AS June 2022 Q2
2.
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
\[\mathbf{M} = \begin{pmatrix} 4 & 2 \\ 3 & -1 \end{pmatrix}\]Find a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that
\[\mathbf{P}^{-1}\mathbf{M}\mathbf{P} = \mathbf{D}\](7)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} 4-\lambda & 2 \\ 3 & -1-\lambda \end{vmatrix} = 0 \Rightarrow (4-\lambda)(-1-\lambda) - 6 = 0\) \(\Rightarrow \lambda^2 - 3\lambda - 10 = 0 \Rightarrow \lambda = \ldots\) | M1 | 3.1a |
| \((\Rightarrow (\lambda - 5)(\lambda + 2) = 0)\) so eigenvalues are \(-2\) and 5 | A1 | 1.1b |
| For \(\lambda = -2\) eigenvector equations are \(\left\{\begin{matrix} 4x + 2y = -2x \\ 3x - y = -2y \end{matrix}\right. \Rightarrow x, y = \ldots\) OR For \(\lambda = 5\) eigenvector equations are \(\left\{\begin{matrix} 4x + 2y = 5x \\ 3x - y = 5y \end{matrix}\right. \Rightarrow x, y = \ldots\) | M1 | 2.1 |
| (For \(\lambda = -2\), \(3x + y = 0\), for \(\lambda = 5\), \(x - 2y = 0\) so eigenvectors are) One of \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\) | A1 | 1.1b |
| Both of \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\) | A1 | 1.1b |
| Hence e.g. \(\mathbf{P} = \begin{pmatrix} 1 & 2 \\ -3 & 1 \end{pmatrix}\) or \(\mathbf{P} = \begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}\) | B1ft | 1.1b |
| and \(\mathbf{D} = \begin{pmatrix} -2 & 0 \\ 0 & 5 \end{pmatrix}\) or \(\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix}\) Note: If \(\mathbf{P}\) is given, their \(\mathbf{D}\) must be consistent with it to award this mark. | B1ft | 2.2a |
| (7) | ||
| (7 marks) |
Notes
M1: Begins the process of finding suitable matrices by attempting the eigenvalues of \(\mathbf{M}\).
A1: Correct eigenvalues.
M1: Correct method to find an eigenvector for either of their eigenvalues.
A1: One correct eigenvector – accept any non-zero multiples of their eigenvectors.
A1: A correct eigenvector for each of their eigenvalues.
B1ft: Gives \(\mathbf{P}\) as a matrix with their columns as the eigenvectors found.
B1ft: Gives \(\mathbf{D}\) as the matrix with eigenvalues on the diagonal – which must be in the correct order for their \(\mathbf{P}\) (if it is given).