AS October 2020 Paper 1 Q6
6.
where \(a\) and \(b\) are non-zero constants.
Given that the matrix \(\mathbf{A}\) is self-inverse,
The matrix \(\mathbf{A}\) represents a linear transformation \(M\).
Using the smaller value of \(a\) from part (a),
where \(p\) is a positive constant.
The matrix \(\mathbf{P}\) represents a linear transformation \(U\).
The triangle \(T\) has vertices at the points with coordinates (1, 2), (3, 2) and (2, 5).
The area of the image of \(T\) under the linear transformation \(U\) is 15
The transformation \(V\) consists of a stretch scale factor 3 parallel to the \(x\)-axis with the \(y\)-axis invariant followed by a stretch scale factor \(-2\) parallel to the \(y\)-axis with the \(x\)-axis invariant. The transformation \(V\) is represented by the matrix \(\mathbf{Q}\).
Given that \(U\) followed by \(V\) is the transformation \(W\), which is represented by the matrix \(\mathbf{R}\),
| Scheme | Marks | AO |
|---|---|---|
| Multiplies the matrix \(\mathbf{A}\) by itself and sets equal to \(\mathbf{I}\) to form one equation in \(a\) only and another equation involving both \(a\) and \(b\). \(\begin{pmatrix}2 & a\\ a - 4 & b\end{pmatrix}\begin{pmatrix}2 & a\\ a - 4 & b\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} \Rightarrow 4 + a(a - 4) = 1\) and either \(2a + ab = 0\) or \(2(a - 4) + b(a - 4) = 0\) or \(a(a - 4) + b^2 = 1\) | M1 | 3.1a |
| Solves a 3TQ involving only the constant \(a\). This could come after a value of \(b\) is found and this value substituted into an equation involving both \(a\) and \(b\) \(a^2 - 4a + 3 = 0 \Rightarrow (a - 3)(a - 1) = 0 \Rightarrow a = \ldots\) | dM1 | 1.1b |
| \(a = 1, a = 3\) | A1 | 1.1b |
| Substitutes a value for \(a\) into an equation involving both \(a\) and \(b\) and solves for \(b\). e.g. \(2(1) + (1)b \Rightarrow b = \ldots\) \(2(1 - 4) + b(1 - 4) = 0 \Rightarrow b = \ldots\) \((1)(1 - 4) + b^2 = 1 \Rightarrow b = \ldots\) Alternatively uses \(2a + ab = 0\) \(a(2 + b) = 0\) As \(a \neq 0\quad 2 + b = 0 \Rightarrow b = \ldots\) | dM1 | 1.1b |
| \(b = -2\) | A1 | 1.1b |
| (5) |
Notes
(Corrected from the printed mark scheme: the printed example is \(2(1 - 4)b + (1 - 4) = 0\); from \(2(a - 4) + b(a - 4) = 0\) with \(a = 1\) it is \(2(1 - 4) + b(1 - 4) = 0\).)
M1: Forming two equations, one involving \(a\) only and one involving \(a\) and \(b\)
dM1: Dependent on previous mark, solves a 3TQ involving \(a\)
A1: Correct values for \(a\)
dM1: Dependent on first method mark Substitutes one of their values of \(a\) into an equation involving \(a\) and \(b\) and solve to find a value for \(b\). Alternatively factorises either \(2a + ab = 0\) and uses \(a \neq 0\) to find a value for \(b\).
A1: Correct value for \(b\)
Alternative (i)(a)
| Scheme | Marks | AO |
|---|---|---|
| Finds \(\mathbf{A}^{-1}\) in terms of \(a\) and \(b\), sets equal to \(\mathbf{A}\) and attempts to find at least two different equations. Allow a single sign slip\[\frac{1}{2b - a(a - 4)}\begin{pmatrix}b & -a\\ -(a - 4) & 2\end{pmatrix} = \begin{pmatrix}2 & a\\ a - 4 & b\end{pmatrix}\]One equation from \(\dfrac{b}{2b - a(a - 4)} = 2,\ \dfrac{2}{2b - a(a - 4)} = b\) One equation from \(\dfrac{-a}{2b - a(a - 4)} = a,\ \dfrac{-(a - 4)}{2b - a(a - 4)} = a - 4\) | M1 | 3.1a |
| Uses their value of \(b\) and their value of the determinant to form and solve a 3TQ involving only the constant \(a\) \(a^2 - 4a + 3 = 0 \Rightarrow (a - 3)(a - 1) = 0 \Rightarrow a = \ldots\) or Eliminates \(b\) from their equations and solve a 3TQ involving only the constant \(a\) \(a^2 - 4a + 3 = 0 \Rightarrow (a - 3)(a - 1) = 0 \Rightarrow a = \ldots\) | dM1 | 1.1b |
| \(a = 1, a = 3\) | A1 | 1.1b |
| \(\dfrac{-a}{2b - a(a - 4)} = a \Rightarrow 2b - a(a - 4) = -1 \Rightarrow \dfrac{b}{-1} = 2\) Or \(\dfrac{-(a - 4)}{2b - a(a - 4)} = a - 4 \Rightarrow 2b - a(a - 4) = -1 \Rightarrow \dfrac{2}{-1} = b\) or Substitutes a value for \(a\) into an equation to find a value for \(b\) | dM1 | 1.1b |
| \(b = -2\) | A1 | 1.1b |
| (5) |
M1: Finds \(\mathbf{A}^{-1}\) and sets equal to \(\mathbf{A}\) and forms two different equations
dM1: Dependent on previous mark. Eliminates \(b\) from their equations and solves a 3TQ involving only the constant \(a\). Alternatively if the value of \(b\) is found first substitutes their value for \(b\) into their determinant = –1 to form and solve a 3TQ for \(a\)
A1: Correct value for \(a\)
dM1: Dependent on first method mark. Substitutes a value for \(a\) into an equation to find a value for \(b\). Alternatively uses one equation to find the determinant = –1 and uses this to find a value of \(b\).
A1: Correct values for \(b\)
| Scheme | Marks | AO |
|---|---|---|
| Uses their smallest value of \(a\) and their value for \(b\) to form two equations \(\begin{pmatrix}2 & \text{'}a\text{'}\\ \text{'}a - 4\text{'} & \text{'}b\text{'}\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow 2x + ay = x\) and \((a - 4)x + by = y\) \(\begin{pmatrix}2 & 1\\ -3 & -2\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow 2x + y = x\) and \(-3x - 2y = y\) | M1 | 3.1a |
| \(2x + y = x \Rightarrow x + y = 0\) o.e. and \(-3x - 2y = y \Rightarrow x + y = 0\) o.e. | M1 | 1.1b |
| \(x + y = 0\) o.e. | A1 | 2.1 |
| (3) |
Notes
M1: Extracts simultaneous equations using their matrix A with their smaller value of \(a\).
M1: Gathers terms from their two equations.
A1: Achieves the correct equations and deduces the correct line. Accept equivalent equations as long as both have been shown to be the same.
| Scheme | Marks | AO |
|---|---|---|
| Area of the triangle \(T = 3\) | B1 | 1.1b |
| Complete method to find a value for \(p\). Need to see an attempt at the determinant and setting equal to 15 divided by their area of \(T\). The resulting 3TQ needs to be solved to find a value of \(p\). Determinant \(3p \times p - (-1) \times 2p = \dfrac{15}{\text{‘their area’}} \Rightarrow p = \ldots\) | M1 | 3.1a |
| \(3p^2 + 2p - 5\,(= 0)\) | A1 | 1.1b |
| \(p = 1\) must reject \(p = -\dfrac{5}{3}\) | A1 | 1.1b |
| (4) |
Notes
B1: Area of the triangle \(T = 3\)
M1: Full method. Finds the determinant, sets equal to 15/their area and solves the resulting 3TQ
A1: Correct quadratic
A1: \(p = 1\) only
| Scheme | Marks | AO |
|---|---|---|
| \[\begin{pmatrix}3 & 0\\ 0 & -2\end{pmatrix}\] | B1 B1 | 1.1b 1.1b |
| (2) |
Notes
B1 One correct row or column
B1: All correct
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\text{their matrix found in part (b)}\right)\begin{pmatrix}\text{'}p\text{'} & 2\text{'}p\text{'}\\ -1 & 3\text{'}p\text{'}\end{pmatrix} = \begin{pmatrix}\ldots & \ldots\\ \ldots & \ldots\end{pmatrix}\) \(\begin{pmatrix}3 & 0\\ 0 & -2\end{pmatrix}\begin{pmatrix}1 & 2\\ -1 & 3\end{pmatrix} = \begin{pmatrix}\ldots & \ldots\\ \ldots & \ldots\end{pmatrix}\) | M1 | 1.1b |
| \(\begin{pmatrix}3 & 6\\ 2 & -6\end{pmatrix}\) | A1ft | 1.1b |
| (2) | ||
| (16 marks) |
Notes
M1: Multiplies the matrices QP in the correct order (if answer only then evidence can be taken from 3 correct or 3 correct ft elements)
A1ft: Correct matrix following through on their answer to part (b) and their value of \(p\) as long as it is a positive constant