| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Geometric interpretation of matrices |
| Difficulty | Standard +0.3 This is a standard FP1 matrices question covering routine techniques: determinant calculation (trivial), solving simultaneous equations from matrix multiplication, triangle area using coordinates, applying the determinant-area relationship, identifying a rotation matrix, and finding a matrix from a composition. All parts are textbook exercises requiring no novel insight, though the multi-part structure and matrix composition in (f) elevate it slightly above average difficulty. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{M}: \begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} \rightarrow \begin{pmatrix} 25 \\ -14 \end{pmatrix}\) | ||
| \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}\begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} = \begin{pmatrix} 25 \\ -14 \end{pmatrix}\) or \(\begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}^{-1}\begin{pmatrix} 25 \\ -14 \end{pmatrix}\) | M1 | Using the information to form the matrix equation. Can be implied by any of the correct equations below. |
| \(\begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} = \frac{1}{(-23)}\begin{pmatrix} -5 & -4 \\ -2 & 3 \end{pmatrix}\begin{pmatrix} 25 \\ -14 \end{pmatrix} = \frac{1}{(-23)}\begin{pmatrix} -125+56 \\ -50-42 \end{pmatrix}\) | ||
| Either \((2a-7) = 3\) or \((a-1)=4\) | A1 | Any one correct equation. |
| \(a = 5\) | A1 | \(a=5\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(ORS = \frac{1}{2}\begin{vmatrix} 6 & 3 & 0 & 6 \\ 0 & 4 & 0 & 0 \end{vmatrix} = \frac{1}{2}\ | (6\times4 - 3\times0 + 0 - 0 + 0 - 0)\ | \) |
| \(= 12\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(ORS = \frac{1}{2}\begin{vmatrix} 18 & 25 & 0 & 18 \\ 12 & -14 & 0 & 12 \end{vmatrix} = \frac{1}{2}\ | (18\times{-14} - 12\times25 + 0 - 0 + 0 - 0)\ | \) |
| \(= 276\) | A1\(\checkmark\) | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{M} = \mathbf{BA}\) | M1 | \(\mathbf{M} = \mathbf{BA}\), seen or implied. |
| \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) | A1 | \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\), with constants to be found. |
| \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} b & -a \\ d & -c \end{pmatrix}\) | M1 | \(\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} =\) their \(\begin{pmatrix} b & -a \\ d & -c \end{pmatrix}\) with at least two elements correct on RHS. |
| \(\mathbf{B} = \begin{pmatrix} -4 & 3 \\ 5 & 2 \end{pmatrix}\) | A1 | Correct matrix for \(\mathbf{B}\); or \(a=-4, b=3, c=5, d=2\) |
| [4] |
## Question 9(b) – Aliter Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M}: \begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} \rightarrow \begin{pmatrix} 25 \\ -14 \end{pmatrix}$ | | |
| $\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}\begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} = \begin{pmatrix} 25 \\ -14 \end{pmatrix}$ or $\begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix}^{-1}\begin{pmatrix} 25 \\ -14 \end{pmatrix}$ | M1 | Using the information to form the matrix equation. Can be implied by any of the correct equations below. |
| $\begin{pmatrix} 2a-7 \\ a-1 \end{pmatrix} = \frac{1}{(-23)}\begin{pmatrix} -5 & -4 \\ -2 & 3 \end{pmatrix}\begin{pmatrix} 25 \\ -14 \end{pmatrix} = \frac{1}{(-23)}\begin{pmatrix} -125+56 \\ -50-42 \end{pmatrix}$ | | |
| Either $(2a-7) = 3$ or $(a-1)=4$ | A1 | Any one correct equation. |
| $a = 5$ | A1 | $a=5$ |
| | **[3]** | |
---
## Question 9(c) – Aliter Way 2 (Determinant):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $ORS = \frac{1}{2}\begin{vmatrix} 6 & 3 & 0 & 6 \\ 0 & 4 & 0 & 0 \end{vmatrix} = \frac{1}{2}\|(6\times4 - 3\times0 + 0 - 0 + 0 - 0)\|$ | M1 | Correct calculation |
| $= 12$ | A1 | |
| | **[2]** | |
---
## Question 9(d) – Aliter Way 2 (Determinant):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $ORS = \frac{1}{2}\begin{vmatrix} 18 & 25 & 0 & 18 \\ 12 & -14 & 0 & 12 \end{vmatrix} = \frac{1}{2}\|(18\times{-14} - 12\times25 + 0 - 0 + 0 - 0)\|$ | M1 | Correct calculation |
| $= 276$ | A1$\checkmark$ | |
| | **[2]** | |
---
## Question 9(f) – Aliter Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M} = \mathbf{BA}$ | M1 | $\mathbf{M} = \mathbf{BA}$, seen or implied. |
| $\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ | A1 | $\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, with constants to be found. |
| $\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} b & -a \\ d & -c \end{pmatrix}$ | M1 | $\begin{pmatrix} 3 & 4 \\ 2 & -5 \end{pmatrix} =$ their $\begin{pmatrix} b & -a \\ d & -c \end{pmatrix}$ with at least two elements correct on RHS. |
| $\mathbf{B} = \begin{pmatrix} -4 & 3 \\ 5 & 2 \end{pmatrix}$ | A1 | Correct matrix for $\mathbf{B}$; or $a=-4, b=3, c=5, d=2$ |
| | **[4]** | |
---9.
$$\mathbf { M } = \left( \begin{array} { r r }
3 & 4 \\
2 & - 5
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find $\operatorname { det } \mathbf { M }$.
The transformation represented by $\mathbf { M }$ maps the point $S ( 2 a - 7 , a - 1 )$, where $a$ is a constant, onto the point $S ^ { \prime } ( 25 , - 14 )$.
\item Find the value of $a$.
The point $R$ has coordinates $( 6,0 )$.
Given that $O$ is the origin,
\item find the area of triangle $O R S$.
Triangle $O R S$ is mapped onto triangle $O R ^ { \prime } S ^ { \prime }$ by the transformation represented by $\mathbf { M }$.
\item Find the area of triangle $O R ^ { \prime } S ^ { \prime }$.
Given that
$$\mathbf { A } = \left( \begin{array} { r r }
0 & - 1 \\
1 & 0
\end{array} \right)$$
\item describe fully the single geometrical transformation represented by $\mathbf { A }$.
The transformation represented by $\mathbf { A }$ followed by the transformation represented by $\mathbf { B }$ is equivalent to the transformation represented by $\mathbf { M }$.
\item Find B.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q9 [14]}}