| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A = PDP⁻¹ |
| Difficulty | Standard +0.3 This is a structured, multi-part eigenvalue question with significant scaffolding. Part (a) is straightforward matrix multiplication, (b) uses the result to find p, (c) requires finding remaining eigenvalues/eigenvectors (standard FP2), and (d) asks for diagonalization. The scaffolding and step-by-step guidance make this easier than average, though it covers substantial content. The use of PDP^T (rather than PDP^{-1}) indicates a symmetric matrix, simplifying the work. |
| Spec | 4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2\times4 - 1\times2 + 2\times0 = \lambda\times2 \Rightarrow \lambda = \ldots\) or \(2\times0-1\times(-2)+2\times2=\lambda\times2\Rightarrow\lambda=\ldots\); Alt 1: Matrix-vector product gives \(\lambda=\ldots\); Alt 2: \((4-\lambda)(2-\lambda)(-2+\ldots)=0\) approach | M1 | Attempts to find \(\lambda\) |
| \(\lambda = 3\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2\times2 - p - 2\times2 = 3\times(-1) \Rightarrow 4-p-4=-3 \Rightarrow p=3\); Alt 1: \((4-3)((p-3)(2-3)-4)-2(2(2-3)-0)+0=0\Rightarrow p=3\); Alt 2 (using Alt 2 from (a)): \(4-(p-\lambda)-4=0\), uses \(\lambda=3\) to show \(p=3\) | B1* | Dependent on correct \(\lambda=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((4-\lambda)\bigl((3-\lambda)(2-\lambda)-4\bigr) - 2\bigl(2(2-\lambda)-0\bigr)+0=0\) | M1 | Expands characteristic equation |
| \(\Rightarrow (4-\lambda)(\lambda^2-5\lambda+2)-8+4\lambda=0 \Rightarrow \lambda^3-9\lambda^2+18\lambda=0\) \(\Rightarrow \lambda(\lambda-3)(\lambda-6)=0 \Rightarrow \lambda=\ldots\) | M1 | Simplifies to factorised cubic |
| Remaining eigenvalues are 0 and 6 | A1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For \(\lambda=0\): \(\begin{cases}4x+2y=0\\2x+3y-2z=0\\-2y+2z=0\end{cases}\) or for \(\lambda=6\): \(\begin{cases}-2x+2y=0\\2x-3y-2z=0\\-2y-4z=0\end{cases}\) \(\Rightarrow x=\ldots,\,y=\ldots,\,z=\ldots\) | M1 | Sets up and solves correct system |
| \(c\begin{pmatrix}-1\\2\\2\end{pmatrix}\) or \(d\begin{pmatrix}2\\2\\-1\end{pmatrix}\) (o.e.) | A1 | Correct eigenvectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(c\begin{pmatrix}-1\\2\\2\end{pmatrix}\) and \(d\begin{pmatrix}2\\2\\-1\end{pmatrix}\) o.e. | A1 | 1.1b |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mathbf{D}=\begin{pmatrix}3&0&0\\0&6&0\\0&0&0\end{pmatrix}\) or \(\mathbf{D}=\begin{pmatrix}3&0&0\\0&0&0\\0&0&6\end{pmatrix}\) or equivalent | B1ft | 2.2a |
| \(\sqrt{2^2+(-1)^2+2^2}=\sqrt{9}=3 \Rightarrow \mathbf{v}=\ldots\) | M1 | 3.1a |
| \(\mathbf{P}=\frac{1}{3}\begin{pmatrix}2&2&-1\\-1&2&2\\2&-1&2\end{pmatrix}\) or \(\mathbf{P}=\frac{1}{3}\begin{pmatrix}2&-1&2\\-1&2&2\\2&2&-1\end{pmatrix}\) | A1ft | 1.1b |
| The order must correspond to their D | ||
| (3) | ||
| (12 marks) |
## Question 4(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\times4 - 1\times2 + 2\times0 = \lambda\times2 \Rightarrow \lambda = \ldots$ or $2\times0-1\times(-2)+2\times2=\lambda\times2\Rightarrow\lambda=\ldots$; **Alt 1:** Matrix-vector product gives $\lambda=\ldots$; **Alt 2:** $(4-\lambda)(2-\lambda)(-2+\ldots)=0$ approach | M1 | Attempts to find $\lambda$ |
| $\lambda = 3$ | A1 | Correct value |
---
## Question 4(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\times2 - p - 2\times2 = 3\times(-1) \Rightarrow 4-p-4=-3 \Rightarrow p=3$; **Alt 1:** $(4-3)((p-3)(2-3)-4)-2(2(2-3)-0)+0=0\Rightarrow p=3$; **Alt 2 (using Alt 2 from (a)):** $4-(p-\lambda)-4=0$, uses $\lambda=3$ to show $p=3$ | B1* | Dependent on correct $\lambda=3$ |
---
## Question 4(c)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(4-\lambda)\bigl((3-\lambda)(2-\lambda)-4\bigr) - 2\bigl(2(2-\lambda)-0\bigr)+0=0$ | M1 | Expands characteristic equation |
| $\Rightarrow (4-\lambda)(\lambda^2-5\lambda+2)-8+4\lambda=0 \Rightarrow \lambda^3-9\lambda^2+18\lambda=0$ $\Rightarrow \lambda(\lambda-3)(\lambda-6)=0 \Rightarrow \lambda=\ldots$ | M1 | Simplifies to factorised cubic |
| Remaining eigenvalues are 0 and 6 | A1 | Both correct |
---
## Question 4(c)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For $\lambda=0$: $\begin{cases}4x+2y=0\\2x+3y-2z=0\\-2y+2z=0\end{cases}$ or for $\lambda=6$: $\begin{cases}-2x+2y=0\\2x-3y-2z=0\\-2y-4z=0\end{cases}$ $\Rightarrow x=\ldots,\,y=\ldots,\,z=\ldots$ | M1 | Sets up and solves correct system |
| $c\begin{pmatrix}-1\\2\\2\end{pmatrix}$ or $d\begin{pmatrix}2\\2\\-1\end{pmatrix}$ (o.e.) | A1 | Correct eigenvectors |
# Question 4 (Eigenvectors/Matrices):
## Part (c)(ii) - Eigenvectors:
| Working | Mark | Guidance |
|---------|------|----------|
| $c\begin{pmatrix}-1\\2\\2\end{pmatrix}$ and $d\begin{pmatrix}2\\2\\-1\end{pmatrix}$ o.e. | **A1** | 1.1b |
| | **(3)** | |
## Part (d) - Diagonal and Modal Matrices:
| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{D}=\begin{pmatrix}3&0&0\\0&6&0\\0&0&0\end{pmatrix}$ or $\mathbf{D}=\begin{pmatrix}3&0&0\\0&0&0\\0&0&6\end{pmatrix}$ or equivalent | **B1ft** | 2.2a |
| $\sqrt{2^2+(-1)^2+2^2}=\sqrt{9}=3 \Rightarrow \mathbf{v}=\ldots$ | **M1** | 3.1a |
| $\mathbf{P}=\frac{1}{3}\begin{pmatrix}2&2&-1\\-1&2&2\\2&-1&2\end{pmatrix}$ or $\mathbf{P}=\frac{1}{3}\begin{pmatrix}2&-1&2\\-1&2&2\\2&2&-1\end{pmatrix}$ | **A1ft** | 1.1b |
| The order must correspond to their **D** | | |
| | **(3)** | |
| | **(12 marks)** | |
**Notes:**
- **B1ft:** Correct diagonal matrix following through on their eigenvalues
- **M1:** Normalises the eigenvectors
- **A1ft:** Forms correct matrix **P** with columns in appropriate order for their **D**, following through on their eigenvectors
---4.
$$\mathbf { A } = \left( \begin{array} { r r r }
4 & 2 & 0 \\
2 & p & - 2 \\
0 & - 2 & 2
\end{array} \right) \quad \text { where } p \text { is a constant }$$
Given that $\left( \begin{array} { r } 2 \\ - 1 \\ 2 \end{array} \right)$ is an eigenvector of $\mathbf { A }$,
\begin{enumerate}[label=(\alph*)]
\item determine the eigenvalue corresponding to this eigenvector.
\item Hence show that $p = 3$
\item Determine
\begin{enumerate}[label=(\roman*)]
\item the remaining eigenvalues of $\mathbf { A }$,
\item corresponding eigenvectors for these eigenvalues.
\end{enumerate}\item Hence determine a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } = \mathbf { P D P } ^ { \mathrm { T } }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2024 Q4 [12]}}