AS June 2025 Q4
4.
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
\[\mathbf{A} = \begin{pmatrix} 6 & -1 \\ 2 & 3 \end{pmatrix}\]Determine a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that
\[\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}\](7)
| Scheme | Marks | AO |
|---|---|---|
| \(|\mathbf{A} - \lambda\mathbf{I}| = 0 \Rightarrow \begin{vmatrix} 6-\lambda & -1 \\ 2 & 3-\lambda \end{vmatrix} = 0 \Rightarrow (6-\lambda)(3-\lambda) + 2 = 0\) \(\Rightarrow \lambda^2 - 9\lambda + 20 = 0 \Rightarrow \lambda = \ldots\) | M1 | 3.1a |
| \(\lambda = 5,\ 4\) | A1 | 1.1b |
| \(\lambda = 5 \Rightarrow \begin{matrix} 6x - y = 5x \\ 2x + 3y = 5y \end{matrix} \Rightarrow x = \ldots,\ y = \ldots\) or \(\lambda = 4 \Rightarrow \begin{matrix} 6x - y = 4x \\ 2x + 3y = 4y \end{matrix} \Rightarrow x = \ldots,\ y = \ldots\) | M1 | 2.1 |
| \(\lambda = 5 \to \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) or \(\lambda = 4 \to \begin{pmatrix} 1 \\ 2 \end{pmatrix}\) | A1 | 1.1b |
| \(\lambda = 5 \to \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\lambda = 4 \to \begin{pmatrix} 1 \\ 2 \end{pmatrix}\) | A1 | 1.1b |
| \(\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\) or \(\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}\) | B1ft | 1.1b |
| \(\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & 4 \end{pmatrix}\) or \(\mathbf{D} = \begin{pmatrix} 4 & 0 \\ 0 & 5 \end{pmatrix}\) | B1ft | 2.2a |
| (7) | ||
| (7 marks) |
Notes
M1: Makes a start to the problem by finding the eigenvalues of \(\mathbf{A}\)
A1: Correct eigenvalues
M1: Uses a correct method to find an eigenvector for any eigenvalue
A1: One correct eigenvector (allow any integer multiple)
A1: Both correct eigenvectors (allow any integer multiples)
B1ft: For a matrix \(\mathbf{P}\) with their eigenvectors as columns
B1ft: For \(\mathbf{D}\) as a matrix with their eigenvalues on the leading diagonal. Must be consistent with their \(\mathbf{P}\) if \(\mathbf{P}\) is attempted.
Note method must be shown for full marks. Answer that obtain eigenvectors from a calculator can score M1A1 (if method shown for eigenvalues) M0A0A0 B1ftB1ft max.