FP3 June 2016 Q1
1. \[\mathbf{A} = \begin{pmatrix}-2 & 1 & -3\\ k & 1 & 3\\ 2 & -1 & k\end{pmatrix}, \quad \text{where } k \text{ is a constant}\]
Given that the matrix \(\mathbf{A}\) is singular, find the possible values of \(k\). (4)
| Scheme | Marks |
|---|---|
| \(\mathbf{A} = \begin{pmatrix}-2 & 1 & -3\\ k & 1 & 3\\ 2 & -1 & k\end{pmatrix}\) | |
| \(\det\mathbf{A} = -2(k + 3) - (k^2 - 6) - 3(-k - 2)\ \ row1\) or e.g. \(\det\mathbf{A} = -k(k - 3) + (-2k + 6) - 3(2 - 2)\ \ row2\) \(\det\mathbf{A} = 2(3 + 3) + (-6 + 3k) + k(-2 - k)\ \ row3\) \(\det\mathbf{A} = -2(k + 3) - k(k - 3) + 2(3 + 3)\ \ col1\) \(\det\mathbf{A} = -(k^2 - 6) + (-2k + 6) + (-6 + 3k)\ \ col2\) \(\det\mathbf{A} = -3(-2 - k) - 3(2 - 2) + k(-2 - k)\ \ col3\) | M1A1 |
| \(-2(k + 3) - (k^2 - 6) - 3(-k - 2) = 0 \Rightarrow k = \ldots\) | M1 |
| \((k + 2)(k - 3) = 0 \Rightarrow k = -2,\ 3\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: Correct attempt at determinant (3 ‘elements’ (may be implied if one is zero) with at least two elements correct). Note that there are various alternatives depending on the choice of row or column.
A1: Correct determinant in any form
Note that e.g. \(\det\mathbf{A} = -2\begin{vmatrix}1 & 3\\ -1 & k\end{vmatrix} - \begin{vmatrix}k & 3\\ 2 & k\end{vmatrix} - 3\begin{vmatrix}k & 1\\ 2 & -1\end{vmatrix}\) scores no marks until the determinants are ‘extracted’.
M1: Sets their \(\det\mathbf{A} = 0\) (= 0 may be implied) and attempts to solve a 3 term quadratic (see general guidance) as far as \(k = \ldots\) NB Correct quadratic is \(k^2 - k - 6 = 0\)
A1: Both values correct