A2 June 2022 Paper 2 Q3
3.
\[\mathbf{M} = \begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix} \qquad \text{where } a \text{ is a constant}\]Triangle \(T\) has vertices \(A\), \(B\) and \(C\).
Triangle \(T\) is transformed to triangle \(T^{\prime}\) by the transformation represented by \(\mathbf{M}^n\) where \(n \in \mathbb{N}\)
Given that
- triangle \(T\) has an area of \(5\,\text{cm}^2\)
- triangle \(T^{\prime}\) has an area of \(1215\,\text{cm}^2\)
- vertex \(A(2, -2)\) is transformed to vertex \(A^\prime(123, -2)\)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1 \Rightarrow \mathbf{M}^1 = \begin{pmatrix}3^1 & \dfrac{a}{2}\left(3^1 - 1\right)\\ 0 & 1\end{pmatrix} = \begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}\) {So the result is true for \(n = 1\)} | B1 | 2.2a |
| Assume true for \(n = k\) Or assume \(\mathbf{M}^n\) or \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}^k = \begin{pmatrix}3^{k} & \dfrac{a}{2}\left(3^{k} - 1\right)\\ 0 & 1\end{pmatrix}\) | M1 | 2.4 |
| A correct method to find an expression for \(n = k + 1\) \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}^{k+1} = \begin{pmatrix}3^{k} & \dfrac{a}{2}\left(3^{k} - 1\right)\\ 0 & 1\end{pmatrix}\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}\) or \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}^{k+1} = \begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}\begin{pmatrix}3^{k} & \dfrac{a}{2}\left(3^{k} - 1\right)\\ 0 & 1\end{pmatrix}\) | M1 | 1.1b |
| \(\begin{pmatrix}3\left(3^k\right) & a\left(3^k\right) + \dfrac{a}{2}\left(3^k - 1\right)\\ 0 & 1\end{pmatrix}\) or \(\begin{pmatrix}3\left(3^k\right) & 3 \times \dfrac{a}{2}\left(3^k - 1\right) + a\\ 0 & 1\end{pmatrix}\) | A1 | 1.1b |
| \(\begin{pmatrix}3^{k+1} & \dfrac{a}{2}\left[2\left(3^k\right) + \left(3^k - 1\right)\right]\\ 0 & 1\end{pmatrix} = \begin{pmatrix}3^{k+1} & \dfrac{a}{2}\left[3\left(3^k\right) - 1\right]\\ 0 & 1\end{pmatrix} = \begin{pmatrix}3^{k+1} & \dfrac{a}{2}\left[3^{k+1} - 1\right]\\ 0 & 1\end{pmatrix}\) \(\begin{pmatrix}3\left(3^k\right) & 3 \times \dfrac{a}{2}\left(3^k - 1\right) + a\\ 0 & 1\end{pmatrix} = \begin{pmatrix}3^{k+1} & \dfrac{a}{2}\left(3\left(3^k - 1\right) + 2\right)\\ 0 & 1\end{pmatrix} = \begin{pmatrix}3^{k+1} & \dfrac{a}{2}\left(3^{k+1} - 1\right)\\ 0 & 1\end{pmatrix}\) | A1 | 2.1 |
| If true for \(n = k\) then true for \(n = k + 1\) and as it is true for \(n = 1\) the statement is true for all (positive integers) \(\boldsymbol{n}\) | A1 | 2.4 |
| (6) |
Notes
B1: Shows that the result holds for \(n = 1\). Must see substitution in the RHS minimum required is \(\begin{pmatrix}3 & \frac{a}{2}(3 - 1)\\ 0 & 1\end{pmatrix}\) and reaches \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}\)
M1: Assumes the result is true for some value of \(n = k\). Assume (true for) \(n = k\) is sufficient. Alternatively states assume \(\mathbf{M}^n\) or \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}^k = \begin{pmatrix}3^{k} & \dfrac{a}{2}\left(3^{k} - 1\right)\\ 0 & 1\end{pmatrix}\)
M1: Sets up a matrix multiplication of their assumed result multiplied by the original matrix, either way round. Allow a slip as long as the intention is clear.
A1: Achieves a correct un-simplified matrix
A1: Reaches a correct simplified matrix with no errors, the correct un-simplified matrix seen previously and at least one intermediate line which must be correct.
A1: Correct conclusion. This mark is dependent on all previous marks except B mark but \(n = 1\) must have been attempted. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Condone \(n \in \mathbb{Z}\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\det\left(\mathbf{M}^n\right) = 3^n\) or \(\det(\mathbf{M}) = 3\) | B1 | 1.1b |
| Uses \(5 \times \det\left(\mathbf{M}^n\right) = 1215 \Rightarrow p^n = q \Rightarrow n = \ldots\) \(5 \times 3^n = 1215 \Rightarrow 3^n = 243 \Rightarrow n = \ldots\) | M1 | 3.1a |
| \(n = 5\) | A1 | 1.1b |
| (ii) \(\begin{pmatrix}3^{n} & \dfrac{a}{2}\left(3^{n} - 1\right)\\ 0 & 1\end{pmatrix}\begin{pmatrix}2\\ -2\end{pmatrix} = \begin{pmatrix}123\\ -2\end{pmatrix} \Rightarrow 2\left(3^n\right) - 2\dfrac{a}{2}\left(3^n - 1\right) = 123 \Rightarrow a = \ldots\) \(\begin{pmatrix}243 & \dfrac{a}{2}(243 - 1)\\ 0 & 1\end{pmatrix}\begin{pmatrix}2\\ -2\end{pmatrix} = \begin{pmatrix}123\\ -2\end{pmatrix} \Rightarrow 2(243) - 2\dfrac{a}{2}(243 - 1) = 123 \Rightarrow a = \ldots\) \(\dfrac{1}{243}\begin{pmatrix}1 & -\dfrac{a}{2}(243 - 1)\\ 0 & 243\end{pmatrix}\begin{pmatrix}123\\ -2\end{pmatrix} = \begin{pmatrix}2\\ -2\end{pmatrix} \Rightarrow \dfrac{123 + 2\dfrac{a}{2}(243 - 1)}{243} = 2 \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(a = 1.5\) | A1 | 1.1b |
| (5) | ||
| (11 marks) |
Notes
(i) B1: States correct determinant. This can be implied by a correct equation
M1: Correct method to find a value of \(n\) using \(5 \times \text{`their } \det\left(\mathbf{M}^n\right)\text{'} = 1215\) which involves solving an index equation of the form \(p^n = q\) where \(n \gt 1\)
A1: \(n = 5\)
(ii) M1: Sets up an equation by multiplying the matrix \(\mathbf{M}^n\) by \(\begin{pmatrix}2\\ -2\end{pmatrix}\) setting equal to \(\begin{pmatrix}123\\ -2\end{pmatrix}\) and reaches a value for \(a\). You may just see \(2\left(3^n\right) - 2\dfrac{a}{2}\left(3^n - 1\right) = 123 \Rightarrow a = \ldots\)
Follow through on their value for \(n\).
A1: \(a = 1.5\)
(Corrected from the printed mark scheme: in the inverse-matrix line the printed scheme has \(a\) in place of 243 in the bottom right of the inverse matrix, and the equation as \(\dfrac{123 - 2\frac{a}{2}(243 - 1)}{243} = -2\).)