| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvectors given eigenvalue |
| Difficulty | Standard +0.3 This is a standard Further Maths eigenvalue/eigenvector question with routine calculations: finding an eigenvalue by matrix multiplication, determining k from the eigenvector equation, showing repeated eigenvalues via characteristic equation, and applying a linear transformation to a line. All parts follow textbook procedures with no novel insight required, making it slightly easier than average for Further Maths content. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}1 & 0 & 3\\0 & -2 & 1\\k & 0 & 1\end{pmatrix}\begin{pmatrix}6\\1\\6\end{pmatrix} = \lambda\begin{pmatrix}6\\1\\6\end{pmatrix}\) giving \(\begin{pmatrix}24\\4\\6k+6\end{pmatrix} = \begin{pmatrix}6\lambda\\\lambda\\6\lambda\end{pmatrix}\) | M1 A1 | |
| Uses first or second row to obtain \(\lambda = 4\) | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses third row with \(\lambda = 4\): \(6k + 6 = 24 \Rightarrow k = 3\) \(\ast\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{vmatrix}1-\lambda & 0 & 3\\0 & -2-\lambda & 1\\3 & 0 & 1-\lambda\end{vmatrix} = 0\) | ||
| \((1-\lambda)((-2-\lambda)(1-\lambda)-0) - 0 + 3(0 - 3(-2-\lambda)) = 0\) | M1 A1 | Expanding determinant |
| \((1-\lambda)(-2-\lambda)(1-\lambda) + 9(2+\lambda) = 0 \Rightarrow (\lambda^3 - 12\lambda - 16 = 0)\) | ||
| \((\lambda+2)(\lambda^2 - 2\lambda - 8) = 0\) | ||
| \((\lambda+2)(\lambda+2)(\lambda-4) = 0\) | M1 | Factorising |
| \(\lambda = -2,\ 4\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Parametric form of \(l_1\): \((t+2,\ -3t,\ 4t-1)\) | M1 | |
| \(\begin{pmatrix}1 & 0 & 3\\0 & -2 & 1\\3 & 0 & 1\end{pmatrix}\begin{pmatrix}t+2\\-3t\\4t-1\end{pmatrix} = \begin{pmatrix}13t-1\\10t-1\\7t+5\end{pmatrix}\) | M1 A1 | |
| Cartesian equations of \(l_2\): \(\dfrac{x+1}{13} = \dfrac{y+1}{10} = \dfrac{z-5}{7}\) | ddM1 A1 | (5) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1 & 0 & 3\\0 & -2 & 1\\k & 0 & 1\end{pmatrix}\begin{pmatrix}6\\1\\6\end{pmatrix} = \lambda\begin{pmatrix}6\\1\\6\end{pmatrix}$ giving $\begin{pmatrix}24\\4\\6k+6\end{pmatrix} = \begin{pmatrix}6\lambda\\\lambda\\6\lambda\end{pmatrix}$ | M1 A1 | |
| Uses first or second row to obtain $\lambda = 4$ | | (2) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses third row with $\lambda = 4$: $6k + 6 = 24 \Rightarrow k = 3$ $\ast$ | M1 A1 | (2) | Completion to given result |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}1-\lambda & 0 & 3\\0 & -2-\lambda & 1\\3 & 0 & 1-\lambda\end{vmatrix} = 0$ | | |
| $(1-\lambda)((-2-\lambda)(1-\lambda)-0) - 0 + 3(0 - 3(-2-\lambda)) = 0$ | M1 A1 | Expanding determinant |
| $(1-\lambda)(-2-\lambda)(1-\lambda) + 9(2+\lambda) = 0 \Rightarrow (\lambda^3 - 12\lambda - 16 = 0)$ | | |
| $(\lambda+2)(\lambda^2 - 2\lambda - 8) = 0$ | | |
| $(\lambda+2)(\lambda+2)(\lambda-4) = 0$ | M1 | Factorising |
| $\lambda = -2,\ 4$ | A1 | (4) | |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Parametric form of $l_1$: $(t+2,\ -3t,\ 4t-1)$ | M1 | |
| $\begin{pmatrix}1 & 0 & 3\\0 & -2 & 1\\3 & 0 & 1\end{pmatrix}\begin{pmatrix}t+2\\-3t\\4t-1\end{pmatrix} = \begin{pmatrix}13t-1\\10t-1\\7t+5\end{pmatrix}$ | M1 A1 | |
| Cartesian equations of $l_2$: $\dfrac{x+1}{13} = \dfrac{y+1}{10} = \dfrac{z-5}{7}$ | ddM1 A1 | (5) | |6. $\mathbf { M } = \left( \begin{array} { c c c } 1 & 0 & 3 \\ 0 & - 2 & 1 \\ k & 0 & 1 \end{array} \right)$, where $k$ is a constant.\\
Given that $\left( \begin{array} { l } 6 \\ 1 \\ 6 \end{array} \right)$ is an eigenvector of $\mathbf { M }$ ,
\begin{enumerate}[label=(\alph*)]
\item find the eigenvalue of $\mathbf { M }$ corresponding to $\left( \begin{array} { l } 6 \\ 1 \\ 6 \end{array} \right)$ ,
\item show that $k = 3$ ,
\item show that $\mathbf { M }$ has exactly two eigenvalues.
A transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ is represented by $\mathbf { M }$ .\\
The transformation $T$ maps the line $l _ { 1 }$ ,with cartesian equations $\frac { x - 2 } { 1 } = \frac { y } { - 3 } = \frac { z + 1 } { 4 }$ ,onto the line $l _ { 2 }$ .\\
6. $\mathbf { M } = \left( \begin{array} { c c c } 0 & - 2 & 1 \\ k & 0 & 1 \end{array} \right)$, where $k$ is a constant.
Given that $\left( \begin{array} { l } 6 \\ 1 \\ 6 \end{array} \right)$ is an eigenvector of $\mathbf { M }$ ,\\
(a)find the eigenvalue of $\mathbf { M }$ corresponding to $\left( \begin{array} { l } 6 \\ 1 \\ 6 \end{array} \right)$
\item Taking $k = 3$ ,find cartesian equations of $l _ { 2 }$ .
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 Q6 [13]}}