| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Combined transformation matrix product |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing basic matrix multiplication, determinants for area scaling, and matrix inverses. Part (a) requires simple 2×2 matrix multiplication, part (b) uses the standard result that area scales by |det(P)|, and part (c) asks for the inverse matrix using a routine formula. All three parts are standard textbook exercises with no problem-solving insight required, making this slightly easier than average even for Further Maths content. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03i Determinant: area scale factor and orientation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P = AB = \left\{ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \right\}\) | M1 | \(P = AB\), seen or implied. |
| \(P = \begin{pmatrix} 1 & 4 \\ -2 & -3 \end{pmatrix}\) | A1 | Correct answer. |
| (b) \(\det P = 1(-3) - (4)(-2) = \{-3 + 8 = 5\}\) | M1 | Applies "\(ad - bc\)". |
| \(\text{Area}(T) = \frac{24}{5}\) (units)\(^2\) | A1 ft | \(\frac{24}{5}\) or 4.8, dependent on previous M |
| (c) \(QP = I \Rightarrow QPP^{-1} = IP^{-1} \Rightarrow Q = P^{-1}\) | M1 | Q = P\(^{-1}\) stated or an attempt to find P\(^{-1}\). |
| \(Q = P^{-1} = \frac{1}{5} \begin{pmatrix} -3 & -4 \\ 2 & 1 \end{pmatrix}\) | A1 ft | Correct ft inverse matrix. |
| Using \(BA\), area is the same in (b) and inverse is \(\frac{1}{5}\begin{pmatrix} 1 & -2 \\ 4 & -3 \end{pmatrix}\) in (c) and could gain it marks. |
**(a)** $P = AB = \left\{ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \right\}$ | M1 | $P = AB$, seen or implied.
$P = \begin{pmatrix} 1 & 4 \\ -2 & -3 \end{pmatrix}$ | A1 | Correct answer. | [2]
**(b)** $\det P = 1(-3) - (4)(-2) = \{-3 + 8 = 5\}$ | M1 | Applies "$ad - bc$".
$\text{Area}(T) = \frac{24}{5}$ (units)$^2$ | A1 ft | $\frac{24}{5}$ or 4.8, dependent on previous M | [3]
**(c)** $QP = I \Rightarrow QPP^{-1} = IP^{-1} \Rightarrow Q = P^{-1}$ | M1 | Q = P$^{-1}$ stated or an attempt to find P$^{-1}$.
$Q = P^{-1} = \frac{1}{5} \begin{pmatrix} -3 & -4 \\ 2 & 1 \end{pmatrix}$ | A1 ft | Correct ft inverse matrix. | [2]
Using $BA$, area is the same in (b) and inverse is $\frac{1}{5}\begin{pmatrix} 1 & -2 \\ 4 & -3 \end{pmatrix}$ in (c) and could gain it marks. | |
**Total: [7]**
---$$\mathbf{A} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \quad \mathbf{B} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$$
The transformation represented by $\mathbf{B}$ followed by the transformation represented by $\mathbf{A}$ is equivalent to the transformation represented by $\mathbf{P}$.
\begin{enumerate}[label=(\alph*)]
\item Find the matrix $\mathbf{P}$. [2]
\end{enumerate}
Triangle $T$ is transformed to the triangle $T'$ by the transformation represented by $\mathbf{P}$.
Given that the area of triangle $T'$ is 24 square units,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the area of triangle $T$. [3]
\end{enumerate}
Triangle $T'$ is transformed to the original triangle $T$ by the matrix represented by $\mathbf{Q}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the matrix $\mathbf{Q}$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q6 [7]}}