A2 June 2024 Paper 2 Q8
8.
\[\mathbf{A} = \begin{pmatrix}3 & 1 & -1\\ 1 & 1 & 1\\ k & 3 & 6\end{pmatrix} \qquad k \neq 0\]| Scheme | Marks | AO |
|---|---|---|
| \(|\mathbf{A}| = 3(6 - 3) - 1(6 - k) - 1(3 - k) = 2k\) | B1 | 1.1b |
| Cofactors \(\begin{pmatrix}3 & k - 6 & 3 - k\\ -9 & 18 + k & k - 9\\ 2 & -4 & 2\end{pmatrix}\) or Transpose of matrix of minors \(\begin{pmatrix}3 & 9 & 2\\ 6 - k & 18 + k & 4\\ 3 - k & 9 - k & 2\end{pmatrix}\) | M1 | 2.1 |
| \(\mathbf{A}^{-1} = \dfrac{1}{2k}\begin{pmatrix}3 & -9 & 2\\ k - 6 & 18 + k & -4\\ 3 - k & k - 9 & 2\end{pmatrix}\) | dM1 A1 | 1.1b 1.1b |
| (4) |
Notes
If no attempt at part (a) has been made you may award marks for work seen in part (b) for finding the inverse.
B1(M1 on epen): Correct determinant of \(2k\)
M1: Starts the process of finding the inverse and obtains at least 6 correct elements of cofactors
Alternatively transposes their matrix of minors and obtains at least 6 correct elements.
dM1: A complete recognisable method to find the inverse including dividing by the determinant
Allow minor slips if the process is clearly correct.
A1: Correct inverse.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \dfrac{1}{2k}\begin{pmatrix}3 & -9 & 2\\ k - 6 & 18 + k & -4\\ 3 - k & k - 9 & 2\end{pmatrix}\begin{pmatrix}3\\ 1\\ 6\end{pmatrix} = \ldots\) | M1 | 1.1a |
| Any 2 of \(\ x = \dfrac{6}{k},\ \ y = \dfrac{2k - 12}{k},\ \ z = \dfrac{6 - k}{k}\) | A1 | 2.1 |
| \(\left(\dfrac{6}{k}, \dfrac{2k - 12}{k}, \dfrac{6 - k}{k}\right)\) or \(x = \dfrac{6}{k},\ y = \dfrac{2k - 12}{k},\ z = \dfrac{6 - k}{k}\) or \(x = \dfrac{6}{k},\ y = 2 - \dfrac{12}{k},\ z = \dfrac{6}{k} - 1\) | A1 | 2.5 |
| (3) | ||
| (7 marks) |
Notes
M1: Attempts \(\mathbf{A}^{-1}\begin{pmatrix}3\\ 1\\ 6\end{pmatrix}\) with their \(\mathbf{A}^{-1}\) which must be in terms of \(k\), to obtain at least one of \(x =\), \(y =\) or \(z =\) which may be seen embedded in a column vector, simplified or unsimplified, the determinant may be outside their column vector.
Condone a slip in copying \(\begin{pmatrix}3\\ 1\\ 6\end{pmatrix}\) e.g. \(\begin{pmatrix}1\\ 3\\ 6\end{pmatrix}\) or \(\begin{pmatrix}3\\ 1\\ 2\end{pmatrix}\)
A1(M1 on epen): Two correct expressions for \(x\), \(y\) or \(z\) simplified or unsimplified, which may appear in a column vector, determinant cannot be outside the vector.
A1: Correct coordinates in simplest form. Allow e.g. \(y = 2 - \dfrac{12}{k}\)
Their final answer must be written as coordinates and not as a column vector but can be written as \(x =\), \(y =\) and \(z =\)
Alternative (using algebraic method for simultaneous equations):
M1: Solves simultaneously to obtain at least one of \(x =\), \(y =\) or \(z =\) must be in terms of \(k\).
A1(M1 on epen): Two correct expressions for \(x\), \(y\) or \(z\) simplified or unsimplified.
A1: Correct coordinates in simplest form. Allow e.g. \(y = 2 - \dfrac{12}{k}\)
Their final answer must be written as coordinates and not as a column vector but can be written as \(x =\), \(y =\) and \(z =\)
e.g.
Eliminates \(z\) and achieves \(4x + 2y = 4\) and \((6 - k)x + 3y = 0\)
Uses \(12x + 6y = 12\) and \((12 - 2k)x + 6y = 0\) to produce \(2kx = 12 \Rightarrow x = \dfrac{6}{k}\)