| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| 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 Pure 3 eigenvector question with routine calculations. Part (a) requires simple matrix-vector multiplication, (b) uses the eigenvector definition to find k, (c) involves finding the characteristic equation (straightforward for this matrix structure), and (d) applies the transformation to a line using a general point. All parts follow textbook procedures with no novel insight required, making it slightly easier than average even for FP3. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03c Matrix multiplication: properties (associative, not commutative)4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix4.04a Line equations: 2D and 3D, cartesian and vector forms |
| 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 \(\begin{pmatrix}24\\4\\6k+6\end{pmatrix}=\begin{pmatrix}6\lambda\\\lambda\\6\lambda\end{pmatrix}\) | Matrix-vector multiplication | |
| Uses first or second row to obtain \(\lambda = 4\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses third row and \(\lambda = 4\): \(6k+6=24 \Rightarrow k=3\) ✽ | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{vmatrix}1-\lambda&0&3\\0&-2-\lambda&1\\3&0&1-\lambda\end{vmatrix}=0\) | Setting up characteristic equation | |
| \((1-\lambda)((-2-\lambda)(1-\lambda)-0)-0+3(0-3(-2-\lambda))=0\) | M1 A1 | Expanding determinant |
| \((\lambda^3-12\lambda-16=0)\) | ||
| \(\Rightarrow (\lambda+2)(\lambda^2-2\lambda-8)=0\) | ||
| \(\Rightarrow (\lambda+2)(\lambda+2)(\lambda-4)=0\) | 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 | Matrix multiplication |
| 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 $\begin{pmatrix}24\\4\\6k+6\end{pmatrix}=\begin{pmatrix}6\lambda\\\lambda\\6\lambda\end{pmatrix}$ | | Matrix-vector multiplication |
| Uses first or second row to obtain $\lambda = 4$ | M1 A1 | (2) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses third row and $\lambda = 4$: $6k+6=24 \Rightarrow k=3$ ✽ | M1 A1 | (2) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}1-\lambda&0&3\\0&-2-\lambda&1\\3&0&1-\lambda\end{vmatrix}=0$ | | Setting up characteristic equation |
| $(1-\lambda)((-2-\lambda)(1-\lambda)-0)-0+3(0-3(-2-\lambda))=0$ | M1 A1 | Expanding determinant |
| $(\lambda^3-12\lambda-16=0)$ | | |
| $\Rightarrow (\lambda+2)(\lambda^2-2\lambda-8)=0$ | | |
| $\Rightarrow (\lambda+2)(\lambda+2)(\lambda-4)=0$ | 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 | Matrix multiplication |
| 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 $l _ { 2 }$.
\item Taking $k = 3$, find cartesian equations of $l _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2010 Q6 [13]}}