4p-10\\-15+6p & 12-5p\end{pmatrix}$ | M1 | At least 2 elements correct |
| | A1 | Correct matrix |
## Part (i)(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB + 2A = kI$ gives $\begin{pmatrix}-3p+12 & 4p-6\\-9+6p & 12-3p\end{pmatrix} = \begin{pmatrix}k & 0\\0 & k\end{pmatrix}$ | M1 | Forms an equation in $p$ |
| $p = \frac{3}{2}$ | A1 | |
| Substitutes $p = \frac{3}{2}$ into $(-5p+12)+2p$ to find $k$ | M1 | |
| $k = \frac{15}{2}$ | A1 | |
## Part (ii) Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\frac{270}{15}\ \{= \pm 18\}$ | B1 | Can be implied from calculations |
| $\det M = (a)(2) - (-9)(1)$ | M1 | Applies $ad - bc$ to $M$. Require clear evidence of correct formula |
| $2a + 9 = 18$ or $2a + 9 = -18$ | M1 | Equates their $\det A$ to either $18$ or $-18$ |
| $a = 4.5$ or $a = -13.5$ | A1 | At least one correct |
| Both $a = 4.5$ and $a = -13.5$ | A1 | |
## Part (ii) Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| Consider vertices of triangle with area 15 units e.g. $(0,0), (15,0)$ and $(0,2)$ | B1 | |
| Pre-multiplies matrix by $M$ and obtains single matrix | M1 | |
| Equates their determinant to $270$ and attempts to solve | M1 | |
| $a = 4.5$ or $a = -13.5$ | A1 | At least one correct |
| Both $a = 4.5$ and $a = -13.5$ | A1 | |5. (i)
$$\mathbf { A } = \left( \begin{array} { l l }
p & 2 \\
3 & p
\end{array} \right) , \quad \mathbf { B } = \left( \begin{array} { r r }
- 5 & 4 \\
6 & - 5
\end{array} \right)$$
where $p$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $p$, the matrix $\mathbf { A B }$
Given that
$$\mathbf { A B } + 2 \mathbf { A } = k \mathbf { I }$$
where $k$ is a constant and $\mathbf { I }$ is the $2 \times 2$ identity matrix,
\item find the value of $p$ and the value of $k$.\\
(ii)
$$\mathbf { M } = \left( \begin{array} { r r }
a & - 9 \\
1 & 2
\end{array} \right) , \text { where } a \text { is a real constant }$$
Triangle $T$ has an area of 15 square units.\\
Triangle $T$ is transformed to the triangle $T ^ { \prime }$ by the transformation represented by the matrix M.
Given that the area of triangle $T ^ { \prime }$ is 270 square units, find the possible values of $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2017 Q5 [11]}}