AS June 2018 Paper 1 Q1
1.
\[\mathbf{M} = \begin{pmatrix}2 & 1 & -3\\ 4 & -2 & 1\\ 3 & 5 & -2\end{pmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| \[\mathbf{M}^{-1} = \frac{1}{69}\begin{pmatrix}1 & 13 & 5\\ -11 & -5 & 14\\ -26 & 7 & 8\end{pmatrix}\] | B1 B1 | 1.1b 1.1b |
| (2) |
Notes
B1: Evidence that the determinant is \(\pm 69\) (may be implied by their matrix e.g. where entries are not in exact form: \(\pm\begin{pmatrix}0.014 & 0.188 & 0.072\\ -0.159 & -0.072 & 0.203\\ -0.377 & 0.101 & 0.116\end{pmatrix}\)) (Should be mostly correct)
Must be seen in part (a).
B1: Fully correct inverse with all elements in exact form
| Scheme | Marks | AO |
|---|---|---|
| \[\frac{1}{69}\begin{pmatrix}1 & 13 & 5\\ -11 & -5 & 14\\ -26 & 7 & 8\end{pmatrix}\begin{pmatrix}-4\\ 9\\ 5\end{pmatrix} = \ldots\] | M1 | 1.1b |
| \(x = 2,\ y = 1,\ z = 3\) or \((2, 1, 3)\) or \(2\mathbf{i} + \mathbf{j} + 3\mathbf{k}\) or \(\begin{pmatrix}2\\ 1\\ 3\end{pmatrix}\) | A1 | 1.1b |
| (2) |
Notes
M1: Any complete method to find the values of \(x\), \(y\) and \(z\) (Must be using their inverse if using the method in the main scheme)
A1: Correct coordinates
A solution not using the inverse requires a complete method to find values for \(x\), \(y\) and \(z\) for the method mark.
Correct coordinates only scores both marks.
| Scheme | Marks | AO |
|---|---|---|
| The point where three planes meet | B1ft | 2.2a |
| (1) | ||
| (5 marks) |
Notes
B1: Describes the correct geometrical configuration.
Must include the two ideas of planes and meet in a point with no contradictory statements.
This is dependent on having obtained a unique point in part (b)