A2 June 2025 Paper 2 Q3
3. Given
\[\mathbf{A} = \begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}\]Given
\[\mathbf{B} = \begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}\]The transformation \(Q\) is represented by the matrix \(\mathbf{A}^n\)
The transformation \(P\) followed by the transformation \(Q\) is the transformation \(R\), which is represented by the matrix \(\mathbf{C}\)
Given that, for a particular value of \(n\), the transformation \(R\) maps the point with coordinates \((27,\ 1)\) to the point with coordinates \((a,\ a)\), where \(a\) is a constant,
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1,\ \text{LHS} = \begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}^1 = \begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix},\ \text{RHS} = \begin{pmatrix}1 & 5\left(2^1 - 1\right)\\ 0 & 2^1\end{pmatrix} = \begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}\) So the result is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k,\ \begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}^k = \begin{pmatrix}1 & 5\left(2^k - 1\right)\\ 0 & 2^k\end{pmatrix}\) then for \(n = k + 1\) \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}^{k+1} = \begin{pmatrix}1 & 5\left(2^k - 1\right)\\ 0 & 2^k\end{pmatrix}\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}\) or \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}^{k+1} = \begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}\begin{pmatrix}1 & 5\left(2^k - 1\right)\\ 0 & 2^k\end{pmatrix}\) | M1 | 2.4 |
| \(= \begin{pmatrix}1 & 5 + 2 \times 5\left(2^k - 1\right)\\ 0 & 2\left(2^k\right)\end{pmatrix}\) or \(\begin{pmatrix}1 & 5\left(2^k - 1\right) + 5\left(2^k\right)\\ 0 & 2\left(2^k\right)\end{pmatrix}\) | M1 A1 | 1.1b 1.1b |
| \(\begin{pmatrix}1 & 10\left(2^k\right) - 5\\ 0 & 2^{k+1}\end{pmatrix} = \begin{pmatrix}1 & 5\left[2\left(2^k\right) - 1\right]\\ 0 & 2^{k+1}\end{pmatrix} = \begin{pmatrix}1 & 5\left(2^{k+1} - 1\right)\\ 0 & 2^{k+1}\end{pmatrix}\) | 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) n” | A1 | 2.4 |
| (6) |
Notes
B1: Shows that the result holds for \(n = 1\). Must see substitution in the RHS minimum required is \(\begin{pmatrix}1 & 5(2 - 1)\\ 0 & 2\end{pmatrix}\) and reaches \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}\) No need to state “true for \(n = 1\)” for this mark.
M1: Assumes the result is true for some value of \(n = k\), and sets up a matrix multiplication of the assumed result multiplied by the original matrix, either way round (no need to carry out for this mark, it is for essentially explaining the procedure). “Assume (true for) \(n = k\)” (oe) is sufficient for the assumption and may even be part of the conclusion and accept any alternative wording that indicates the assumption has been made. Allow for setting up the multiplication in reverse (ie working from \(n = k + 1\) towards \(n = k\), \(\mathbf{A}^k = \mathbf{A}^{k+1}\mathbf{A}^{-1}\) oe)
M1: Carries out the multiplication. (Allow the reverse case.)
A1: Achieves a correct un-simplified matrix. Accept in working in reverse.
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. If working from both sides all steps in showing the two sides are equal must be seen. If working in reverse they must return to \(\mathbf{A}^{k+1} = \ldots\) for this mark.
A1: Correct formal conclusion. This mark is dependent on the M and previous A marks having been scored and an attempt at the check for \(n = 1\) (if e.g. they didn’t show sufficient detail). It is gained by conveying the ideas of all three bold points at the end of their solution.
Note: Some cases may use \(n = 0\) as the base case. These can score full marks if dealt with correctly, but the conclusion must be consistent with their initial check to score the final A.
For the B mark minimum \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}^0 = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) ( or \(\mathbf{I}\)) and \(\begin{pmatrix}1 & 5(1 - 1)\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) should be seen. (No need to state true for \(n = 0\) for this mark – but must have consistent conclusion for the final A as noted above). If unsure send to review.
| Scheme | Marks | AO |
|---|---|---|
| Reflection | B1 | 1.1b |
| Reflection in the \(y\)-axis or line \(x = 0\) | B1 | 1.1b |
| (2) |
Notes
B1: Identifies the transformation as a reflection. Accept stretch with scale factor \(-1\).
B1: Identifies reflection and the correct line of reflection. Condone phrasing such as “across the \(y\)-axis”. Must be a single transformation – B0 if they state something else also happens.
NB Allow B1B0 for answers such as “flip in the y-axis” that convey the correct transformation in imprecise language.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}1 & 5\left(2^n - 1\right)\\ 0 & 2^n\end{pmatrix}\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix} = \begin{pmatrix}-1 & 5\left(2^n - 1\right)\\ 0 & 2^n\end{pmatrix}\) | B1 | 1.1b |
| (1) |
Notes
B1: Correct matrix
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}-1 & 5\left(2^n - 1\right)\\ 0 & 2^n\end{pmatrix}\begin{pmatrix}27\\ 1\end{pmatrix} = \begin{pmatrix}-27 + 5\left(2^n - 1\right)\\ 2^n\end{pmatrix}\) Sets \(-27 + 5\left(2^n - 1\right) = 2^n \Rightarrow 2^n = \ldots\) | M1 | 3.1a |
| \(\left(\mathbf{A}^n =\right)\ \begin{pmatrix}1 & 5(\text{`}8\text{'} - 1)\\ 0 & \text{`}8\text{'}\end{pmatrix} = \begin{pmatrix}1 & 35\\ 0 & 8\end{pmatrix}\) | M1 A1 | 1.1b 1.1b |
| (3) | ||
| (12 marks) |
Notes
M1: A complete method to find a value for \(2^n\) (or \(n\)). Multiplies the coordinates \((27,\ 1)\) by their matrix \(\mathbf{C}\) and sets the \(x\) and \(y\) coordinates equal to reach a value for \(2^n\) or \(n\) (condone negative values for \(2^n\)). Note you may allow this for reach a value of \(a\) if \(a = 2^n\) is stated or clearly implied by their working.
M1: Uses their value of \(2^n\) or \(n\), to find the matrix \(\mathbf{A}^n\). Note substituting into \(\mathbf{C}\) is M0 without further work to find \(\mathbf{A}^n\).
A1: Correct matrix.