| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find constant from eigenvalue condition |
| Difficulty | Standard +0.3 This is a straightforward eigenvalue problem requiring students to use the fact that diagonalization preserves eigenvalues. Part (a) involves finding k by equating trace or determinant (or characteristic polynomial), and part (b) requires finding eigenvectors—both standard FP2 procedures with no novel insight needed. Slightly easier than average due to the given diagonal form making the eigenvalues explicit. |
| Spec | 4.03g Invariant points and lines4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Eigenvalues are \(-3\) and \(8\) from diagonal matrix entries | B1 | Recalls eigenvalues form entries of diagonal matrix; may be implied by working |
| \(\begin{vmatrix} 3-\lambda & k \\ -5 & 2-\lambda \end{vmatrix} = 0 \Rightarrow (3-\lambda)(2-\lambda)+5k=0 \Rightarrow \lambda^2 - 5\lambda + 6 + 5k = 0 \Rightarrow 6+5k = -24\) | M1 | Full method to find \(k\) via characteristic equation using eigenvalue |
| \(k = -6\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(\lambda = -3\): \(\begin{cases} 6x - \text{"6"}y = 0 \\ -5x + 5y = 0 \end{cases}\) or for \(\lambda = 8\): \(\begin{cases} -5x - \text{"6"}y = 0 \\ -5x - 6y = 0 \end{cases}\) | M1 | Correct method to find eigenvector for either eigenvalue |
| One of \(\begin{pmatrix}1\\1\end{pmatrix}\) and \(\begin{pmatrix}6\\-5\end{pmatrix}\) | A1 | One correct eigenvector (any non-zero multiple) |
| Both of \(\begin{pmatrix}1\\1\end{pmatrix}\) and \(\begin{pmatrix}6\\-5\end{pmatrix}\) | A1 | Both correct eigenvectors |
| \(\mathbf{P} = \begin{pmatrix} 6 & 1 \\ -5 & 1 \end{pmatrix}\) | B1ft | Columns in correct order matching given diagonal matrix |
# Question 3:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Eigenvalues are $-3$ and $8$ from diagonal matrix entries | B1 | Recalls eigenvalues form entries of diagonal matrix; may be implied by working |
| $\begin{vmatrix} 3-\lambda & k \\ -5 & 2-\lambda \end{vmatrix} = 0 \Rightarrow (3-\lambda)(2-\lambda)+5k=0 \Rightarrow \lambda^2 - 5\lambda + 6 + 5k = 0 \Rightarrow 6+5k = -24$ | M1 | Full method to find $k$ via characteristic equation using eigenvalue |
| $k = -6$ | A1 | Correct value |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| For $\lambda = -3$: $\begin{cases} 6x - \text{"6"}y = 0 \\ -5x + 5y = 0 \end{cases}$ or for $\lambda = 8$: $\begin{cases} -5x - \text{"6"}y = 0 \\ -5x - 6y = 0 \end{cases}$ | M1 | Correct method to find eigenvector for either eigenvalue |
| One of $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}6\\-5\end{pmatrix}$ | A1 | One correct eigenvector (any non-zero multiple) |
| Both of $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}6\\-5\end{pmatrix}$ | A1 | Both correct eigenvectors |
| $\mathbf{P} = \begin{pmatrix} 6 & 1 \\ -5 & 1 \end{pmatrix}$ | B1ft | Columns in correct order matching given diagonal matrix |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
$$\mathbf { A } = \left( \begin{array} { r r }
3 & k \\
- 5 & 2
\end{array} \right)$$
where $k$ is a constant.\\
Given that there exists a matrix $\mathbf { P }$ such that $\mathbf { P } ^ { - 1 } \mathbf { A P }$ is a diagonal matrix where
$$\mathbf { P } ^ { - 1 } \mathbf { A } \mathbf { P } = \left( \begin{array} { r r }
8 & 0 \\
0 & - 3
\end{array} \right)$$
(a) show that $k = - 6$\\
(b) determine a suitable matrix $\mathbf { P }$
\hfill \mbox{\textit{Edexcel FP2 AS 2024 Q3 [7]}}