Standard +0.8 This question requires understanding of invariant lines and eigenvalues/eigenvectors. Students must recognize that lines y=kx through the origin are invariant if and only if they correspond to eigenvectors, then show the characteristic equation has no real solutions (discriminant is negative). While conceptually demanding for Further Maths, it's a standard proof technique once the connection is understood.
The matrix \(\mathbf{C}\) is defined by
$$\mathbf{C} = \begin{bmatrix} 3 & 2 \\ -4 & 5 \end{bmatrix}$$
Prove that the transformation represented by \(\mathbf{C}\) has no invariant lines of the form \(y = kx\)
[4 marks]
Question 10:
10 | Obtains 3x+2kxand −4x+5kx
Accept any letter for k
Condone use of y =kx+c
Or
Uses det (C – λI) | 1.1a | M1 | For an invariant line y = kx
3 2 x x′
=
–4 5kx y′
3x+2kx= x′
–4x+5kx= y′
y′=kx′
–4x+5kx=k ( 3x+2kx )
–4 + 5k = 2k2 + 3k
2k2 – 2k + 4 = 0
∆ = 22 – 32 = –28
∆ < 0
The equation has no real roots,
so there is no invariant line of the
form y = kx.
Substitutes their x′ and y′ in
y′=kx′
Accept any letter for k
Condone use of y =kx+c
Or
Expands det (C – λI) | 1.1a | M1
Deduces
2k2 – 2k + 4 = 0 OE
Or
λ2 −8λ+23=0 | 2.2a | A1
Completes a reasoned
argument justifying that the
quadratic equation has no real
roots to prove that the
transformation represented by C
has no invariant lines of the form
y = kx
If y =kx+cis used then must
state and use c = 0 | 2.1 | R1
Question total | 4
Q | Marking instructions | AO | Marks | Typical solution
The matrix $\mathbf{C}$ is defined by
$$\mathbf{C} = \begin{bmatrix} 3 & 2 \\ -4 & 5 \end{bmatrix}$$
Prove that the transformation represented by $\mathbf{C}$ has no invariant lines of the form $y = kx$
[4 marks]
\hfill \mbox{\textit{AQA Further Paper 2 2024 Q10 [4]}}