AQA Further Paper 1 2022 June — Question 4 1 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2022
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
DifficultyModerate -0.5 This is a straightforward application of eigenvector properties requiring students to recognize that M²v = 9v and Nv = 4v, then multiply: NM²v = N(9v) = 9Nv = 36v, so λ = 36. While it's a Further Maths topic, it's a single-step conceptual question worth only 1 mark with multiple choice answers, making it easier than average.
Spec4.03b Matrix operations: addition, multiplication, scalar

The vector \(\mathbf{v}\) is an eigenvector of the matrix \(\mathbf{N}\) with corresponding eigenvalue 4 The vector \(\mathbf{v}\) is also an eigenvector of the matrix \(\mathbf{M}\) with corresponding eigenvalue 3 Given that $$\mathbf{N}\mathbf{M}^2\mathbf{v} = \lambda\mathbf{v}$$ find the value of \(\lambda\) Circle your answer. [1 mark] 10 \(\quad\) 24 \(\quad\) 36 \(\quad\) 144

Question 4:
AnswerMarks Guidance
4Circles correct answer 2.2a
Total1
QMarking instructions AO
Question 4:
4 | Circles correct answer | 2.2a | B1 | 36
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
The vector $\mathbf{v}$ is an eigenvector of the matrix $\mathbf{N}$ with corresponding eigenvalue 4

The vector $\mathbf{v}$ is also an eigenvector of the matrix $\mathbf{M}$ with corresponding eigenvalue 3

Given that
$$\mathbf{N}\mathbf{M}^2\mathbf{v} = \lambda\mathbf{v}$$

find the value of $\lambda$

Circle your answer.
[1 mark]

10 $\quad$ 24 $\quad$ 36 $\quad$ 144

\hfill \mbox{\textit{AQA Further Paper 1 2022 Q4 [1]}}