| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A = PDP⁻¹ |
| Difficulty | Standard +0.3 This is a standard diagonalization problem requiring finding eigenvalues (via characteristic equation), eigenvectors, and constructing P and D. While it involves multiple steps, it's a routine textbook exercise with a 2×2 matrix requiring no novel insight—slightly easier than average for Further Maths students who have practiced this technique. |
| Spec | 4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\begin{vmatrix} 4-\lambda & 2 \\ 3 & -1-\lambda \end{vmatrix} = 0 \Rightarrow (4-\lambda)(-1-\lambda)-6=0\) | M1 | 3.1a |
| \(\Rightarrow \lambda^2 - 3\lambda - 10 = 0\) | ||
| \((\Rightarrow (\lambda-5)(\lambda+2)=0)\) so eigenvalues are \(-2\) and \(5\) | A1 | 1.1b |
| For \(\lambda=-2\): \(\begin{cases} 4x+2y=-2x \\ 3x-y=-2y \end{cases} \Rightarrow x,y=\ldots\) OR For \(\lambda=5\): \(\begin{cases} 4x+2y=5x \\ 3x-y=5y \end{cases} \Rightarrow x,y=\ldots\) | M1 | 2.1 |
| For \(\lambda=-2\): \(3x+y=0\); for \(\lambda=5\): \(x-2y=0\). One of \(\begin{pmatrix}1\\-3\end{pmatrix}\) and \(\begin{pmatrix}2\\1\end{pmatrix}\) | A1 | 1.1b |
| Both of \(\begin{pmatrix}1\\-3\end{pmatrix}\) and \(\begin{pmatrix}2\\1\end{pmatrix}\) | A1 | 1.1b |
| \(\mathbf{P}=\begin{pmatrix}1&2\\-3&1\end{pmatrix}\) or \(\mathbf{P}=\begin{pmatrix}2&1\\1&-3\end{pmatrix}\) | B1ft | 1.1b |
| \(\mathbf{D}=\begin{pmatrix}-2&0\\0&5\end{pmatrix}\) or \(\mathbf{D}=\begin{pmatrix}5&0\\0&-2\end{pmatrix}\). Note: if \(\mathbf{P}\) is given, \(\mathbf{D}\) must be consistent with it | B1ft | 2.2a |
## Question 2:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix} 4-\lambda & 2 \\ 3 & -1-\lambda \end{vmatrix} = 0 \Rightarrow (4-\lambda)(-1-\lambda)-6=0$ | M1 | 3.1a |
| $\Rightarrow \lambda^2 - 3\lambda - 10 = 0$ | | |
| $(\Rightarrow (\lambda-5)(\lambda+2)=0)$ so eigenvalues are $-2$ and $5$ | A1 | 1.1b |
| For $\lambda=-2$: $\begin{cases} 4x+2y=-2x \\ 3x-y=-2y \end{cases} \Rightarrow x,y=\ldots$ OR For $\lambda=5$: $\begin{cases} 4x+2y=5x \\ 3x-y=5y \end{cases} \Rightarrow x,y=\ldots$ | M1 | 2.1 |
| For $\lambda=-2$: $3x+y=0$; for $\lambda=5$: $x-2y=0$. One of $\begin{pmatrix}1\\-3\end{pmatrix}$ and $\begin{pmatrix}2\\1\end{pmatrix}$ | A1 | 1.1b |
| Both of $\begin{pmatrix}1\\-3\end{pmatrix}$ and $\begin{pmatrix}2\\1\end{pmatrix}$ | A1 | 1.1b |
| $\mathbf{P}=\begin{pmatrix}1&2\\-3&1\end{pmatrix}$ or $\mathbf{P}=\begin{pmatrix}2&1\\1&-3\end{pmatrix}$ | B1ft | 1.1b |
| $\mathbf{D}=\begin{pmatrix}-2&0\\0&5\end{pmatrix}$ or $\mathbf{D}=\begin{pmatrix}5&0\\0&-2\end{pmatrix}$. Note: if $\mathbf{P}$ is given, $\mathbf{D}$ must be consistent with it | B1ft | 2.2a |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
$$\mathbf { M } = \left( \begin{array} { r r }
4 & 2 \\
3 & - 1
\end{array} \right)$$
Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that
$$\mathbf { P } ^ { - 1 } \mathbf { M P } = \mathbf { D }$$
\hfill \mbox{\textit{Edexcel FP2 AS 2022 Q2 [7]}}