| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Orthogonal matrix diagonalization |
| Difficulty | Standard +0.8 This is a structured Further Maths question on orthogonal diagonalization. Part (a) is routine matrix multiplication, part (b) requires solving (M + 3I)v = 0, but part (c) requires understanding that P must be orthogonal (requiring normalization of eigenvectors and verification of orthogonality), which is more sophisticated than standard diagonalization. The scaffolding helps, but the orthogonal matrix construction elevates this above typical eigenvalue questions. |
| Spec | 4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{M}x = \lambda x \Rightarrow \begin{pmatrix}0&-1&3\\-1&4&-1\\3&-1&0\end{pmatrix}\begin{pmatrix}1\\-2\\1\end{pmatrix} = \begin{pmatrix}\lambda\\-2\lambda\\\lambda\end{pmatrix} \Rightarrow 2+3=\lambda \Rightarrow \lambda=5\) | M1 A1 | M1: Correct method leading to value for \(\lambda\); A1: Correct value. Correct answer only scores both marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}0&-1&3\\-1&4&-1\\3&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = -3\begin{pmatrix}x\\y\\z\end{pmatrix}\), obtains \(x=\ldots, y=\ldots, z=\ldots\) | M1 | Uses \(\mathbf{M}x=-3x\) or \((\mathbf{M}-(-3)\mathbf{I})x=\mathbf{0}\) to produce simultaneous equations and obtains values for \(x,y,z\) (not all 0), or uses suitable vector product |
| \(k\begin{pmatrix}1\\0\\-1\end{pmatrix}\) | A1 | Any correct eigenvector; allow \(x=\ldots, y=\ldots, z=\ldots\) and apply isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda^3 - 4\lambda^2 - 11\lambda + 30 = 0\); \(\det\mathbf{M} = -30 = \lambda_1\lambda_2\lambda_3 = -15\lambda\); \(\lambda = 2\) | B1 | Correct value; may be seen in D |
| \((\mathbf{D}=)\begin{pmatrix}-3&0&0\\0&2&0\\0&0&5\end{pmatrix}\) | B1ft | Diagonal matrix with \(-3\) and eigenvalues on leading diagonal, 0's elsewhere; ignore labelling |
| Normalise eigenvectors: \(\begin{pmatrix}1\\-2\\1\end{pmatrix}\rightarrow\begin{pmatrix}\frac{\sqrt{6}}{6}\\-\frac{\sqrt{6}}{3}\\\frac{\sqrt{6}}{6}\end{pmatrix}\) etc. | M1 | Correct method to normalise at least one eigenvector |
| \(\mathbf{D}=\begin{pmatrix}-3&0&0\\0&2&0\\0&0&5\end{pmatrix}\) and \(\mathbf{P}=\begin{pmatrix}\frac{\sqrt{2}}{2}&\frac{\sqrt{3}}{3}&\frac{\sqrt{6}}{6}\\0&\frac{\sqrt{3}}{3}&-\frac{\sqrt{6}}{3}\\-\frac{\sqrt{2}}{2}&\frac{\sqrt{3}}{3}&\frac{\sqrt{6}}{6}\end{pmatrix}\) | A1 | Both fully correct, consistent and labelled matrices; columns of P may be in opposite direction |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M}x = \lambda x \Rightarrow \begin{pmatrix}0&-1&3\\-1&4&-1\\3&-1&0\end{pmatrix}\begin{pmatrix}1\\-2\\1\end{pmatrix} = \begin{pmatrix}\lambda\\-2\lambda\\\lambda\end{pmatrix} \Rightarrow 2+3=\lambda \Rightarrow \lambda=5$ | M1 A1 | M1: Correct method leading to value for $\lambda$; A1: Correct value. Correct answer only scores both marks. |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}0&-1&3\\-1&4&-1\\3&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = -3\begin{pmatrix}x\\y\\z\end{pmatrix}$, obtains $x=\ldots, y=\ldots, z=\ldots$ | M1 | Uses $\mathbf{M}x=-3x$ or $(\mathbf{M}-(-3)\mathbf{I})x=\mathbf{0}$ to produce simultaneous equations and obtains values for $x,y,z$ (not all 0), or uses suitable vector product |
| $k\begin{pmatrix}1\\0\\-1\end{pmatrix}$ | A1 | Any correct eigenvector; allow $x=\ldots, y=\ldots, z=\ldots$ and apply isw |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda^3 - 4\lambda^2 - 11\lambda + 30 = 0$; $\det\mathbf{M} = -30 = \lambda_1\lambda_2\lambda_3 = -15\lambda$; $\lambda = 2$ | B1 | Correct value; may be seen in **D** |
| $(\mathbf{D}=)\begin{pmatrix}-3&0&0\\0&2&0\\0&0&5\end{pmatrix}$ | B1ft | Diagonal matrix with $-3$ and eigenvalues on leading diagonal, 0's elsewhere; ignore labelling |
| Normalise eigenvectors: $\begin{pmatrix}1\\-2\\1\end{pmatrix}\rightarrow\begin{pmatrix}\frac{\sqrt{6}}{6}\\-\frac{\sqrt{6}}{3}\\\frac{\sqrt{6}}{6}\end{pmatrix}$ etc. | M1 | Correct method to normalise at least one eigenvector |
| $\mathbf{D}=\begin{pmatrix}-3&0&0\\0&2&0\\0&0&5\end{pmatrix}$ and $\mathbf{P}=\begin{pmatrix}\frac{\sqrt{2}}{2}&\frac{\sqrt{3}}{3}&\frac{\sqrt{6}}{6}\\0&\frac{\sqrt{3}}{3}&-\frac{\sqrt{6}}{3}\\-\frac{\sqrt{2}}{2}&\frac{\sqrt{3}}{3}&\frac{\sqrt{6}}{6}\end{pmatrix}$ | A1 | Both fully correct, consistent and labelled matrices; columns of **P** may be in opposite direction |
---4.
$$\mathbf { M } = \left( \begin{array} { r r r }
0 & - 1 & 3 \\
- 1 & 4 & - 1 \\
3 & - 1 & 0
\end{array} \right)$$
Given that $\left( \begin{array} { r } 1 \\ - 2 \\ 1 \end{array} \right)$ is an eigenvector of $\mathbf { M }$
\begin{enumerate}[label=(\alph*)]
\item determine its corresponding eigenvalue.
Given that - 3 is an eigenvalue of $\mathbf { M }$
\item determine a corresponding eigenvector.
Hence, given that $\left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right)$ is also an eigenvector of $\mathbf { M }$
\item determine a diagonal matrix $\mathbf { D }$ and an orthogonal matrix $\mathbf { P }$ such that $\mathbf { D } = \mathbf { P } ^ { \mathrm { T } } \mathbf { M P }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2024 Q4 [8]}}