| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Orthogonal matrix diagonalization |
| Difficulty | Challenging +1.2 This is a structured multi-part eigenvalue/eigenvector question with significant scaffolding. Part (a) is straightforward matrix-vector multiplication given the eigenvector. Part (b) requires solving (M - 8I)v = 0, which is routine. Part (c) involves finding the third eigenvalue (via trace), its eigenvector, normalizing all three eigenvectors, and constructing P—this is methodical but involves multiple standard steps. The orthogonal diagonalization is a core Further Maths topic, making this slightly above average difficulty but well within expected scope for F3. |
| Spec | 4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3\) | B1 | Correct value seen in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}-2 & 5 & 0\\5 & 1 & -3\\0 & -3 & 6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}8x\\8y\\8z\end{pmatrix}\) giving \(-2x+5y=8x\), \(5x+y-3z=8y\), \(-3y+6z=8z\) | M1 | Correct method for eigenvector (making a variable equal to 0 is not a correct method) |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\-3\end{pmatrix}\) | A1 | Any correct eigenvector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ | \mathbf{M} - \lambda\mathbf{I}\ | = \begin{vmatrix}-2-\lambda & 5 & 0\\5 & 1-\lambda & -3\\0 & -3 & 6-\lambda\end{vmatrix} = 0 \Rightarrow (-2-\lambda)\left[(1-\lambda)(6-\lambda)-9\right] - 5\left[5(6-\lambda)\right] = 0\) |
| \(\lambda = -6\) | A1 | Correct third eigenvalue. Work for these 2 marks may be seen in (a). Correct third eigenvalue by different method – send to review |
| \(\mathbf{D} = \begin{pmatrix}3 & 0 & 0\\0 & 8 & 0\\0 & 0 & -6\end{pmatrix}\) | A1ft | Correct D following through their third eigenvalue |
| \(\begin{pmatrix}-2 & 5 & 0\\5 & 1 & -3\\0 & -3 & 6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-6x\\-6y\\-6z\end{pmatrix} \Rightarrow \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-5\\4\\1\end{pmatrix}\) | M1 | Correct strategy for 3rd eigenvector |
| \(\mathbf{P} = \begin{pmatrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{14}} & \frac{-5}{\sqrt{42}}\\ \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{14}} & \frac{4}{\sqrt{42}}\\ \frac{1}{\sqrt{3}} & \frac{-3}{\sqrt{14}} & \frac{1}{\sqrt{42}}\end{pmatrix}\) | A1 | Fully correct matrix consistent with their D. May have \(\frac{\sqrt{3}}{3}\) etc |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3$ | B1 | Correct value seen in (a) |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-2 & 5 & 0\\5 & 1 & -3\\0 & -3 & 6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}8x\\8y\\8z\end{pmatrix}$ giving $-2x+5y=8x$, $5x+y-3z=8y$, $-3y+6z=8z$ | M1 | Correct method for eigenvector (making a variable equal to 0 is not a correct method) |
| $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\-3\end{pmatrix}$ | A1 | Any correct eigenvector |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|\mathbf{M} - \lambda\mathbf{I}\| = \begin{vmatrix}-2-\lambda & 5 & 0\\5 & 1-\lambda & -3\\0 & -3 & 6-\lambda\end{vmatrix} = 0 \Rightarrow (-2-\lambda)\left[(1-\lambda)(6-\lambda)-9\right] - 5\left[5(6-\lambda)\right] = 0$ | M1 | NB characteristic equation is $\lambda^3 - 5\lambda^2 - 42\lambda + 144 = 0$ but may only find constant term |
| $\lambda = -6$ | A1 | Correct third eigenvalue. Work for these 2 marks may be seen in (a). Correct third eigenvalue by different method – send to review |
| $\mathbf{D} = \begin{pmatrix}3 & 0 & 0\\0 & 8 & 0\\0 & 0 & -6\end{pmatrix}$ | A1ft | Correct **D** following through their third eigenvalue |
| $\begin{pmatrix}-2 & 5 & 0\\5 & 1 & -3\\0 & -3 & 6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-6x\\-6y\\-6z\end{pmatrix} \Rightarrow \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-5\\4\\1\end{pmatrix}$ | M1 | Correct strategy for 3rd eigenvector |
| $\mathbf{P} = \begin{pmatrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{14}} & \frac{-5}{\sqrt{42}}\\ \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{14}} & \frac{4}{\sqrt{42}}\\ \frac{1}{\sqrt{3}} & \frac{-3}{\sqrt{14}} & \frac{1}{\sqrt{42}}\end{pmatrix}$ | A1 | Fully correct matrix consistent with their **D**. May have $\frac{\sqrt{3}}{3}$ etc |3.
$$\mathbf { M } = \left( \begin{array} { r r r }
- 2 & 5 & 0 \\
5 & 1 & - 3 \\
0 & - 3 & 6
\end{array} \right)$$
Given that $\mathbf { i } + \mathbf { j } + \mathbf { k }$ is an eigenvector of $\mathbf { M }$,
\begin{enumerate}[label=(\alph*)]
\item determine the corresponding eigenvalue.
Given that 8 is an eigenvalue of $\mathbf { M }$,
\item determine a corresponding eigenvector.
\item Determine a diagonal matrix $\mathbf { D }$ and an orthogonal matrix $\mathbf { P }$ such that
$$\mathbf { D } = \mathbf { P } ^ { \mathrm { T } } \mathbf { M P }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2022 Q3 [8]}}