| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 12 |
| 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 straightforward Further Maths question with routine procedures: (a) uses the definition of eigenvector (matrix multiplication and solving for k), (b) requires finding eigenvalues via characteristic equation (standard 3×3 determinant), and (c) involves applying the transformation and verifying a linear constraint. All parts follow standard algorithms 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, scalar |
| Answer | Marks |
|---|---|
| \(\begin{pmatrix} 2 & k & 0 \\ 1 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}\begin{pmatrix} 9 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3k+18 \\ 12 \\ -8 \end{pmatrix} = \lambda\begin{pmatrix} 9 \\ 3 \\ -2 \end{pmatrix}\) | M1 |
| \(\lambda = 4 \Rightarrow 3k + 18 = 36 \Rightarrow k = 6\) | M1 A1 (3) |
| Answer | Marks |
|---|---|
| \(\lambda = 4\) is an eigenvalue | B1 |
| \(\begin{vmatrix} 2-\lambda & 6 & 0 \\ 1 & 1-\lambda & 0 \\ 0 & -2 & 1-\lambda \end{vmatrix} = (1-\lambda)[(2-\lambda)(1-\lambda)-6]\) | M1 |
| \((\lambda-1)(\lambda^2-3\lambda-4) = (\lambda-1)(\lambda-4)(\lambda+1)\) | M1 |
| \(\lambda = (4,) 1, -1\) | A1 (4) |
| Answer | Marks |
|---|---|
| \(\begin{pmatrix} 2 & 6 & 0 \\ 1 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}\begin{pmatrix} t-2 \\ t \\ 2t \end{pmatrix} = \begin{pmatrix} 8t-4 \\ 2t-2 \\ 0 \end{pmatrix}\) | M1 A2,1,0 |
| Answer | Marks |
|---|---|
| \(x - 4y - 4 = 0\) | M1 A1 (5) |
**(a)**
$\begin{pmatrix} 2 & k & 0 \\ 1 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}\begin{pmatrix} 9 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3k+18 \\ 12 \\ -8 \end{pmatrix} = \lambda\begin{pmatrix} 9 \\ 3 \\ -2 \end{pmatrix}$ | M1 |
$\lambda = 4 \Rightarrow 3k + 18 = 36 \Rightarrow k = 6$ | M1 A1 (3) |
**(b)**
$\lambda = 4$ is an eigenvalue | B1 |
$\begin{vmatrix} 2-\lambda & 6 & 0 \\ 1 & 1-\lambda & 0 \\ 0 & -2 & 1-\lambda \end{vmatrix} = (1-\lambda)[(2-\lambda)(1-\lambda)-6]$ | M1 |
$(\lambda-1)(\lambda^2-3\lambda-4) = (\lambda-1)(\lambda-4)(\lambda+1)$ | M1 |
$\lambda = (4,) 1, -1$ | A1 (4) |
**(c)**
$\begin{pmatrix} 2 & 6 & 0 \\ 1 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}\begin{pmatrix} t-2 \\ t \\ 2t \end{pmatrix} = \begin{pmatrix} 8t-4 \\ 2t-2 \\ 0 \end{pmatrix}$ | M1 A2,1,0 |
$x = 8t - 4, y = 2t - 2, z = 0$
$x - 4y - 4 = 0$ | M1 A1 (5) |
(12 marks)7. $\quad \mathbf { A } = \left( \begin{array} { c c c } 2 & \mathrm { k } & 0 \\ 1 & 1 & 0 \\ 0 & - 2 & 1 \end{array} \right)$, where k is a constant.
Given that $\left( \begin{array} { c } 9 \\ 3 \\ - 2 \end{array} \right)$ is an eigenvector of $\mathbf { A }$,
\begin{enumerate}[label=(\alph*)]
\item show that $\mathrm { k } = 6$,
\item find the eigenvalues of $\mathbf { A }$.
A transformation $\mathrm { T } : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ is represented by the matrix $\mathbf { A }$.\\
The point P has coordinates $( \mathrm { t } - 2 , \mathrm { t } , 2 \mathrm { t } )$ where t is a parameter.
\item Show that, for any value of $t$, the transformation $T$ maps $P$ onto a point on the line with equation $x - 4 y - 4 = 0$\\
(5)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 Q7 [12]}}