AS October 2020 Paper 1 Q2
2 P, Q and T are three transformations in 2-D.
P is a reflection in the \(x\)-axis. \(\mathbf{A}\) is the matrix that represents P.
Q is a shear in which the \(y\)-axis is invariant and the point \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) is transformed to the point \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\). \(\mathbf{B}\) is the matrix that represents Q.
T is P followed by Q. \(\mathbf{C}\) is the matrix that represents T.
\(L\) is the line whose equation is \(y = x\).
An object parallelogram, \(M\), is transformed under T to an image parallelogram, \(N\).
- the area of \(N\) compared to the area of \(M\),
- the orientation of \(N\) compared to the orientation of \(M\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) | B1 | 1.2 |
| [1] |
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} . & 0 \\ . & 1 \end{pmatrix}\) | B1 | 1.1 |
| \(\begin{pmatrix} 1 & . \\ 2 & . \end{pmatrix}\) | B1 | 1.1 |
| [2] |
Notes
B1: (1st) \(\begin{pmatrix} 0 \\ 1 \end{pmatrix} \to \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) ie \(y\)-axis invariant
\(\begin{pmatrix} 1 & 0 \\ . & . \end{pmatrix}\) or \(\begin{pmatrix} . & . \\ 2 & 1 \end{pmatrix}\) insufficient for B1
B1: (2nd) \(\begin{pmatrix} 1 \\ 0 \end{pmatrix} \to \begin{pmatrix} 1 \\ 2 \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{C} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
M1: Their \(\mathbf{A}\) and \(\mathbf{B}\) but must be the correct way round (ie \(\mathbf{BA}\), not \(\mathbf{AB}\)).
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} x \\ x \end{pmatrix} = \begin{pmatrix} x \\ x \end{pmatrix}\) | M1 | 3.1a |
| Since each point gets mapped to itself it is a line of invariant points | A1 | 2.2a |
| [2] |
Notes
M1: Multiplying a correct vector (\(\begin{pmatrix} x \\ x \end{pmatrix}\) or \(\begin{pmatrix} y \\ y \end{pmatrix}\)) correctly into C.
If a particular point (eg \((1, 1)\)) is used then supporting statement required for M1 that eg this must therefore apply on the line through \((0, 0)\) and \((1, 1)\).
A1: If M0 then SC1 for use of a particular point leading to correct conclusion (This can follow from an incorrect matrix, possibly to show that \(y = x\) is not a line of invariant points)
Can use \(\begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) to deduce that \(x = y\).
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{C} = 1 \times -1 - 2 \times 0 = -1\) | B1ft | 1.1 |
| So area of \(N\) is the same as area of \(M\) oe | B1ft | 2.2a |
| But the orientation of \(N\) is the reverse of the orientation of \(M\). | B1ft | 2.2a |
| [3] |
Notes
B1ft: (2nd) Do not award if implication of statement is that area of \(N\) is negative.
B1ft: (3rd) ft for \(\det\mathbf{C} \lt 0\).
Allow “orientation changed”