A2 June 2025 Q2
2.
\[\mathbf{A} = \begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} 1-\lambda & -2 \\ -2 & 4-\lambda \end{vmatrix} = 0\) leading to \((1-\lambda)(4-\lambda) - 4 = 0\) | M1 | 1.1b |
| Solves \(\lambda^2 - 5\lambda = 0 \Rightarrow \lambda = \ldots\) | dM1 | 1.1b |
| \(\lambda = 5,\ 0\) | A1 | 1.1b |
| (3) |
Notes
M1: Attempts \(\begin{vmatrix} 1-\lambda & -2 \\ -2 & 4-\lambda \end{vmatrix} = 0\)
dM1: Dependent on the previous method mark. Solves their 3TQ to find a value for \(\lambda\)
A1: Correct values for \(\lambda\)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{their } \text{‘}0\text{’}\begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(x - 2y = 0 \Rightarrow x = 2y\) or \(\begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{their } \text{‘}5\text{’}\begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(x - 2y = 5x \Rightarrow 2y = -4x\) or \(-2x + 4y = 5y \Rightarrow y = -2x\) | M1 | 3.1a |
| Eigenvectors \(\lambda = 0: \begin{pmatrix} 2 \\ 1 \end{pmatrix}\) and \(\lambda = 5: \begin{pmatrix} 1 \\ -2 \end{pmatrix}\) oe | A1 A1 | 2.2a 2.2a |
| eg \(\mathbf{P} = \begin{pmatrix} \dfrac{2}{\sqrt{5}} & \dfrac{1}{\sqrt{5}} \\[4pt] \dfrac{1}{\sqrt{5}} & \dfrac{-2}{\sqrt{5}} \end{pmatrix}\) or \(\mathbf{D} = \begin{pmatrix} 0 & 0 \\ 0 & 5 \end{pmatrix}\) | B1ft | 2.2a |
| eg \(\mathbf{P} = \begin{pmatrix} \dfrac{1}{\sqrt{5}} & \dfrac{2}{\sqrt{5}} \\[4pt] \dfrac{-2}{\sqrt{5}} & \dfrac{1}{\sqrt{5}} \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & 0 \end{pmatrix}\) | A1ft | 2.5 |
| (5) | ||
| (8 marks) |
Notes
Note this appears on epen as M1A1A1A1A1 but is being marked as M1A1A1B1A1
M1: Uses \(\begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{‘their } \lambda\text{’}\begin{pmatrix} x \\ y \end{pmatrix}\) to form an equation of the form \(ax = by\) for at least one of their eigenvalues. May be implied by a correct eigenvector.
A1: Deduces one correct eigenvector for one of the (correct) eigenvalues. Accept any non-zero multiple.
A1: Deduces both correct eigenvectors for the correct eigenvalues. Accept any non-zero multiples.
B1ft: Note: not dependent on the method mark. Deduces a correct matrix \(\mathbf{P}\) or \(\mathbf{D}\), following through on their eigenvalues or non-zero eigenvectors. This may be scored for a correct \(\mathbf{D}\) even if no attempt at the eigenvectors has been made, or a correct f.t. \(\mathbf{P}\) even if the method for eigenvectors was incorrect.
A1ft: Depends on the M having been scored. Correct matrices \(\mathbf{D}\) and \(\mathbf{P}\) which are consistent. Follow through on their eigenvalues and non-zero eigenvectors.
Note if they assume the eigenvector for \(\lambda = 0\) is 0 then do not allow this for the follow through mark(s) (though the first may be gained for correct D).
SC: If they mislabel P and D then allow A1ftA0ft for both correct but the wrong order, but A0A0 if only one is “correct” but wrong order.