| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 10 |
| 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 standard Further Maths diagonalization question with straightforward structure: verify a given eigenvalue, find remaining eigenvalues from a 3×3 matrix with a convenient zero column, find eigenvectors, and construct P. The block structure (with zeros) simplifies calculations significantly. While it requires multiple techniques, each step follows routine procedures without requiring insight or problem-solving beyond textbook methods. |
| Spec | 4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Characteristic equation: \((2-\lambda)^2(4-\lambda) - (4-\lambda) = 0\) | M1 | Attempts characteristic equation (one slip allowed) |
| \((4-\lambda)(\lambda^2 - 4\lambda + 3) = 0\), so \(\lambda = 4\) | A1* | Deduces \(\lambda = 4\) is solution |
| Solves quadratic to give \(\lambda = 1\) and \(\lambda = 3\) | M1 | Solves quadratic |
| \(\lambda = 1\) and \(\lambda = 3\) | A1 | Both correct answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses correct method to find eigenvector | M1 | Correct method used |
| Vector parallel to one of \(\begin{pmatrix}0\\0\\1\end{pmatrix}\) or \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) or \(\begin{pmatrix}3\\-3\\1\end{pmatrix}\) | A1 | One correct vector (multiples accepted) |
| Obtains two correct vectors | A1 | Two correct vectors |
| Obtains all three correct vectors | A1 | All three correct vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses three vectors to form a matrix | M1 | Forms matrix with vectors as columns |
| \(\begin{pmatrix}0 & 1 & 3\\0 & 1 & -3\\1 & 1 & 1\end{pmatrix}\) or columns in different order | A1 | Correct matrix or valid alternative column ordering |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Characteristic equation: $(2-\lambda)^2(4-\lambda) - (4-\lambda) = 0$ | M1 | Attempts characteristic equation (one slip allowed) |
| $(4-\lambda)(\lambda^2 - 4\lambda + 3) = 0$, so $\lambda = 4$ | A1* | Deduces $\lambda = 4$ is solution |
| Solves quadratic to give $\lambda = 1$ and $\lambda = 3$ | M1 | Solves quadratic |
| $\lambda = 1$ and $\lambda = 3$ | A1 | Both correct answers |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses correct method to find eigenvector | M1 | Correct method used |
| Vector parallel to one of $\begin{pmatrix}0\\0\\1\end{pmatrix}$ or $\begin{pmatrix}1\\1\\1\end{pmatrix}$ or $\begin{pmatrix}3\\-3\\1\end{pmatrix}$ | A1 | One correct vector (multiples accepted) |
| Obtains two correct vectors | A1 | Two correct vectors |
| Obtains all three correct vectors | A1 | All three correct vectors |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses three vectors to form a matrix | M1 | Forms matrix with vectors as columns |
| $\begin{pmatrix}0 & 1 & 3\\0 & 1 & -3\\1 & 1 & 1\end{pmatrix}$ or columns in different order | A1 | Correct matrix or valid alternative column ordering |
---\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 each of the eigenvalues find a corresponding eigenvector.\\
(c) Find a matrix $\mathbf { P }$ such that $\mathbf { P } ^ { - 1 } \mathbf { M P }$ is a diagonal matrix.
\hfill \mbox{\textit{Edexcel FP2 Q3 [10]}}