AS June 2019 Paper 1 Q1
1.
\[\mathbf{M} = \begin{pmatrix}4 & -5\\ 2 & -7\end{pmatrix}\]The transformation \(T\) of the plane is represented by the matrix \(\mathbf{M}\).
The triangle \(R\) is transformed to the triangle \(S\) by the transformation \(T\).
Given that the area of \(S\) is 63 square units,
| Scheme | Marks | AO |
|---|---|---|
| \((\det(\mathbf{M}) =)\ (4)(-7) - (2)(-5)\) | M1 | 1.1a |
| \(\mathbf{M}\) is non-singular because \(\det(\mathbf{M}) = -18\) and so \(\det(\mathbf{M}) \ne 0\) | A1 | 2.4 |
| (2) |
Notes
M1: An attempt to find \(\det(\mathbf{M})\). Just the calculation is sufficient. Sight of \(-18\) implies this mark, which may be embedded in an attempt at the inverse.
A1: \(\det(\mathbf{M}) = -18\) and reference to zero, e.g. \(-18 \ne 0\) and conclusion. The conclusion may precede finding the determinant (e.g. “Non-singular if \(\det(\mathbf{M}) \ne 0\), \(\det(\mathbf{M}) = -18 \ne 0\)” is sufficient or accept “Non-singular if \(\det(\mathbf{M}) \ne 0\), \(\det(\mathbf{M}) = -18\), therefore non-singular” or some other indication of conclusion.) Need not mention “\(\det(\mathbf{M})\)” to gain both marks here, a correct calculation, statement \(-18 \ne 0\), and conclusion hence \(\mathbf{M}\) is non-singular can gain M1A1.
| Scheme | Marks | AO |
|---|---|---|
| Area \(R = \dfrac{\text{Area } S}{(\pm)|\det \mathbf{M}|} = \ldots\) | M1 | 1.2 |
| Area\((R) = \dfrac{63}{|{-18}|} = \dfrac{7}{2}\) oe | A1ft | 1.1b |
| (2) |
Notes
M1: Recalls determinant is needed for area scale factor by dividing 63 by \(\pm\)their determinant.
A1ft: \(\dfrac{7}{2}\) or follow through \(\dfrac{63}{|\text{their det}|}\). Must be positive and should be simplified to single fraction or exact decimal. (Allow if made positive following division by a negative determinant.)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}4 & -5\\ 2 & -7\end{pmatrix}\begin{pmatrix}x\\ 2x\end{pmatrix} = \begin{pmatrix}4x - 10x\\ 2x - 14x\end{pmatrix}\) | M1 | 1.1b |
| \(= \begin{pmatrix}-6x\\ -12x\end{pmatrix}\) and so all points on \(y = 2x\) map to points on \(y = 2x\), hence the line is invariant. OR \(= -6\begin{pmatrix}x\\ 2x\end{pmatrix}\) hence \(y = 2x\) is invariant. | A1 | 2.1 |
| (2) | ||
| (6 marks) |
Notes
M1: Attempts the matrix multiplication shown or with equivalent, e.g. \(\begin{pmatrix}\frac{1}{2}y\\ y\end{pmatrix}\). May use \(\begin{pmatrix}x\\ y\end{pmatrix}\) and substitute \(y = 2x\) later and this is fine for the method.
A1: Correct multiplication and working leading to conclusion that the line is invariant. If the \(-6\) is not extracted, they must make reference to image points being on line \(y = 2x\). If the \(-6\) is extracted to show it is a multiple of \(\begin{pmatrix}x\\ 2x\end{pmatrix}\) followed by a conclusion “invariant” as minimum.
Alternative (Alt)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}a\\ b\end{pmatrix} = \dfrac{1}{-18}\begin{pmatrix}-7 & 5\\ -2 & 4\end{pmatrix}\begin{pmatrix}x\\ 2x\end{pmatrix} = -\dfrac{1}{18}\begin{pmatrix}-7x + 10x\\ -2x + 8x\end{pmatrix}\) | M1 | 1.1b |
| \(= -\dfrac{1}{18}\begin{pmatrix}3x\\ 6x\end{pmatrix}\left(= -\dfrac{1}{6}\begin{pmatrix}x\\ 2x\end{pmatrix}\right) \Rightarrow b = 2a\) so points on line \(y = 2x\) map to points on \(y = 2x\), hence it is invariant. | A1 | 2.1 |
Marks as per main scheme.
Alternative (Alt 2)
| Scheme | Marks | AO |
|---|---|---|
| (Since linear transformations map straight lines to straight lines…) E.g. \((1, 2)\) is on line \(y = 2x\), and \(\begin{pmatrix}4 & -5\\ 2 & -7\end{pmatrix}\begin{pmatrix}1\\ 2\end{pmatrix} = \begin{pmatrix}4 - 10\\ 2 - 14\end{pmatrix}\) | M1 | 1.1b |
| \(= \begin{pmatrix}-6\\ -12\end{pmatrix}\), which is also on the line \(y = 2x\), hence as \((0, 0)\) and \((1, 2)\) both map to points on \(y = 2x\) (and transformation is linear) then \(y = 2x\) is invariant. | A1 | 2.1 |
M1: Identifies a point on the line \(y = 2x\) and finds its image under \(T\). If \((0, 0)\) is used there must be a clear statement it is because this is on the line, but for other points accept with any line on \(y = 2x\) without statement.
A1: Shows the image and another point, which may be \((0, 0)\), on \(y = 2x\) both map to points on \(y = 2x\), concludes line is invariant. Need not reference transformation being linear for either mark here.
Alternative (Alt 3)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}4 & -5\\ 2 & -7\end{pmatrix}\begin{pmatrix}x\\ mx + c\end{pmatrix} = \begin{pmatrix}X\\ mX + c\end{pmatrix}\)\[\Rightarrow 4x - 5(mx + c) = X,\quad 2x - 7(mx + c) = mX + c\]\[\Rightarrow 2x - 7(mx + c) = m\bigl(4x - 5(mx + c)\bigr) + c\]\[\Rightarrow (5m^2 - 11m + 2)x + (5m - 8)c = 0 \Rightarrow (5m - 1)(m - 2) = 0 \Rightarrow m = \ldots\]Or similar work with \(c = 0\) throughout. | M1 | 2.1 |
| \((5m - 8 \ne 0 \Rightarrow c = 0)\) Hence \(m = 2\) gives an invariant line (with \(c = 0\)), so \(y = 2x\) is invariant. | A1 | 1.1b |
M1: Attempts to find the equation of a general invariant line, or general invariant line through the origin (so may have \(c = 0\) throughout). To gain the method mark they must progress from finding the simultaneous equations to forming a quadratic in \(m\) and solving to a value of \(m\).
A1: Correct quadratic in \(m\) found, with \(m = 2\) as solution (ignore the other) and deduction that hence \(y = 2x\) is an invariant line. Ignore errors in the \((5m - 8)\) here as \(c = 0\) is always a possible solution. No need to see \(c = 0\) derived.