AS June 2018 Paper 1 Q7
7 Find two invariant points under the transformation given by \(\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\) [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains two equations in \(x\) and \(y\). May be seen as a single vector equation. At least one equation must be correct. Accept a pair of letters other than \(x\) and \(y\). Ignore any subsequent incorrect working. | M1 | 1.1a |
| Obtains any two correct invariant points, with no incorrect points. Condone correct points given as position vectors. NMS: Correct answer scores 2/2. NMS: One correct invariant point and only one incorrect point scores SC1. | A1 | 1.1b |
| (2 marks) |
Typical solution
\[\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}\]\[\begin{bmatrix} 2x + 3y \\ x + 4y \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}\]\[2x + 3y = x \quad \text{and} \quad x + 4y = y\]\[x = -3y\]\(\therefore\) two invariant points are \((0, 0)\) and \((-3, 1)\)