AS June 2023 Q2
2. A linear transformation \(T : \mathbb{R}^2 \to \mathbb{R}^2\) is represented by the matrix
\[\mathbf{M} = \begin{pmatrix} 5 & 1 \\ k & -3 \end{pmatrix}\]where \(k\) is a constant.
Given that matrix \(\mathbf{M}\) has a repeated eigenvalue,
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} 5-\lambda & 1 \\ k & -3-\lambda \end{vmatrix} = (5-\lambda)(-3-\lambda) - k = 0\) | M1 | 2.1 |
| \(\lambda^2 - 2\lambda - k - 15 = 0\) | A1 | 1.1b |
| \(b^2 - 4ac = (-2)^2 - 4(1)(-k - 15) = 0 \Rightarrow k = \ldots\) or \(-k - 15 = 1 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k = -16\) | A1 | 1.1b |
| (4) |
Notes
M1: Starts the process by finding the determinant of \(\mathbf{M} - \lambda\mathbf{I}\) and sets = 0.
A1: Correct quadratic equation.
M1: Sets the discriminant of their 3TQ = 0 to find a value for \(k\). Alternatively identifies that for a perfect square the constant term must equal 1 and finds a value for \(k\).
A1: Correct value for \(k\).
| Scheme | Marks | AO |
|---|---|---|
| \(\lambda^2 - 2\lambda + 1 = 0 \Rightarrow \lambda = \ldots\) | M1 | 1.1b |
| \(\lambda = 1\) | A1 | 1.1b |
| (2) |
Notes
M1: Uses their value of \(k\) to form and solve their 3TQ to find the eigenvalue. This mark can be implied by a correct eigenvalue.
A1: Correct eigenvalue
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 5 & 1 \\ \text{their } k & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(ax + by = 0\) or \(ax = by\) \(\begin{pmatrix} 5 & 1 \\ \text{their } k & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{their } \lambda\begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(ax + by = 0\) or \(ax = by\) Alternative approaches \(\begin{pmatrix} 5 & 1 \\ -16 & -3 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} x \\ mx \end{pmatrix}\) leading to \(\left.\begin{matrix} 5x + mx = x \\ -16x - 3mx = mx \end{matrix}\right\}; {-16x} - 3mx = m(5x + mx)\) Leading to a value for \(m\) \(\{m^2 + 8m + 16 = 0\}\) \(\begin{pmatrix} 5 & 1 \\ -16 & -3 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} x \\ mx + c \end{pmatrix}\) leading to \(\left.\begin{matrix} 5x + mx + c = x \\ -16x - 3mx - 3c = mx + c \end{matrix}\right\}; {-16x} - 3mx - 3c = m(5x + mx + c) + c\) Leading to a value for \(m\) \(\{-3c = mc + c\ \text{ or }\ m^2 + 8m + 16 = 0\}\) | M1 | 2.1 |
| \(y = -4x\) o.e. | A1 | 1.1b |
| (2) | ||
| (8 marks) |
Notes
M1: Constructs a rigorous argument using their eigenvalue to find the Cartesian equation of the invariant line.
A1: Correct equation.