A2 October 2021 Q8
8.
\[\mathbf{A} = \begin{pmatrix} 5 & -2 & 5 \\ 0 & 3 & p \\ -6 & 6 & -4 \end{pmatrix} \qquad \text{where } p \text{ is a constant}\]Given that \(\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}\) is an eigenvector for \(\mathbf{A}\)
With respect to a fixed origin \(O\), the velocity of a particle moving through space is modelled by
\[\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \mathbf{A}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\]By considering \(\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \mathbf{P}^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\) so that \(\begin{pmatrix} \dot{u} \\ \dot{v} \\ \dot{w} \end{pmatrix} = \mathbf{P}^{-1}\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 5 & -2 & 5 \\ 0 & 3 & p \\ -6 & 6 & -4 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 - 2p \\ 2 \end{pmatrix} = -1 \times \begin{pmatrix} 2 \\ 2p - 3 \\ -2 \end{pmatrix}\) | ||
| Corresponding eigenvalue is \(-1\) | B1 | 1.1b |
| (1) |
Notes
B1: For the correct eigenvalue of \(-1\)
| Scheme | Marks | AO |
|---|---|---|
| \(2p - 3 = 1 \Rightarrow p = \ldots\) | M1 | 1.1b |
| \(p = 2\) * | A1* | 1.1b |
| (2) |
Notes
M1: Correct equation with their eigenvalue set up – need only see middle equation for this.
A1*: Correct proof (full matrix calculation not necessary).
| Scheme | Marks | AO |
|---|---|---|
| \(\det\begin{pmatrix} 5 - \lambda & -2 & 5 \\ 0 & 3 - \lambda & p \\ -6 & 6 & -4 - \lambda \end{pmatrix} = 0\) \(\Rightarrow (5 - \lambda)\left((3 - \lambda)(-4 - \lambda) - 12\right) - (-2)(12) + 5\left(6(3 - \lambda)\right) = 0\) | M1 | 1.1b |
| \(\Rightarrow \lambda^3 - 4\lambda^2 + \lambda + 6 = 0\) | A1 | 1.1b |
| \(\left(\Rightarrow (\lambda + 1)(\lambda^2 - 5\lambda + 6) = 0 \Rightarrow (\lambda + 1)(\lambda - 2)(\lambda - 3) = 0\right)\) Eigenvalues are \((-1)\), 2 and 3 | A1 | 1.1b |
| Either \(\left.\begin{matrix} 3x - 2y + 5z = 0 \\ y + 2z = 0 \\ -6x + 6y - 6z = 0 \end{matrix}\right\}\) or \(\left.\begin{matrix} 2x - 2y + 5z = 0 \\ 2z = 0 \\ -6x + 6y - 7z = 0 \end{matrix}\right\} \Rightarrow x / y / z = \ldots\) | M1 | 2.1 |
| Either \(k\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}\) (for \(\lambda = 2\)) or \(m\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) (for \(\lambda = 3\)) | A1 | 1.1b |
| Both \(\left.\begin{matrix} 3x - 2y + 5z = 0 \\ y + 2z = 0 \\ -6x + 6y - 6z = 0 \end{matrix}\right\}\) and \(\left.\begin{matrix} 2x - 2y + 5z = 0 \\ 2z = 0 \\ -6x + 6y - 7z = 0 \end{matrix}\right\} \Rightarrow x / y / z = \ldots\) | M1 | 2.1 |
| Both \(k\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}\) (for \(\lambda = 2\)) and \(m\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) (for \(\lambda = 3\)) | A1 | 1.1b |
| (7) |
Notes
M1: Applies \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) to achieve a cubic in \(\lambda\) (or other variable, simplification not required). Allow with \(p\) used instead of 2, and look for two correct “terms” in the expansion leading to a cubic as evidence of the expansion.
A1: Correct simplified cubic. Note this may be implied by correct answers from a calculator following a correct expansion seen for the M.
A1: Correct eigenvalues
M1: Forms and solves eigenvector equations for at least one (other than \(-1\)) eigenvalue.
A1: One correct (other) eigenvector
M1: Both eigenvectors attempted.
A1: Both (other) eigenvectors correct.
| Scheme | Marks | AO |
|---|---|---|
| E.g. \(\mathbf{P} = \begin{pmatrix} 2 & 3 & 1 \\ 1 & 2 & 1 \\ -2 & -1 & 0 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\) | B1ft | 2.2a |
| (1) |
Notes
B1ft: A correct corresponding P and D, follow through on their answer to (a). Columns may be in different order, but should be consistent for their P and D.
| Scheme | Marks | AO |
|---|---|---|
| \(\dot{u} = ku \Rightarrow \displaystyle\int \frac{1}{u}\,\mathrm{d}u = k\int \mathrm{d}t \Rightarrow \ln u = kt (+c)\) | M1 | 1.1b |
| So \(u = A\mathrm{e}^{kt}\) or \(u = \mathrm{e}^{kt + c}\) | A1 | 1.1b |
| (2) |
Notes
M1: Separates variables and attempts the integration (constant not required).
A1: Correct answer for \(u = \ldots\), either form, including constant of integration
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \mathbf{PDP}^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix} \Rightarrow \begin{pmatrix} \dot{u} \\ \dot{v} \\ \dot{w} \end{pmatrix} = \mathbf{P}^{-1}\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \mathbf{D}\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} -u \\ 2v \\ 3w \end{pmatrix}\) | M1 | 3.1b |
| \(\Rightarrow \begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} A\mathrm{e}^{-t} \\ B\mathrm{e}^{2t} \\ C\mathrm{e}^{3t} \end{pmatrix}\) | M1 | 2.2a |
| \(\Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{P}\begin{pmatrix} A\mathrm{e}^{-t} \\ B\mathrm{e}^{2t} \\ C\mathrm{e}^{3t} \end{pmatrix} = \ldots\) | M1 | 3.4 |
| \(\Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2A\mathrm{e}^{-t} + 3B\mathrm{e}^{2t} + C\mathrm{e}^{3t} \\ A\mathrm{e}^{-t} + 2B\mathrm{e}^{2t} + C\mathrm{e}^{3t} \\ -2A\mathrm{e}^{-t} - B\mathrm{e}^{2t} \end{pmatrix}\) | A1 | 1.1b |
| (4) | ||
| (17 marks) |
Notes
NB different orderings of the columns of P and D will give the terms in different orders here.
M1: Uses their P and D to transform system into equation in \(u\), \(v\) and \(w\) (may be implied).
M1: Forms the solution for \(u\), \(v\) and \(w\) using their eigenvalues.
M1: Reverses the substitution (multiplies by their P) to get solution for \(x\), \(y\) and \(z\).
A1: Correct answer, in matrix form or as separate equations – award when first seen and isw.