A2 October 2020 Q3
3.
\[\mathbf{M} = \begin{pmatrix} 1 & k & -2 \\ 2 & -4 & 1 \\ 1 & 2 & 3 \end{pmatrix}\]where \(k\) is a constant.
Given that \(\det \mathbf{M} = 5\)
| Scheme | Marks | AO |
|---|---|---|
| Sight of \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\) | B1 | 1.1a |
| \(\begin{vmatrix} 1 - \lambda & k & -2 \\ 2 & -4 - \lambda & 1 \\ 1 & 2 & 3 - \lambda \end{vmatrix} = 0 \Rightarrow\) \((1 - \lambda)\big[(-4 - \lambda)(3 - \lambda) - 2\big] - k\big[2(3 - \lambda) - 1\big] + (-2)\big[4 - (-4 - \lambda)\big] = 0\) | M1 | 1.1b |
| \(\Rightarrow (1 - \lambda)(\lambda^2 + \lambda - 14) - k(5 - 2\lambda) - 16 - 2\lambda = 0\) \(\Rightarrow \lambda^3 - (2k + 13)\lambda + 5(k + 6) = 0\) * | A1* | 2.1 |
| (3) |
Notes
B1: Recalls characteristic equation is found using \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\)
M1: Attempts to expand the determinant.
A1*: Achieves the correct equation with no errors and at least one intermediate step following the expansion.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\pm 5(k + 6) = 5 \Rightarrow k = \ldots\) or \((-12 - 2) - k(6 - 1) - 2(4 + 4) = 5 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k = -7\) | A1 | 2.2a |
| (ii) Hence by the C-H theorem \(\mathbf{M}^3 + \mathbf{M} - 5\mathbf{I} = \mathbf{0}\) | M1 | 2.1 |
| Multiplying by \(\mathbf{M}^{-1}\) gives \(\mathbf{M}^2 + \mathbf{I} - 5\mathbf{M}^{-1} = \mathbf{0} \Rightarrow \mathbf{M}^{-1} = \ldots\) | M1 | 3.1a |
| So \(\mathbf{M}^{-1} = \dfrac{1}{5}\left(\mathbf{M}^2 + \mathbf{I}\right)\) | A1 | 1.1b |
| \(= \dfrac{1}{5}\left(\begin{pmatrix} -15 & 17 & -15 \\ -5 & 4 & -5 \\ 8 & -9 & 9 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\right) = \ldots\) | M1 | 1.1b |
| \(= \dfrac{1}{5}\begin{pmatrix} -14 & 17 & -15 \\ -5 & 5 & -5 \\ 8 & -9 & 10 \end{pmatrix}\) or \(\begin{pmatrix} -\dfrac{14}{5} & \dfrac{17}{5} & -3 \\ -1 & 1 & -1 \\ \dfrac{8}{5} & -\dfrac{9}{5} & 2 \end{pmatrix}\) | A1 | 1.1b |
| (7) | ||
| (10 marks) |
Notes
(i)
M1: Attempts to use determinant equals 5 to find \(k\). May be attempted by finding determinant from original matrix, or attempt at using the \(\text{“}{-5}(k + 6)\text{”}\) from the expansion in (a) (allow \(\pm\) for the method mark).
A1: \(k = -7\)
(ii)
M1: Attempts to use the Cayley-Hamilton theorem to set up a matrix equation. The equation should be correct for their \(k\), including correct use of \(\mathbf{I}\).
M1: Realises the need to multiply the equation through (either side) by \(\mathbf{M}^{-1}\) and rearrange to make \(\mathbf{M}^{-1}\) the subject.
A1: \(\mathbf{M}^{-1} = \dfrac{1}{5}\left(\mathbf{M}^2 + \mathbf{I}\right)\)
M1: Proceeds to find \(\mathbf{M}^{-1}\) from their equation.
A1: Correct answer.