A2 June 2019 Paper 2 Q9
9
(a) Find the eigenvalues and corresponding eigenvectors of the matrix\[\mathbf{M} = \begin{bmatrix} \dfrac{1}{5} & \dfrac{2}{5} \\[12pt] \dfrac{-3}{5} & \dfrac{13}{10} \end{bmatrix}\] [5 marks]
(b) Find matrices \(\mathbf{U}\) and \(\mathbf{D}\) such that \(\mathbf{D}\) is a diagonal matrix and \(\mathbf{M} = \mathbf{UDU}^{-1}\) [2 marks]
(c) Given that \(\mathbf{M}^n \to \mathbf{L}\) as \(n \to \infty\), find the matrix \(\mathbf{L}\). [4 marks]
(d) The transformation represented by \(\mathbf{L}\) maps all points onto a line.
Find the equation of this line. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms correct characteristic equation and solves. (PI) Condone one error | M1 | 1.1a |
| Obtains the correct eigenvalues | A1 | 1.1b |
| Uses correct equation to find eigenvector for either \(\lambda = 1\) or \(\lambda = 0.5\) (PI) | M1 | 1.1a |
| Obtains correct eigenvector for \(\lambda = 1\) Allow any scalar multiple. | A1 | 1.1b |
| Obtains correct eigenvector for \(\lambda = 0.5\) Allow any scalar multiple. | A1 | 1.1b |
Typical solution
\[\left(\frac{1}{5} - \lambda\right)\left(\frac{13}{10} - \lambda\right) + \frac{6}{25} = 0\]\[0.5 - 1.5\lambda + \lambda^2 = 0\]\[\lambda = 1 \ \&\ \lambda = 0.5\]\[0 = \begin{bmatrix} \frac{-4}{5} & \frac{2}{5} \\[4pt] \frac{-3}{5} & \frac{3}{10} \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\]or
\[0 = \begin{bmatrix} \frac{-3}{10} & \frac{2}{5} \\[4pt] \frac{-3}{5} & \frac{4}{5} \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\]\[\lambda = 1 : \ \begin{bmatrix} 1 \\ 2 \end{bmatrix}\]\[\lambda = 0.5 : \ \begin{bmatrix} 4 \\ 3 \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Finds their correct \(\mathbf{U}\) with no zero column. | B1F | 1.1b |
| Finds their correct \(\mathbf{D}\) (must be consistent with their \(\mathbf{U}\)) | B1F | 1.1b |
Typical solution
\[\mathbf{U} = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} \text{ and } \mathbf{D} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix}\]or
\[\mathbf{U} = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} \text{ and } \mathbf{D} = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Finds correct \(\mathbf{U}^{-1}\) – CAO | B1 | 1.1b |
| Multiplies their matrices in correct order with powers taken inside \(\mathbf{D}^n\) | M1 | 1.1a |
| Correctly takes \(n \to \infty\) limit of their \(\mathbf{D}^n\), (\(\mathbf{D}\) must not contain only ones and zeros) | M1 | 2.2a |
| Obtains correct \(\mathbf{L}\) Correct answer scores 4/4 | R1 | 2.1 |
Typical solution
\[\mathbf{U}^{-1} = \frac{1}{5}\begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix}\]\[\mathbf{M}^n = \frac{1}{5}\begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} \left(\tfrac{1}{2}\right)^n & 0 \\ 0 & 1^n \end{bmatrix}\begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix}\]\[\mathbf{L} = \frac{1}{5}\begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix}\]\[\mathbf{L} = \begin{bmatrix} -0.6 & 0.8 \\ -1.2 & 1.6 \end{bmatrix}\]or
\[\mathbf{U}^{-1} = \frac{-1}{5}\begin{bmatrix} 3 & -4 \\ -2 & 1 \end{bmatrix}\]\[\mathbf{M}^n = \frac{-1}{5}\begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1^n & 0 \\ 0 & \left(\tfrac{1}{2}\right)^n \end{bmatrix}\begin{bmatrix} 3 & -4 \\ -2 & 1 \end{bmatrix}\]\[\mathbf{L} = \frac{-1}{5}\begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ -2 & 1 \end{bmatrix}\]\[\mathbf{L} = \begin{bmatrix} -0.6 & 0.8 \\ -1.2 & 1.6 \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains correct equations for image points from their \(\mathbf{L}\) PI by \(y = 2x\) | M1 | 3.1a |
| Obtains correct equation for the line Condone \(y^{\prime} = 2x^{\prime}\) | A1 | 3.2a |
| (13 marks) |