| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Eigenvalues and eigenvectors |
| Difficulty | Standard +0.3 This is a straightforward Further Maths eigenvalue/eigenvector question with a helpful hint (given that λ=7). Part (a) requires verifying λ=7 satisfies the characteristic equation and finding two other eigenvalues by solving a quadratic. Part (b) requires solving (M-7I)v=0. The matrix structure (with a zero row element) makes calculations cleaner than typical. While Further Maths content is inherently harder than Core, this is a standard textbook exercise requiring only routine application of the eigenvalue algorithm with no novel insight. |
| Spec | 4.03j Determinant 3x3: calculation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{vmatrix} 6-\lambda & 1 & -1 \\ 0 & 7-\lambda & 0 \\ 3 & -1 & 2-\lambda \end{vmatrix} = 0 \therefore (6-\lambda)(7-\lambda)(2-\lambda)+3(7-\lambda)=0\) | M1 | |
| \((7-\lambda)=0\) verifies \(\lambda=7\) is an eigenvalue | M1 | Can be seen anywhere |
| \(\therefore (7-\lambda)\{12-8\lambda+\lambda^2+3\}=0 \therefore (7-\lambda)\{\lambda^2-8\lambda+15\}=0\) | A1 | |
| \(\therefore (7-\lambda)(\lambda-5)(\lambda-3)=0\) and 3 and 5 are the other two eigenvalues | M1 A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Set \(\begin{pmatrix}6&1&-1\\0&7&0\\3&-1&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=7\begin{pmatrix}x\\y\\z\end{pmatrix}\) or \(\begin{pmatrix}-1&1&-1\\0&0&0\\3&-1&-5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\) | M1 | |
| Solve \(-x+y-z=0\) and \(3x-y-5z=0\) to obtain \(x=3z\) or \(y=4z\) and a second equation which can contain 3 variables | M1 A1 | |
| Obtain eigenvector as \(3\mathbf{i}+4\mathbf{j}+\mathbf{k}\) (or multiple) | A1 | (4) [9] |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix} 6-\lambda & 1 & -1 \\ 0 & 7-\lambda & 0 \\ 3 & -1 & 2-\lambda \end{vmatrix} = 0 \therefore (6-\lambda)(7-\lambda)(2-\lambda)+3(7-\lambda)=0$ | M1 | |
| $(7-\lambda)=0$ verifies $\lambda=7$ is an eigenvalue | M1 | Can be seen anywhere |
| $\therefore (7-\lambda)\{12-8\lambda+\lambda^2+3\}=0 \therefore (7-\lambda)\{\lambda^2-8\lambda+15\}=0$ | A1 | |
| $\therefore (7-\lambda)(\lambda-5)(\lambda-3)=0$ and 3 and 5 are the other two eigenvalues | M1 A1 | (5) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Set $\begin{pmatrix}6&1&-1\\0&7&0\\3&-1&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=7\begin{pmatrix}x\\y\\z\end{pmatrix}$ or $\begin{pmatrix}-1&1&-1\\0&0&0\\3&-1&-5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ | M1 | |
| Solve $-x+y-z=0$ and $3x-y-5z=0$ to obtain $x=3z$ or $y=4z$ and a second equation which can contain 3 variables | M1 A1 | |
| Obtain eigenvector as $3\mathbf{i}+4\mathbf{j}+\mathbf{k}$ (or multiple) | A1 | (4) **[9]** |
---3.
$$\mathbf { M } = \left( \begin{array} { r r r }
6 & 1 & - 1 \\
0 & 7 & 0 \\
3 & - 1 & 2
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that 7 is an eigenvalue of the matrix $\mathbf { M }$ and find the other two eigenvalues of $\mathbf { M }$.
\item Find an eigenvector corresponding to the eigenvalue 7.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2009 Q3 [9]}}