| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Normalized eigenvectors |
| Difficulty | Standard +0.3 This is a straightforward Further Maths linear algebra question requiring standard techniques: verifying an eigenvalue by solving det(M - 6I) = 0, finding remaining eigenvalues from the characteristic equation, and normalizing an eigenvector. While it involves 3×3 matrices (more computational work than A-level Core), the methods are routine and algorithmic with no conceptual challenges or novel insights required. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{vmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \end{vmatrix} = 0\) | M1 | Attempt characteristic equation with at least 2 'elements' correct. Elements are \((1-\lambda)\{(5-\lambda)(1-\lambda)-1\}\), \(-(1-\lambda-3)\), \(+3(1-3(5-\lambda))\) |
| \(\lambda=6 \Rightarrow -5\times4+8+12=0\) or \(\lambda^3-7\lambda^2+36=(\lambda-6)(\lambda^2-\lambda-6) \Rightarrow \lambda=6\) | A1 | Verifies \(\lambda=6\) is an eigenvalue or factorises cubic to \((\lambda-6)\times\) quadratic and extracts \(\lambda=6\) |
| \((\lambda^2-\lambda-6)=0 \Rightarrow \lambda=3,-2\) | M1A1 | M1: Solves their 3 term quadratic or cubic \((\lambda-6)(\lambda^2+k\lambda\pm6)\). A1: Two other correct eigenvalues |
| ALT: Sub \(\lambda=6\) into \( | \mathbf{M}-\lambda\mathbf{I} | \) and shows this \(=0\) with no further work scores M1A1M0A0. For factor theorem solution, one further correct value scores M1A1M1A0. Both further correct values scores 4/4. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}1&1&3\\1&5&1\\3&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=6\begin{pmatrix}x\\y\\z\end{pmatrix}\) or \(\begin{pmatrix}-5&1&3\\1&-1&1\\3&1&-5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\) | M1 | Either statement sufficient or equivalent in equation form |
| \(x+y+3z=6x,\ x+5y+z=6y,\ 3x+y+z=6z \Rightarrow x=\ldots\) or \(y=\ldots\) or \(z=\ldots\) | dM1 | Solves two equations to obtain one variable in terms of another |
| \(\begin{pmatrix}1\\2\\1\end{pmatrix}\) or \(x=k,\ y=2k,\ z=k,\ k\neq0\) | A1 | Any multiple |
| \(\pm\begin{pmatrix}\frac{1}{\sqrt{6}}\\\frac{2}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\end{pmatrix}\) | A1 | Correct normalised vector. Can be positive or negative. Can be written in i,j,k form |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \end{vmatrix} = 0$ | M1 | Attempt characteristic equation with at least 2 'elements' correct. Elements are $(1-\lambda)\{(5-\lambda)(1-\lambda)-1\}$, $-(1-\lambda-3)$, $+3(1-3(5-\lambda))$ |
| $\lambda=6 \Rightarrow -5\times4+8+12=0$ or $\lambda^3-7\lambda^2+36=(\lambda-6)(\lambda^2-\lambda-6) \Rightarrow \lambda=6$ | A1 | Verifies $\lambda=6$ is an eigenvalue or factorises cubic to $(\lambda-6)\times$ quadratic and extracts $\lambda=6$ |
| $(\lambda^2-\lambda-6)=0 \Rightarrow \lambda=3,-2$ | M1A1 | M1: Solves their 3 term quadratic or cubic $(\lambda-6)(\lambda^2+k\lambda\pm6)$. A1: Two other correct eigenvalues |
**ALT:** Sub $\lambda=6$ into $|\mathbf{M}-\lambda\mathbf{I}|$ and shows this $=0$ with no further work scores M1A1M0A0. For factor theorem solution, one further correct value scores M1A1M1A0. Both further correct values scores 4/4.
---
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}1&1&3\\1&5&1\\3&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=6\begin{pmatrix}x\\y\\z\end{pmatrix}$ or $\begin{pmatrix}-5&1&3\\1&-1&1\\3&1&-5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ | M1 | Either statement sufficient or equivalent in equation form |
| $x+y+3z=6x,\ x+5y+z=6y,\ 3x+y+z=6z \Rightarrow x=\ldots$ or $y=\ldots$ or $z=\ldots$ | dM1 | Solves two equations to obtain one variable in terms of another |
| $\begin{pmatrix}1\\2\\1\end{pmatrix}$ or $x=k,\ y=2k,\ z=k,\ k\neq0$ | A1 | Any multiple |
| $\pm\begin{pmatrix}\frac{1}{\sqrt{6}}\\\frac{2}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\end{pmatrix}$ | A1 | Correct normalised vector. Can be positive or negative. Can be written in **i,j,k** form |
---4.
$$\mathbf { M } = \left( \begin{array} { l l l }
1 & 1 & 3 \\
1 & 5 & 1 \\
3 & 1 & 1
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that 6 is an eigenvalue of the matrix $\mathbf { M }$ and find the other two eigenvalues of $\mathbf { M }$.
\item Find a normalised eigenvector corresponding to the eigenvalue 6
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2017 Q4 [8]}}