| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Transformation mapping problems |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question combining eigenvalues/eigenvectors with 3D line transformations. Part (a) involves standard characteristic equation work with a block-diagonal structure making it easier. Part (b) is routine eigenvector calculation. Part (c) requires understanding that lines map to lines under linear transformations and finding two transformed points, which is conceptually more demanding but still follows established techniques. The question is harder than average A-level due to being Further Maths content, but it's a relatively standard FP3 question without requiring novel insights. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\begin{vmatrix} 2-\lambda & 1 & 0 \\ 1 & 2-\lambda & 0 \\ -1 & 0 & 4-\lambda \end{vmatrix} = 0 \therefore (2-\lambda)(2-\lambda)(4-\lambda) - (4-\lambda) = 0\) | M1 | Condone missing \(=0\); may expand using any row or column |
| \((4-\lambda) = 0\) verifies \(\lambda = 4\) is an eigenvalue | M1 | Shows \(\lambda=4\) is an eigenvalue, some working needed |
| \(\therefore (4-\lambda)\{4 - 4\lambda + \lambda^2 - 1\} = 0 \therefore (4-\lambda)\{\lambda^2 - 4\lambda + 3\} = 0\) | A1 | Three term quadratic factor cao (depends on 1st M only) |
| \(\therefore (4-\lambda)(\lambda-1)(\lambda-3) = 0\) and \(3\) and \(1\) are the other two eigenvalues | M1 A1 | Attempt at factorisation; cao — if they state \(\lambda=1\) and \(3\) award marks |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Set \(\begin{pmatrix}2&1&0\\1&2&0\\-1&0&4\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = 4\begin{pmatrix}x\\y\\z\end{pmatrix}\) giving \(\begin{pmatrix}-2&1&0\\1&-2&0\\-1&0&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\) | M1 | Using \(A\mathbf{x}=4\mathbf{x}\) o.e. |
| Solve \(-2x+y=0\) and \(x-2y=0\) and \(-x=0\) to obtain \(x=0,\, y=0,\, z=k\) | M1 | Getting a pair of correct equations |
| Obtain eigenvector as \(\mathbf{k}\) (or multiple) | A1 | cao |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(l_1\) has equation which may be written \(\begin{pmatrix}3+\lambda\\2-\lambda\\-2+2\lambda\end{pmatrix}\) | B1 | Using a and b |
| So \(l_2\) is given by \(\mathbf{r} = \begin{pmatrix}2&1&0\\1&2&0\\-1&0&4\end{pmatrix}\begin{pmatrix}3+\lambda\\2-\lambda\\-2+2\lambda\end{pmatrix}\) | M1 | Using \(\mathbf{r} = M \times\) their matrix in a and b |
| i.e. \(\mathbf{r} = \begin{pmatrix}8+\lambda\\7-\lambda\\-11+7\lambda\end{pmatrix}\) | M1 A1 | Getting an expression for \(l_2\) with at least one component correct; cao all three components correct |
| So \((\mathbf{r}-\mathbf{c})\times\mathbf{d}=\mathbf{0}\) where \(\mathbf{c} = 8\mathbf{i}+7\mathbf{j}-11\mathbf{k}\) and \(\mathbf{d}=\mathbf{i}-\mathbf{j}+7\mathbf{k}\) | A1ft | ft their vector; must have \(\mathbf{r}=\) or \((\mathbf{r}-\mathbf{c})\times\mathbf{d}=0\), need both equation and r |
| (5) |
# Question 8:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\begin{vmatrix} 2-\lambda & 1 & 0 \\ 1 & 2-\lambda & 0 \\ -1 & 0 & 4-\lambda \end{vmatrix} = 0 \therefore (2-\lambda)(2-\lambda)(4-\lambda) - (4-\lambda) = 0$ | M1 | Condone missing $=0$; may expand using any row or column |
| $(4-\lambda) = 0$ verifies $\lambda = 4$ is an eigenvalue | M1 | Shows $\lambda=4$ is an eigenvalue, some working needed |
| $\therefore (4-\lambda)\{4 - 4\lambda + \lambda^2 - 1\} = 0 \therefore (4-\lambda)\{\lambda^2 - 4\lambda + 3\} = 0$ | A1 | Three term quadratic factor cao (depends on 1st M only) |
| $\therefore (4-\lambda)(\lambda-1)(\lambda-3) = 0$ and $3$ and $1$ are the other two eigenvalues | M1 A1 | Attempt at factorisation; cao — if they state $\lambda=1$ and $3$ award marks |
| | **(5)** | |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Set $\begin{pmatrix}2&1&0\\1&2&0\\-1&0&4\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = 4\begin{pmatrix}x\\y\\z\end{pmatrix}$ giving $\begin{pmatrix}-2&1&0\\1&-2&0\\-1&0&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ | M1 | Using $A\mathbf{x}=4\mathbf{x}$ o.e. |
| Solve $-2x+y=0$ and $x-2y=0$ and $-x=0$ to obtain $x=0,\, y=0,\, z=k$ | M1 | Getting a pair of correct equations |
| Obtain eigenvector as $\mathbf{k}$ (or multiple) | A1 | cao |
| | **(3)** | |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $l_1$ has equation which may be written $\begin{pmatrix}3+\lambda\\2-\lambda\\-2+2\lambda\end{pmatrix}$ | B1 | Using **a** and **b** |
| So $l_2$ is given by $\mathbf{r} = \begin{pmatrix}2&1&0\\1&2&0\\-1&0&4\end{pmatrix}\begin{pmatrix}3+\lambda\\2-\lambda\\-2+2\lambda\end{pmatrix}$ | M1 | Using $\mathbf{r} = M \times$ their matrix in **a** and **b** |
| i.e. $\mathbf{r} = \begin{pmatrix}8+\lambda\\7-\lambda\\-11+7\lambda\end{pmatrix}$ | M1 A1 | Getting an expression for $l_2$ with at least one component correct; cao all three components correct |
| So $(\mathbf{r}-\mathbf{c})\times\mathbf{d}=\mathbf{0}$ where $\mathbf{c} = 8\mathbf{i}+7\mathbf{j}-11\mathbf{k}$ and $\mathbf{d}=\mathbf{i}-\mathbf{j}+7\mathbf{k}$ | A1ft | ft their vector; must have $\mathbf{r}=$ or $(\mathbf{r}-\mathbf{c})\times\mathbf{d}=0$, need both equation and **r** |
| | **(5)** | |\begin{enumerate}
\item The matrix $\mathbf { M }$ is given by
\end{enumerate}
$$\mathbf { M } = \left( \begin{array} { r r r }
2 & 1 & 0 \\
1 & 2 & 0 \\
- 1 & 0 & 4
\end{array} \right)$$
(a) Show that 4 is an eigenvalue of $\mathbf { M }$, and find the other two eigenvalues.\\
(b) For the eigenvalue 4, find a corresponding eigenvector.
The straight line $l _ { 1 }$ is mapped onto the straight line $l _ { 2 }$ by the transformation represented by the matrix $\mathbf { M }$.
The equation of $l _ { 1 }$ is $( \mathbf { r } - \mathbf { a } ) \times \mathbf { b } = 0$, where $\mathbf { a } = 3 \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k }$ and $\mathbf { b } = \mathbf { i } - \mathbf { j } + 2 \mathbf { k }$.\\
(c) Find a vector equation for the line $l _ { 2 }$.\\
\hfill \mbox{\textit{Edexcel FP3 2012 Q8 [13]}}