AS June 2018 Paper 1 Q8
8 The \(2 \times 2\) matrix \(\mathbf{A}\) represents a transformation T which has the following properties.
- The image of the point \((0, 1)\) is the point \((3, 4)\).
- An object shape whose area is 7 is transformed to an image shape whose area is 35.
- T has a line of invariant points.
The transformation S is represented by the matrix \(\mathbf{B}\) where \(\mathbf{B} = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\).
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\) | B1 | 3.1a |
| Determinant \(= ad - bc = 5\) | B1 | 3.1a |
| \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) | B1 | 1.2 |
| \((1 - a)x = by\) or \((1 - d)y = cx\) | M1 | 2.2a |
| \(\dfrac{1 - a}{c} = \dfrac{b}{1 - d}\) Or \(\dfrac{1 - a}{b} = \dfrac{c}{1 - d}\) | M1 | 1.1 |
| \(c = a - 1\) | A1 | 1.1 |
| e.g. \(4a - 3(a - 1) = 5\) | M1 | 1.1 |
| \(\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\) | A1 | 3.2a |
| [8] |
Notes
B1: (1st) \(\Rightarrow b = 3, d = 4\)
B1: (2nd) \(4a - 3c = 5\)
Or det \(= -5\) and follow through
B1: (3rd) Understanding of invariant point seen or implied
M1: (1st) May have \(b = 3\) and/or \(d = 4\) already substituted
\((1 - a)x = 3y\) or \(-3y = cx\)
M1: (2nd) Eliminating \(x\) and \(y\)
M1: (3rd) Attempting to solve their simultaneous equations
If no working and incorrect then M0A0.
A1: (2nd) Condone \(a = 2\), etc as long as Matrix seen as \(\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| Need \(\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x + y \\ 2x + 2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) | M1 | 1.1 |
| \(2x + y = 0\) | A1 | 2.2a |
| [2] |
Notes
M1: Substituting a general point into their matrix, calculating an image point and equating it to the object point.
A1: Final form must be \(y = -2x\) or \(x = -\tfrac{1}{2}y\) or a numerical multiple of \(2x + y = 0\).
Need to have considered both \(x\) and \(y\) coordinates.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ x + c \end{pmatrix} = \begin{pmatrix} 3x + x + c \\ 2x + 2x + 2c \end{pmatrix}\) | M1* | 3.1a |
| So \(\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 4x + c \\ 4x + 2c \end{pmatrix} \ldots\) | M1dep* | 2.2a |
| \(\ldots\)and \(Y = X + c\) | A1 | 1.1 |
| [3] |
Notes
A1: Could see the \(y\) component of the vector written as \(4x + c + c\).
Alternative
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} X \\ mX + c \end{pmatrix}\) \(3x + mx + c = X\) \(2x + 2mx + 2c = mX + c\) \(2x + 2mx + 2c = m(3x + mx + c) + c\) | M1 |
| \(x(m^2 + m - 2) + c(m - 1) = 0\) \(x(m + 2)(m - 1) + c(m - 1) = 0\) | M1 |
| If \(m = 1\), \(c(m - 1) = 0\) satisfied by any \(c\) | A1 |
M1: (1st) Need to eliminate \(X\), i.e. an equation in \(x\), \(m\) and \(c\).