FP1 June 2011 Q5
5. \[\mathbf{A} = \begin{pmatrix} -4 & a \\ b & -2 \end{pmatrix}, \text{ where } a \text{ and } b \text{ are constants.}\]
Given that the matrix \(\mathbf{A}\) maps the point with coordinates \((4,\ 6)\) onto the point with coordinates \((2,\ -8)\),
A quadrilateral \(R\) has area 30 square units.
It is transformed into another quadrilateral \(S\) by the matrix \(\mathbf{A}\).
Using your values of \(a\) and \(b\),
| Scheme | Marks |
|---|---|
| \(\mathbf{A} = \begin{pmatrix} -4 & a \\ b & -2 \end{pmatrix}\), where \(a\) and \(b\) are constants. | |
| \(\mathbf{A}\begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -8 \end{pmatrix}\) | |
| Therefore, \(\begin{pmatrix} -4 & a \\ b & -2 \end{pmatrix}\begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -8 \end{pmatrix}\) Using the information in the question to form the matrix equation. Can be implied by both correct equations below. | M1 |
| So, \(-16 + 6a = 2\) and \(4b - 12 = -8\) Allow \(\begin{pmatrix} -16 + 6a \\ 4b - 12 \end{pmatrix} = \begin{pmatrix} 2 \\ -8 \end{pmatrix}\) Any one correct equation. Any correct horizontal line | M1 |
| giving \(a = 3\) and \(b = 1\). Any one of \(a = 3\) or \(b = 1\). Both \(a = 3\) and \(b = 1\). | A1 A1 |
| (4) |
Notes
Do not allow this mark for other incorrect statements unless interpreted correctly later
e.g. \(\begin{pmatrix} 4 \\ 6 \end{pmatrix}\begin{pmatrix} -4 & a \\ b & -2 \end{pmatrix} = \begin{pmatrix} 2 \\ -8 \end{pmatrix}\) would be M0 unless followed by correct equations or \(\begin{pmatrix} -16 + 6a \\ 4b - 12 \end{pmatrix} = \begin{pmatrix} 2 \\ -8 \end{pmatrix}\)
| Scheme | Marks |
|---|---|
| \(\det\mathbf{A} = 8 - (3)(1) = 5\) Finds determinant by applying \(8 -\) their \(ab\). \(\det\mathbf{A} = 5\) | M1 A1 |
| Area \(S = (\det\mathbf{A})(\text{Area } R)\) | |
| Area \(S = 5 \times 30 = 150\ (\text{units})^2\) \(\dfrac{30}{\text{their }\det\mathbf{A}}\) or \(30 \times (\text{their }\det\mathbf{A})\) 150 or ft answer | M1 A1ft |
| (4) | |
| (8 marks) |
Notes
Special case: The equations \(-16 + 6b = 2\) and \(4a - 12 = -8\) give \(a = 1\) and \(b = 3\). This comes from incorrect matrix multiplication. This will score nothing in (a) but allow all the marks in (b).
Note that \(\det\mathbf{A} = \dfrac{1}{8 - ab}\) scores M0 here but the following 2 marks are available. However, beware \(\det\mathbf{A} = \dfrac{1}{8 - ab} = \dfrac{1}{5} \Rightarrow \textit{area } S = \dfrac{30}{\frac{1}{5}} = 150\)
This scores M0A0 M1A0
If their \(\det\mathbf{A} < 0\) then allow ft provided final answer \(> 0\)
In (b) Candidates may take a more laborious route for the area scale factor and find the area of the unit square, for example, after the transformation represented by \(\mathbf{A}\). This needs to be a complete method to score any marks. Correctly establishing the area scale factor M1. Correct answer 5 A1. Then mark as original scheme.