Moderate -0.3 This is a straightforward application of finding invariant points by solving (M - I)x = 0, requiring only basic matrix algebra and solving a system of linear equations. While it's a Further Maths topic, the execution is mechanical with no conceptual difficulty, making it slightly easier than an average A-level question despite the 2-mark allocation.
Obtains two equations in and . May be seen as a single vector equation.
At least one equation must be correct.
𝑥𝑥 𝑦𝑦
Accept a pair of letters other than x and y.
Answer
Marks
Guidance
Ignore any subsequent incorrect working.
1.1a
M1
� �� �= � �
1 4 𝑦𝑦 𝑦𝑦
2𝑥𝑥+3𝑦𝑦 𝑥𝑥
� � = � �
𝑥𝑥+4𝑦𝑦 𝑦𝑦
2𝑥𝑥+3𝑦𝑦 = 𝑥𝑥 and 𝑥𝑥+4𝑦𝑦 =𝑦𝑦
𝑥𝑥 = −3𝑦𝑦
∴ two invariant points are
Obtains any two correct invariant points, with no incorrect points.
Condone correct points given as position vectors.
NMS: Correct answer scores 2/2.
Answer
Marks
Guidance
NMS: One correct invariant point and only one incorrect point scores SC1.
1.1b
A1
Total
2
(0,0) and (−3,1)
Q
Marking instructions
AO
Question 7:
7 | Obtains two equations in and . May be seen as a single vector equation.
At least one equation must be correct.
𝑥𝑥 𝑦𝑦
Accept a pair of letters other than x and y.
Ignore any subsequent incorrect working. | 1.1a | M1 | 2 3 𝑥𝑥 𝑥𝑥
� �� �= � �
1 4 𝑦𝑦 𝑦𝑦
2𝑥𝑥+3𝑦𝑦 𝑥𝑥
� � = � �
𝑥𝑥+4𝑦𝑦 𝑦𝑦
2𝑥𝑥+3𝑦𝑦 = 𝑥𝑥 and 𝑥𝑥+4𝑦𝑦 =𝑦𝑦
𝑥𝑥 = −3𝑦𝑦
∴ two invariant points are
Obtains any two correct invariant points, with no incorrect points.
Condone correct points given as position vectors.
NMS: Correct answer scores 2/2.
NMS: One correct invariant point and only one incorrect point scores SC1. | 1.1b | A1
Total | 2 | (0,0) and (−3,1)
Q | Marking instructions | AO | Mark | Typical solution
Find two invariant points under the transformation given by $\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$
[2 marks]
\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q7 [2]}}