AS June 2018 Q4
4.
\[\mathbf{A} = \begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}\]Find a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D} = \mathbf{P}^{-1}\mathbf{A}\mathbf{P}\)
(7)
| Scheme | Marks | AO |
|---|---|---|
| \(|\mathbf{A} - \lambda\mathbf{I}| = 0 \Rightarrow \begin{vmatrix} 1-\lambda & 1 \\ -2 & 4-\lambda \end{vmatrix} = 0 \Rightarrow (1-\lambda)(4-\lambda) + 2 = 0\) | M1 | 3.1a |
| \(\lambda_1 = 2, \quad \lambda_2 = 3\) | A1 | 1.1b |
| \(\begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 2\begin{pmatrix} x \\ y \end{pmatrix}\) or \(\begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 3\begin{pmatrix} x \\ y \end{pmatrix}\) | M1 | 2.1 |
| \(2, \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) or \(3, \begin{pmatrix} 1 \\ 2 \end{pmatrix}\) | A1 | 1.1b |
| \(2, \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(3, \begin{pmatrix} 1 \\ 2 \end{pmatrix}\) | A1 | 1.1b |
| \(\mathbf{D} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}\) | B1ft | 1.1b |
| \(\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\) | B1ft | 2.2a |
| (7) | ||
| (7 marks) |
Notes
M1: Correct strategy for finding eigenvalues
A1: Correct eigenvalues
M1: Uses at least one of their eigenvalues correctly to find a corresponding eigenvector
A1: One correct eigenvalue/eigenvector pair
A1: Both pairs correct
B1ft: Correct follow through \(\mathbf{D}\) or \(\mathbf{P}\) clearly identified as \(\mathbf{D}\) or \(\mathbf{P}\)
B1ft: \(\mathbf{P}\) and \(\mathbf{D}\) both correct and consistent and identified as \(\mathbf{D}\) and \(\mathbf{P}\)
Note that the correct matrices may be implied by e.g.\[\begin{pmatrix} * & * \\ * & * \end{pmatrix}\begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}\]