A2 June 2024 Q4
4.
\[\mathbf{A} = \begin{pmatrix} 4 & 2 & 0 \\ 2 & p & -2 \\ 0 & -2 & 2 \end{pmatrix} \qquad \text{where } p \text{ is a constant}\]Given that \(\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\) is an eigenvector of \(\mathbf{A}\),
| Scheme | Marks | AO |
|---|---|---|
| \(2 \times 4 - 1 \times 2 + 2 \times 0 = \lambda \times 2 \Rightarrow \lambda = \ldots\) or \(2 \times 0 - 1 \times -2 + 2 \times 2 = \lambda \times 2 \Rightarrow \lambda = \ldots\) Alternative 1 \(\begin{pmatrix} 4 & 2 & 0 \\ 2 & p & -2 \\ 0 & -2 & 2 \end{pmatrix}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ -p \\ 6 \end{pmatrix} = \lambda\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\) and deduces a value for \(\lambda\) Alternative 2 \(\begin{pmatrix} 4-\lambda & 2 & 0 \\ 2 & p-\lambda & -2 \\ 0 & -2 & 2-\lambda \end{pmatrix}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = 0 \Rightarrow \begin{aligned} 2(4-\lambda) - 2 &= 0 \\ 2 + 2(2-\lambda) &= 0 \end{aligned} \Rightarrow \lambda = \ldots\) | M1 | 1.1b |
| \(\lambda = 3\) | A1 | 2.2a |
| (2) |
Notes
M1: Uses either the first or third row of matrix with the eigenvector to form and solve an equation in \(\lambda\). May see the full matrix equation used, but this is not necessary.
Alternative 1: multiples the matrix by the eigenvector and compares to a multiple of the eigenvector to deduce a value for \(\lambda\)
Alternative 2: Uses \(\begin{pmatrix} 4-\lambda & 2 & 0 \\ 2 & p-\lambda & -2 \\ 0 & -2 & 2-\lambda \end{pmatrix}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = 0\) to find an equation and find a value for \(\lambda\)
A1: For the correct eigenvalue of 3.
| Scheme | Marks | AO |
|---|---|---|
| \(2 \times 2 - p - 2 \times 2 = 3 \times -1 \Rightarrow 4 - p - 4 = -3 \Rightarrow p = 3*\) Alternative 1 \((4-3)\left((p-3)(2-3) - 4\right) - 2\left(2(2-3) - 0\right) + 0 = 0 \Rightarrow p = 3*\) Using Alternative 2 from (a) \(4 - (p - \lambda) - 4 = 0\) uses \(\lambda = 3\) to show \(p = 3*\) | B1* | 1.1b |
| (1) |
Notes
B1*: Uses the second row with eigenvector 3 to set up a correct equation in \(p\) and then solves correctly.
Alternative 1: finds the characteristic equations, use \(\lambda = 3\) and correctly shows that \(p = 3\)
Alternative 2: Uses Alternative 2 from (a) and \(\lambda = 3\) to correctly show that \(p = 3\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \((4-\lambda)\left((3-\lambda)(2-\lambda) - 4\right) - 2\left(2(2-\lambda) - 0\right) + 0 = 0\) | M1 | 1.1b |
| \(\Rightarrow (4-\lambda)\left(\lambda^2 - 5\lambda + 2\right) - 8 + 4\lambda = 0 \Rightarrow \lambda^3 - 9\lambda^2 + 18\lambda = 0\) \(\Rightarrow \lambda(\lambda - 3)(\lambda - 6) = 0 \Rightarrow \lambda = \ldots\) | M1 | 1.1b |
| Remaining eigenvalues are 0 and 6. | A1 | 1.1b |
| (3) | ||
| (ii) \(\left.\begin{aligned} 4x + 2y &= 0 \\ 2x + 3y - 2z &= 0 \\ -2y + 2z &= 0 \end{aligned}\right\}\) or \(\left.\begin{aligned} -2x + 2y &= 0 \\ 2x - 3y - 2z &= 0 \\ -2y - 4z &= 0 \end{aligned}\right\} \Rightarrow x = .., y = .., z = ..\) | M1 | 2.1 |
| \(c\begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}\) or \(d\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\) o.e. | A1 | 1.1b |
| \(c\begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}\) and \(d\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\) o.e. | A1 | 1.1b |
| (3) |
Notes
(c)(i)
M1: Attempts the characteristic equation. Allow sign slips.
M1: Expands, simplifies and factorises to find the values, or equivalent method (e.g. solve by calculator).
A1: Correct remaining values.
(c)(ii)
M1: Correct method for one of the two required eigenvectors.
A1: One correct eigenvector. Allow any non-zero scalar multiple.
A1: Both correct eigenvectors. Allow any non-zero scalar multiple.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{D} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 0 \end{pmatrix}\) or \(\mathbf{D} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 6 \end{pmatrix}\) or equivalent | B1ft | 2.2a |
| \(\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 \Rightarrow \mathbf{v} = \ldots\) | M1 | 3.1a |
| \(\mathbf{P} = \tfrac{1}{3}\begin{pmatrix} 2 & 2 & -1 \\ -1 & 2 & 2 \\ 2 & -1 & 2 \end{pmatrix}\) or \(\mathbf{P} = \tfrac{1}{3}\begin{pmatrix} 2 & -1 & 2 \\ -1 & 2 & 2 \\ 2 & 2 & -1 \end{pmatrix}\) The order must correspond to their D | A1ft | 1.1b |
| (3) | ||
| (12 marks) |
Notes
B1ft: Correct diagonal matrix, following through on their eigenvalues.
M1: Normalises the eigenvectors.
A1ft: Forms correct matrix P with columns in appropriate order for their D, following through on their eigenvectors.