| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Transformation mapping problems |
| Difficulty | Challenging +1.2 This is a standard FP3 question on 3×3 matrices combining determinant calculation, matrix inversion, and applying inverse transformations to lines. Part (a) is routine determinant expansion, part (b) follows the standard cofactor method for finding inverses, and part (c) requires applying M^{-1} to find the pre-image line. While it involves multiple techniques and is Further Maths content (inherently harder), the methods are all standard algorithmic procedures without requiring novel insight or complex problem-solving. |
| Spec | 4.03j Determinant 3x3: calculation4.03o Inverse 3x3 matrix4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\det M = k(0-(-2)) + 1(1+3) + 1(-2-0) = -2k + 4 - 2 = 2 - 2k\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{M}^{-1} = \frac{1}{2-2k}\begin{pmatrix} -2 & -1 & 1 \\ -4 & k-3 & k+1 \\ -2 & 2k-3 & 1 \end{pmatrix}\) | M1, 2M1, M1 A3 | (5 marks) |
| Answer | Marks |
|---|---|
| Answer/Working: Let \((x,y,z)\) be on \(l_1\). Equation of \(l_2\) can be written as \(\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 7 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}\) | B1 |
| Use \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}\) with \(k=2\), i.e. \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-2}\begin{pmatrix} -2 & -1 & 1 \\ -4 & -1 & 3 \\ -2 & 1 & 1 \end{pmatrix}\begin{pmatrix} 4+4\lambda \\ 1+\lambda \\ 7+3\lambda \end{pmatrix}\) | M1 |
| \(\therefore\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3\lambda+1 \\ 4\lambda-2 \\ 2\lambda \end{pmatrix}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| or \(\mathbf{r} = \mathbf{a}+\lambda\mathbf{b}\) where \(\mathbf{a} = \mathbf{i}-2\mathbf{j}\) and \(\mathbf{b} = 3\mathbf{i}+4\mathbf{j}+2\mathbf{k}\) or equivalent | B1ft | (5 marks) |
## Part (a)
**Answer/Working:** $\det M = k(0-(-2)) + 1(1+3) + 1(-2-0) = -2k + 4 - 2 = 2 - 2k$ | **M1 A1** | (2 marks)
## Part (b)
**Answer/Working:** $\mathbf{M}^T = \begin{pmatrix} k & -1 & 1 \\ -1 & 0 & -2 \\ 1 & -1 & 1 \end{pmatrix}$ so cofactors $= \begin{pmatrix} -2 & -1 & 1 \\ -4 & k-3 & k+1 \\ -2 & 2k-3 & 1 \end{pmatrix}$
(-1 A mark for each term wrong)
$\mathbf{M}^{-1} = \frac{1}{2-2k}\begin{pmatrix} -2 & -1 & 1 \\ -4 & k-3 & k+1 \\ -2 & 2k-3 & 1 \end{pmatrix}$ | **M1, 2M1, M1 A3** | (5 marks)
## Part (c)
**Answer/Working:** Let $(x,y,z)$ be on $l_1$. Equation of $l_2$ can be written as $\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 7 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}$ | **B1**
Use $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}$ with $k=2$, i.e. $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-2}\begin{pmatrix} -2 & -1 & 1 \\ -4 & -1 & 3 \\ -2 & 1 & 1 \end{pmatrix}\begin{pmatrix} 4+4\lambda \\ 1+\lambda \\ 7+3\lambda \end{pmatrix}$ | **M1**
$\therefore\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3\lambda+1 \\ 4\lambda-2 \\ 2\lambda \end{pmatrix}$ | **M1 A1**
and so $(\mathbf{r}-\mathbf{a})\times\mathbf{b} = \mathbf{0}$ where $\mathbf{a} = \mathbf{i}-2\mathbf{j}$ and $\mathbf{b} = 3\mathbf{i}+4\mathbf{j}+2\mathbf{k}$ or equivalent
or $\mathbf{r} = \mathbf{a}+\lambda\mathbf{b}$ where $\mathbf{a} = \mathbf{i}-2\mathbf{j}$ and $\mathbf{b} = 3\mathbf{i}+4\mathbf{j}+2\mathbf{k}$ or equivalent | **B1ft** | (5 marks)
**Notes:**
- (a) M1: Finding determinant at least one component correct
- (b) 1M1: Finding matrix of cofactors or its transpose. 2M1: Finding inverse matrix, 1/(det) cofactors + transpose. 1A1: At least seven terms correct (so at most 2 incorrect) condone missing det. 2A1: At least eight terms correct (so at most 1 incorrect) condone missing det. 3A1: All nine terms correct, condone missing det
- (c) 1B1: Equation of $l_2$. 1M1: Using inverse transformation matrix correctly. 2M1: Finding general point in terms of $\lambda$. A1: CAO for general point in terms of one parameter. 2B1: ft for vector equation of their $l_1$
---The matrix $\mathbf{M}$ is given by
$$\mathbf{M} = \begin{pmatrix} k & -1 & 1 \\ 1 & 0 & -1 \\ 3 & -2 & 1 \end{pmatrix}, \quad k \neq 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\det \mathbf{M} = 2 - 2k$.
[2]
\item Find $\mathbf{M}^{-1}$, in terms of $k$.
[5]
The straight line $l_1$ is mapped onto the straight line $l_2$ by the transformation represented by the matrix $\begin{pmatrix} 2 & -1 & 1 \\ 1 & 0 & -1 \\ 3 & -2 & 1 \end{pmatrix}$.
The equation of $l_2$ is $(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$, where $\mathbf{a} = 4\mathbf{i} + \mathbf{j} + 7\mathbf{k}$ and $\mathbf{b} = 4\mathbf{i} + \mathbf{j} + 3\mathbf{k}$.
\item Find a vector equation for the line $l_1$.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2011 Q7 [12]}}