FP3 June 2016 Q6
6. \[\mathbf{M} = \begin{pmatrix}p & -2 & 0\\ -2 & 6 & -2\\ 0 & -2 & q\end{pmatrix}\] where \(p\) and \(q\) are constants.
Given that \(\begin{pmatrix}2\\ -2\\ 1\end{pmatrix}\) is an eigenvector of the matrix \(\mathbf{M}\),
Given that 6 is another eigenvalue of \(\mathbf{M}\),
Given that \(\begin{pmatrix}1\\ 2\\ 2\end{pmatrix}\) is a third eigenvector of \(\mathbf{M}\) with eigenvalue 3
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}p & -2 & 0\\ -2 & 6 & -2\\ 0 & -2 & q\end{pmatrix}\begin{pmatrix}2\\ -2\\ 1\end{pmatrix} = \lambda\begin{pmatrix}2\\ -2\\ 1\end{pmatrix}\) or \(\begin{pmatrix}p - \lambda & -2 & 0\\ -2 & 6 - \lambda & -2\\ 0 & -2 & q - \lambda\end{pmatrix}\begin{pmatrix}2\\ -2\\ 1\end{pmatrix} = \begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\) | M1 |
| \(-4 - 12 - 2 = -2\lambda \Rightarrow \lambda = 9\) | M1A1 |
| (3) |
Notes
M1: This statement is sufficient for this mark. May be implied by one correct equation e.g. \(2p + 4 = 2\lambda,\ -4 - 12 - 2 = -2\lambda,\ 4 + q = \lambda\)
M1: Compares \(y\)-components to obtain a value for \(\lambda\). Note that \(-4 - 12 - 2 = -2\lambda\) leading to a value for \(\lambda\) scores both method marks. If working is not clear, at least 2 terms of \(\text{"}-4 - 12 - 2\text{"}\) should be correct.
A1: Correct eigenvalue
| Scheme | Marks |
|---|---|
| \(\lambda = 9 \Rightarrow 2p + 4 = 18 \Rightarrow p = 7\) \(\lambda = 9 \Rightarrow 4 + q = 9 \Rightarrow q = 5\) | M1A1A1 |
| (3) |
Notes
M1: Uses their eigenvalue to form an equation in \(p\) or \(q\)
A1: Either \(p = 7\) or \(q = 5\)
A1: Both \(p = 7\) and \(q = 5\)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}7 & -2 & 0\\ -2 & 6 & -2\\ 0 & -2 & 5\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = 6\begin{pmatrix}x\\ y\\ z\end{pmatrix} \Rightarrow \begin{aligned}7x - 2y &= 6x\\ -2x + 6y - 2z &= 6y\\ -2y + 5z &= 6z\end{aligned}\) | M1 |
| \(\begin{pmatrix}2\\ 1\\ -2\end{pmatrix}\) or e.g. \(\begin{pmatrix}1\\ \frac{1}{2}\\ -1\end{pmatrix}\) | A1 |
| (2) |
Notes
M1: Uses the eigenvalue 6 and their value of \(p\) or \(q\) correctly to obtain at least 2 equations.
A1: This vector or any multiple of this vector.
Note that an eigenvector can be found from the cross product of any 2 rows of \(\mathbf{M} - 6\mathbf{I}\) e.g. \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -2 & 0 & -2\\ 0 & -2 & -1\end{vmatrix} = \begin{pmatrix}-4\\ -2\\ 4\end{pmatrix}\)
| Scheme | Marks |
|---|---|
| \(\mathbf{P} = \begin{pmatrix}2 & \text{"}2\text{"} & 1\\ -2 & \text{"}1\text{"} & 2\\ 1 & \text{"}-2\text{"} & 2\end{pmatrix}\) | B1ft |
| \(\mathbf{D} = \begin{pmatrix}\text{"}9\text{"} & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 3\end{pmatrix}\) | M1 |
| \(\left(\mathbf{P} = \dfrac{1}{3}\begin{pmatrix}2 & 2 & 1\\ -2 & 1 & 2\\ 1 & -2 & 2\end{pmatrix},\ \mathbf{D} = \begin{pmatrix}9 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 3\end{pmatrix}\right)\) or \(\left(\mathbf{P} = \begin{pmatrix}2 & -2 & 1\\ -2 & -1 & 2\\ 1 & 2 & 2\end{pmatrix},\ \mathbf{D} = \begin{pmatrix}81 & 0 & 0\\ 0 & 54 & 0\\ 0 & 0 & 27\end{pmatrix}\right)\) Fully correct and consistent matrices | A1 |
| (3) | |
| (11 marks) |
Notes
B1ft: Correct ft \(\mathbf{P}\). This should be a matrix of eigenvectors two of which are given in the question together with their eigenvector found from part (c). If an attempt is made to normalise the eigenvectors then allow the ft if slips are made when normalising.
M1: Forms the matrix \(\mathbf{D}\) by writing the eigenvalues 6, 3 and their \(\lambda\) on the leading diagonal and zeros elsewhere or attempts to calculate \(\mathbf{P}^{\mathrm{T}}\mathbf{MP}\) to obtain a single 3 by 3 matrix. Consistency not needed for this mark.
Note that the answers to part (d) may be implied e.g.
\(\mathbf{D} = \mathbf{P}^{\mathrm{T}}\mathbf{MP} = \begin{pmatrix}2 & -2 & 1\\ -2 & -1 & 2\\ 1 & 2 & 2\end{pmatrix}\begin{pmatrix}7 & -2 & 0\\ -2 & 6 & -2\\ 0 & -2 & 5\end{pmatrix}\begin{pmatrix}2 & -2 & 1\\ -2 & -1 & 2\\ 1 & 2 & 2\end{pmatrix} = \begin{pmatrix}81 & 0 & 0\\ 0 & 54 & 0\\ 0 & 0 & 27\end{pmatrix}\)
Would score all 3 marks by implication.