A2 June 2022 Paper 2 Q11
11
(a) Find the eigenvalues and corresponding eigenvectors of the matrix\[\mathbf{M} = \begin{bmatrix} \dfrac{5}{2} & -\dfrac{3}{2} \\[6pt] -\dfrac{3}{2} & \dfrac{13}{2} \end{bmatrix}\] [5 marks]
(b)
(i) Describe how the directions of the invariant lines of the transformation represented by \(\mathbf{M}\) are related to each other.
Fully justify your answer. [2 marks]
(ii) Describe fully the transformation represented by \(\mathbf{M}\) [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms correct characteristic equation and solves. PI by correct eigenvalues. Condone one error. | M1 | 1.1a |
| Obtains the correct eigenvalues | A1 | 1.1b |
| Uses correct equation to find eigenvector for one of their two eigenvalues. PI by a correct eigenvector | M1 | 1.1a |
| Obtains a correct eigenvector for one of their two eigenvalues. Allow any scalar multiple | A1F | 1.1b |
| Obtains both correct eigenvectors paired with the corresponding eigenvalue. Allow any scalar multiple | A1 | 1.1b |
| (5) |
Typical solution
\[0 = \left(\frac{5}{2} - \lambda\right)\left(\frac{13}{2} - \lambda\right) - \frac{9}{4}\]\[0 = \lambda^2 - 9\lambda + 14\]\[\lambda = 2 \ \&\ \lambda = 7\]\[\lambda = 2:\ \mathbf{0} = \begin{bmatrix} \dfrac{1}{2} & \dfrac{-3}{2} \\[6pt] \dfrac{-3}{2} & \dfrac{9}{2} \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}\]\[\lambda = 2:\ \begin{bmatrix}3 \\ 1\end{bmatrix}\]\[\lambda = 7:\ \mathbf{0} = \begin{bmatrix} \dfrac{-9}{2} & \dfrac{-3}{2} \\[6pt] \dfrac{-3}{2} & \dfrac{-1}{2} \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}\]\[\lambda = 7:\ \begin{bmatrix}-1 \\ 3\end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| (i) Compares directions of their eigenvectors eg considers gradients or obtains scalar product | M1 | 2.1 |
| Deduces that the invariant lines are perpendicular | R1 | 2.2a |
| (2) | ||
| (ii) Deduces that it is a two-way stretch | M1 | 2.2a |
| Describes the transformation fully. Follow through their eigenvalues and corresponding eigenvectors | A1F | 1.1b |
| (2) | ||
| (9 marks) |
Typical solution
(i)
\[\begin{bmatrix}3 \\ 1\end{bmatrix} \cdot \begin{bmatrix}-1 \\ 3\end{bmatrix} = 0\]So the invariant lines are perpendicular.
(ii)
Stretch parallel to \(y = \dfrac{1}{3}x\), SF = 2
Stretch parallel to \(y = -3x\), SF = 7