| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Orthogonal matrix diagonalization |
| Difficulty | Standard +0.3 This is a straightforward application of standard eigenvalue/eigenvector techniques for a symmetric 2×2 matrix. The matrix is simple enough that the characteristic equation factors easily, normalization is routine, and writing down P and D requires only knowing the standard diagonalization result for symmetric matrices. Slightly easier than average due to the small matrix size and clean numbers. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\det(\mathbf{A}-\lambda\mathbf{I})=0\) or \(\begin{vmatrix}3-\lambda & 2\\2 & 6-\lambda\end{vmatrix}(=0)\) | M1 | Forms the characteristic equation; \(=0\) may be missing |
| \((3-\lambda)(6-\lambda)-4(=0)\) | M1 | Expands determinant and attempts to solve the equation |
| \(\lambda = 2, 7\) | A1 | Correct eigenvalues obtained |
| \(\begin{pmatrix}3&2\\2&6\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=2\begin{pmatrix}x\\y\end{pmatrix}\) or \(=7\begin{pmatrix}x\\y\end{pmatrix}\) | M1 | Use either of their eigenvalues to obtain at least one pair of non-zero values |
| \(\begin{pmatrix}3-2&2\\2&6-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0\) OR \(\begin{pmatrix}3-7&2\\2&6-7\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0\) | Alt for line above | |
| \(\begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}2\\-1\end{pmatrix}\) or \(x=1,y=2\ /\ x=2,y=-1\) | A1A1 | A1: One correct pair of values (allow any multiples); A1: Both correct pairs (allow any multiples) |
| \(\begin{pmatrix}\frac{1}{\sqrt5}\\\frac{2}{\sqrt5}\end{pmatrix}, \begin{pmatrix}\frac{2}{\sqrt5}\\\frac{-1}{\sqrt5}\end{pmatrix}\) or \(\frac{1}{\sqrt5}\begin{pmatrix}1\\2\end{pmatrix}, \frac{1}{\sqrt5}\begin{pmatrix}2\\-1\end{pmatrix}\) | A1ft | Both correct and normalised; follow through their eigenvectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\mathbf{D}=\begin{pmatrix}7&0\\0&2\end{pmatrix},\ \mathbf{P}=\begin{pmatrix}\frac{1}{\sqrt5}&\frac{2}{\sqrt5}\\\frac{2}{\sqrt5}&\frac{-1}{\sqrt5}\end{pmatrix}=\frac{1}{\sqrt5}\begin{pmatrix}1&2\\2&-1\end{pmatrix}\) | B1ft, B1 | B1ft: One correct ft (must be labelled); B1: Both fully correct and consistent (must both be labelled); order of eigenvalues must be consistent with order of eigenvectors |
| \(\mathbf{D}=\begin{pmatrix}0&7\\2&0\end{pmatrix},\ \mathbf{P}=\begin{pmatrix}\frac{2}{\sqrt5}&\frac{1}{\sqrt5}\\\frac{-1}{\sqrt5}&\frac{2}{\sqrt5}\end{pmatrix}\) | Both can be reversed and multiples allowed. \(\mathbf{D}=k^2\times\) matrix shown; \(\mathbf{P}=k\times\) matrix shown |
# Question 2:
$\mathbf{A} = \begin{pmatrix}3&2\\2&6\end{pmatrix}$
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\det(\mathbf{A}-\lambda\mathbf{I})=0$ or $\begin{vmatrix}3-\lambda & 2\\2 & 6-\lambda\end{vmatrix}(=0)$ | M1 | Forms the characteristic equation; $=0$ may be missing |
| $(3-\lambda)(6-\lambda)-4(=0)$ | M1 | Expands determinant and attempts to solve the equation |
| $\lambda = 2, 7$ | A1 | Correct eigenvalues obtained |
| $\begin{pmatrix}3&2\\2&6\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=2\begin{pmatrix}x\\y\end{pmatrix}$ or $=7\begin{pmatrix}x\\y\end{pmatrix}$ | M1 | Use either of their eigenvalues to obtain at least one pair of non-zero values |
| $\begin{pmatrix}3-2&2\\2&6-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$ OR $\begin{pmatrix}3-7&2\\2&6-7\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$ | | Alt for line above |
| $\begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}2\\-1\end{pmatrix}$ or $x=1,y=2\ /\ x=2,y=-1$ | A1A1 | A1: One correct pair of values (allow any multiples); A1: Both correct pairs (allow any multiples) |
| $\begin{pmatrix}\frac{1}{\sqrt5}\\\frac{2}{\sqrt5}\end{pmatrix}, \begin{pmatrix}\frac{2}{\sqrt5}\\\frac{-1}{\sqrt5}\end{pmatrix}$ or $\frac{1}{\sqrt5}\begin{pmatrix}1\\2\end{pmatrix}, \frac{1}{\sqrt5}\begin{pmatrix}2\\-1\end{pmatrix}$ | A1ft | Both correct and normalised; follow through their eigenvectors |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\mathbf{D}=\begin{pmatrix}7&0\\0&2\end{pmatrix},\ \mathbf{P}=\begin{pmatrix}\frac{1}{\sqrt5}&\frac{2}{\sqrt5}\\\frac{2}{\sqrt5}&\frac{-1}{\sqrt5}\end{pmatrix}=\frac{1}{\sqrt5}\begin{pmatrix}1&2\\2&-1\end{pmatrix}$ | B1ft, B1 | B1ft: One correct ft (must be labelled); B1: Both fully correct and consistent (must both be labelled); order of eigenvalues must be consistent with order of eigenvectors |
| $\mathbf{D}=\begin{pmatrix}0&7\\2&0\end{pmatrix},\ \mathbf{P}=\begin{pmatrix}\frac{2}{\sqrt5}&\frac{1}{\sqrt5}\\\frac{-1}{\sqrt5}&\frac{2}{\sqrt5}\end{pmatrix}$ | | Both can be reversed and multiples allowed. $\mathbf{D}=k^2\times$ matrix shown; $\mathbf{P}=k\times$ matrix shown |
**Total: 9 marks**
---2.
$$\mathbf { A } = \left( \begin{array} { l l }
3 & 2 \\
2 & 6
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find the eigenvalues and corresponding normalised eigenvectors of the matrix $\mathbf { A }$.
\item Write down a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { P } ^ { \mathrm { T } } \mathbf { A P } = \mathbf { D }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q2 [9]}}