| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Orthogonal matrix diagonalization |
| Difficulty | Standard +0.8 This FP3 question requires multiple eigenvalue/eigenvector techniques: finding eigenvalues from given eigenvectors, determining matrix entries from eigenvector conditions, finding eigenvectors from eigenvalues, and constructing an orthogonal diagonalization. While each individual step uses standard methods, the multi-part structure requiring careful algebraic manipulation and the final orthogonal diagonalization (P^T rather than P^{-1}) elevates this above routine exercises. The question tests deep understanding of the relationship between eigenvectors, eigenvalues, and matrix entries, plus knowledge of orthogonal matrices. |
| Spec | 4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}p&-2&0\\-2&6&-2\\0&-2&q\end{pmatrix}\begin{pmatrix}2\\-2\\1\end{pmatrix}=\lambda\begin{pmatrix}2\\-2\\1\end{pmatrix}\) | M1 | Sufficient if implied by one correct equation |
| \(-4-12-2=-2\lambda \Rightarrow \lambda=9\) | M1 A1 | M1: Compares \(y\)-components to obtain \(\lambda\). A1: Correct eigenvalue |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda=9 \Rightarrow 2p+4=18 \Rightarrow p=7\) | M1 A1 A1 | M1: Uses eigenvalue to form equation in \(p\) or \(q\). A1: either \(p=7\) or \(q=5\). A1: both \(p=7\) and \(q=5\) |
| \(\lambda=9 \Rightarrow 4+q=9 \Rightarrow q=5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}7&-2&0\\-2&6&-2\\0&-2&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=6\begin{pmatrix}x\\y\\z\end{pmatrix}\) giving at least 2 equations | M1 | Uses eigenvalue 6 and values of \(p\), \(q\) correctly |
| \(\begin{pmatrix}2\\1\\-2\end{pmatrix}\) or e.g. \(\begin{pmatrix}1\\\frac{1}{2}\\-1\end{pmatrix}\) | A1 | This vector or any multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{P}=\begin{pmatrix}2&2&1\\-2&1&2\\1&-2&2\end{pmatrix}\) | B1ft | Correct P: matrix of eigenvectors, two given in question plus eigenvector from (c). Allow ft if normalising attempted |
| \(\mathbf{D}=\begin{pmatrix}9&0&0\\0&6&0\\0&0&3\end{pmatrix}\) | M1 | Forms D with eigenvalues 6, 3 and their \(\lambda\) on leading diagonal, zeros elsewhere, or attempts \(\mathbf{P}^T\mathbf{MP}\) |
| \(\mathbf{P}=\frac{1}{3}\begin{pmatrix}2&2&1\\-2&1&2\\1&-2&2\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}9&0&0\\0&6&0\\0&0&3\end{pmatrix}\) | A1 | Fully correct and consistent matrices |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}p&-2&0\\-2&6&-2\\0&-2&q\end{pmatrix}\begin{pmatrix}2\\-2\\1\end{pmatrix}=\lambda\begin{pmatrix}2\\-2\\1\end{pmatrix}$ | M1 | Sufficient if implied by one correct equation |
| $-4-12-2=-2\lambda \Rightarrow \lambda=9$ | M1 A1 | M1: Compares $y$-components to obtain $\lambda$. A1: Correct eigenvalue |
**(3 marks)**
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda=9 \Rightarrow 2p+4=18 \Rightarrow p=7$ | M1 A1 A1 | M1: Uses eigenvalue to form equation in $p$ or $q$. A1: either $p=7$ or $q=5$. A1: both $p=7$ and $q=5$ |
| $\lambda=9 \Rightarrow 4+q=9 \Rightarrow q=5$ | | |
**(3 marks)**
---
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}7&-2&0\\-2&6&-2\\0&-2&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=6\begin{pmatrix}x\\y\\z\end{pmatrix}$ giving at least 2 equations | M1 | Uses eigenvalue 6 and values of $p$, $q$ correctly |
| $\begin{pmatrix}2\\1\\-2\end{pmatrix}$ or e.g. $\begin{pmatrix}1\\\frac{1}{2}\\-1\end{pmatrix}$ | A1 | This vector or any multiple |
**(2 marks)**
---
# Question 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{P}=\begin{pmatrix}2&2&1\\-2&1&2\\1&-2&2\end{pmatrix}$ | B1ft | Correct **P**: matrix of eigenvectors, two given in question plus eigenvector from (c). Allow ft if normalising attempted |
| $\mathbf{D}=\begin{pmatrix}9&0&0\\0&6&0\\0&0&3\end{pmatrix}$ | M1 | Forms **D** with eigenvalues 6, 3 and their $\lambda$ on leading diagonal, zeros elsewhere, or attempts $\mathbf{P}^T\mathbf{MP}$ |
| $\mathbf{P}=\frac{1}{3}\begin{pmatrix}2&2&1\\-2&1&2\\1&-2&2\end{pmatrix}$, $\mathbf{D}=\begin{pmatrix}9&0&0\\0&6&0\\0&0&3\end{pmatrix}$ | A1 | Fully correct and consistent matrices |
**(3 marks, Total 11)**6.
$$\mathbf { M } = \left( \begin{array} { r r r }
p & - 2 & 0 \\
- 2 & 6 & - 2 \\
0 & - 2 & q
\end{array} \right)$$
where $p$ and $q$ are constants.\\
Given that $\left( \begin{array} { r } 2 \\ - 2 \\ 1 \end{array} \right)$ is an eigenvector of the matrix $\mathbf { M }$,
\begin{enumerate}[label=(\alph*)]
\item find the eigenvalue corresponding to this eigenvector,
\item find the value of $p$ and the value of $q$.
Given that 6 is another eigenvalue of $\mathbf { M }$,
\item find a corresponding eigenvector.
Given that $\left( \begin{array} { l } 1 \\ 2 \\ 2 \end{array} \right)$ is a third eigenvector of $\mathbf { M }$ with eigenvalue 3
\item find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that
$$\mathbf { P } ^ { \mathrm { T } } \mathbf { M } \mathbf { P } = \mathbf { D }$$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2016 Q6 [11]}}