Standard +0.8 This is a Further Maths question requiring understanding of invariant points and matrix inverses. While the proof is conceptually straightforward (if Mv=v then M^{-1}v=v follows from multiplying both sides by M^{-1}), students must recognize the structure and write a clear algebraic proof. It's above average difficulty due to the proof requirement and abstract reasoning about transformations, but not exceptionally hard as the logic is direct once the approach is identified.
The transformation T is defined by the matrix M. The transformation S is defined by the matrix \(\mathbf{M}^{-1}\). Given that the point \((x, y)\) is invariant under transformation T, prove that \((x, y)\) is also an invariant point under transformation S.
[3 marks]
Question 3:
3 | Commences a proof by correctly
setting up an equation using the
definition of an invariant point | AO2.1 | R1 | For an invariant point
x x
M
y y
Pre-multiply both sides by M–1
x x
M1M M1
y y
M -1 M = I hence
x x
M1
y y
x
Therefore is invariant
y
under S.
Pre-multiplies by M–1. | AO2.1 | R1
Uses M–1M = I and concludes their
rigorous mathematical argument to
deduce that (x, y) is invariant
under S AG | AO2.2a | R1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
The transformation T is defined by the matrix M. The transformation S is defined by the matrix $\mathbf{M}^{-1}$. Given that the point $(x, y)$ is invariant under transformation T, prove that $(x, y)$ is also an invariant point under transformation S.
[3 marks]
\hfill \mbox{\textit{AQA Further Paper 2 Q3 [3]}}