| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| 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 application of the characteristic equation for a 2×2 matrix. Finding k from the repeated eigenvalue condition uses the discriminant being zero (standard technique), then finding the invariant line from the eigenvector is routine. Slightly above average due to being Further Maths content, but mechanically straightforward with no novel insight required. |
| 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 matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{vmatrix} 5-\lambda & 1 \\ k & -3-\lambda \end{vmatrix} = (5-\lambda)(-3-\lambda) - k = 0\) | M1 | 2.1 - Starts the process by finding the determinant of \(\mathbf{M} - l\mathbf{I}\) and sets \(= 0\) |
| \(l^2 - 2l - k - 15 = 0\) | A1 | 1.1b - Correct quadratic equation |
| \(b^2 - 4ac = (-2)^2 - 4(1)(-k-15) = 0 \Rightarrow k = \ldots\) | M1 | 1.1b - Sets discriminant of their \(3TQ = 0\) to find \(k\); or \(-k-15=1\) giving \(k=\ldots\) |
| \(k = -16\) | A1 | 1.1b - Correct value for \(k\) |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(l^2 - 2l + 1 = 0 \Rightarrow l = \ldots\) | M1 | 1.1b - Uses their value of \(k\) to form and solve their \(3TQ\) to find eigenvalue |
| \(l = 1\) | A1 | 1.1b - Correct eigenvalue |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix} 5 & 1 \\ \text{their } k & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(ax + by = 0\) or \(ax = by\); or using \(\begin{pmatrix} x \\ mx \end{pmatrix}\): \(5x + mx = x\), \(-16x - 3mx = mx\) giving \(\{m^2 + 8m + 16 = 0\}\) | M1 | 2.1 - Constructs a rigorous argument using their eigenvalue to find the Cartesian equation of the invariant line |
| \(y = -4x\) o.e. | A1 | 1.1b - Correct equation |
| (2 marks) |
## Question 2:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix} 5-\lambda & 1 \\ k & -3-\lambda \end{vmatrix} = (5-\lambda)(-3-\lambda) - k = 0$ | M1 | 2.1 - Starts the process by finding the determinant of $\mathbf{M} - l\mathbf{I}$ and sets $= 0$ |
| $l^2 - 2l - k - 15 = 0$ | A1 | 1.1b - Correct quadratic equation |
| $b^2 - 4ac = (-2)^2 - 4(1)(-k-15) = 0 \Rightarrow k = \ldots$ | M1 | 1.1b - Sets discriminant of their $3TQ = 0$ to find $k$; or $-k-15=1$ giving $k=\ldots$ |
| $k = -16$ | A1 | 1.1b - Correct value for $k$ |
| **(4 marks)** | | |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $l^2 - 2l + 1 = 0 \Rightarrow l = \ldots$ | M1 | 1.1b - Uses their value of $k$ to form and solve their $3TQ$ to find eigenvalue |
| $l = 1$ | A1 | 1.1b - Correct eigenvalue |
| **(2 marks)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix} 5 & 1 \\ \text{their } k & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$ leading to $ax + by = 0$ or $ax = by$; or using $\begin{pmatrix} x \\ mx \end{pmatrix}$: $5x + mx = x$, $-16x - 3mx = mx$ giving $\{m^2 + 8m + 16 = 0\}$ | M1 | 2.1 - Constructs a rigorous argument using their eigenvalue to find the Cartesian equation of the invariant line |
| $y = -4x$ o.e. | A1 | 1.1b - Correct equation |
| **(2 marks)** | | |
---\begin{enumerate}
\item A linear transformation $T : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ is represented by the matrix
\end{enumerate}
$$\mathbf { M } = \left( \begin{array} { c c }
5 & 1 \\
k & - 3
\end{array} \right)$$
where $k$ is a constant.\\
Given that matrix $\mathbf { M }$ has a repeated eigenvalue,\\
(a) determine\\
(i) the value of $k$\\
(ii) the eigenvalue.\\
(b) Hence determine a Cartesian equation of the invariant line under $T$.
\hfill \mbox{\textit{Edexcel FP2 AS 2023 Q2 [8]}}