FP3 June 2018 Q3
3. \[\mathbf{M} = \begin{pmatrix}3 & k & 2\\ -1 & 0 & 1\\ 1 & k & 1\end{pmatrix}, \quad \text{where } k \text{ is a constant}\]
Given that 3 is an eigenvalue of \(\mathbf{M}\),
| Scheme | Marks |
|---|---|
| \(\mathbf{M} = \begin{pmatrix}3 & k & 2\\ -1 & 0 & 1\\ 1 & k & 1\end{pmatrix}\) | |
| \(\begin{vmatrix}3 - 3 & k & 2\\ -1 & -3 & 1\\ 1 & k & 1 - 3\end{vmatrix} = (3 - 3)\left[-3(1 - 3) - k\right] - k\left[-1(1 - 3) - 1\right] + 2(-k + 3)\ \ (= -3k + 6)\) Attempts determinant of \(\mathbf{M} - 3\mathbf{I}\) or \(\begin{vmatrix}3 - \lambda & k & 2\\ -1 & -\lambda & 1\\ 1 & k & 1 - \lambda\end{vmatrix} = (3 - \lambda)\left[-\lambda(1 - \lambda) - k\right] - k\left[-1(1 - \lambda) - 1\right] + 2(-k + \lambda)\) \(\left(= -\lambda^3 + 4\lambda^2 - \lambda - 3k\right)\) \(\lambda = 3 \Rightarrow \det(\mathbf{M} - \lambda\mathbf{I}) = (3 - 3)\left[-3(1 - 3) - k\right] - k\left[-1(1 - 3) - 1\right] + 2(-k + 3)\) Attempts determinant of \(\mathbf{M} - \lambda\mathbf{I}\) and substitutes \(\lambda = 3\) | M1 |
| \(0 - k + 2(-k + 3) = 0 \Rightarrow k = 2\) | dM1 A1 |
| (3) |
Notes
Should be a recognisable attempt at the determinant (if there is any doubt at least 2 “terms” should be correct)
dM1: Puts = 0 (may be implied) and solves for \(k\). Dependent on the first M
A1: \(k = 2\)
Alternative to part (a)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}3 & k & 2\\ -1 & 0 & 1\\ 1 & k & 1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = 3\begin{pmatrix}x\\ y\\ z\end{pmatrix}\) | |
| \(\begin{aligned}3x + ky + 2z &= 3x\\ -x + z &= 3y\\ x + ky + z &= 3z\end{aligned}\) | M1 |
| Let e.g. \(x = 1\) \(\Rightarrow ky = -2z,\ z = 3y + 1,\ ky + 1 = 2z\) \(\Rightarrow 4z = 1 \Rightarrow z = \dfrac{1}{4}, y = -\dfrac{1}{4}\) \(\Rightarrow k = \ldots\) Or e.g. \(\Rightarrow ky = -2z,\ ky = 2z - x, z = 3y + x\) \(\Rightarrow ky = -6y - 2x,\ 2ky = 12y + 2x \Rightarrow 3ky = 6y\) \(\Rightarrow k = \ldots\) | dM1 |
| \(\Rightarrow k = 2\) | A1 |
| (3) |
M1: Forms 3 equations using the eigenvalue 3
dM1: Allocates a non-zero value to one of \(x\) or \(y\) or \(z\) and solves for the other two variables and finds a value for \(k\) Or Solves to obtain a value for \(k\) Dependent on the first M
| Scheme | Marks |
|---|---|
| \(\det(\mathbf{M} - \lambda\mathbf{I}) = (3 - \lambda)\left[-\lambda(1 - \lambda) - k\right] - k\left[-1(1 - \lambda) - 1\right] + 2(-k + \lambda)\) Attempts determinant of \(\mathbf{M} - \lambda\mathbf{I}\) (may be seen in (a)) but must be seen or used in (b) to score in (b) | M1 |
| \(\det(\mathbf{M} - \lambda\mathbf{I}) = (3 - \lambda)\left[-\lambda(1 - \lambda) - 2\right] - 2\left[-1(1 - \lambda) - 1\right] + 2(-2 + \lambda) = 0\) Uses their \(k\) in their determinant and puts = 0 (May be implied by their work) Dependent on the first M | dM1 |
| \(\left\{(3 - \lambda)\right\}\left(\lambda^2 - \lambda - 2\right) = 0 \Rightarrow \lambda = \ldots\) | ddM1 |
| \(\lambda = -1,\ 2\) | A1 |
| (4) |
Notes
ddM1: Solves 3TQ to find the 2 other eigenvalues (apply usual rules if necessary). Dependent on both previous M’s
If they multiply out they should get \(\lambda^3 - 4\lambda^2 + \lambda + 6 = 0\) and may use a calculator to obtain \(\lambda = -1,\ 2\) (and 3)
A1: Correct eigenvalues. (Must follow \(k = 2\))
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}3 & \text{"}2\text{"} & 2\\ -1 & 0 & 1\\ 1 & \text{"}2\text{"} & 1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = 3\begin{pmatrix}x\\ y\\ z\end{pmatrix} \Rightarrow \begin{aligned}3x + \text{"}2\text{"}y + 2z &= 3x\\ -x + z &= 3y\\ x + \text{"}2\text{"}y + z &= 3z\end{aligned}\) or \(\begin{pmatrix}0 & \text{"}2\text{"} & 2\\ -1 & -3 & 1\\ 1 & \text{"}2\text{"} & -2\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}0\\ 0\\ 0\end{pmatrix} \Rightarrow \begin{aligned}\text{"}2\text{"}y + 2z &= 0\\ -x - 3y + z &= 0\\ x + \text{"}2\text{"}y - 2z &= 0\end{aligned}\) Expands to obtain at least 2 equations. Allow if \(k\) is present. | M1 |
| \(k\begin{pmatrix}4\\ -1\\ 1\end{pmatrix}\) or \(k(4\mathbf{i} - \mathbf{j} + \mathbf{k})\) | A1 |
| (2) | |
| (9 marks) |
Notes
A1: Any non-zero multiple but must be a vector
**Note that the vector product of any 2 rows of \(\mathbf{M} - 3\mathbf{I}\) also gives an eigenvector**
Note on Determinants
Note that determinants can be found using any row or column
And also by applying the rule of Sarrus which is:
\(\begin{vmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{vmatrix} = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{31}a_{22}a_{13} - a_{32}a_{23}a_{11} - a_{33}a_{21}a_{12}\)
Please look out for these alternative approaches in Question 3