SPS SPS FM Pure 2022 February — Question 2 5 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2022
SessionFebruary
Marks5
TopicInvariant lines and eigenvalues and vectors
TypeFind invariant lines through origin
DifficultyChallenging +1.2 This question requires finding eigenvalues of a 2×2 matrix (using the characteristic equation), then interpreting the result geometrically. The calculation is straightforward (determinant of λI - A = 0 gives complex eigenvalues), but connecting complex eigenvalues to 'no invariant lines' requires conceptual understanding beyond routine computation. It's harder than a standard eigenvalue calculation but doesn't require extended problem-solving.
Spec4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines

\(\mathbf{A} = \begin{pmatrix} 4 & -2 \\ 5 & 3 \end{pmatrix}\) The matrix \(\mathbf{A}\) represents the linear transformation \(M\). Prove that, for the linear transformation \(M\), there are no invariant lines. [5]

$\mathbf{A} = \begin{pmatrix} 4 & -2 \\ 5 & 3 \end{pmatrix}$

The matrix $\mathbf{A}$ represents the linear transformation $M$.

Prove that, for the linear transformation $M$, there are no invariant lines. [5]

\hfill \mbox{\textit{SPS SPS FM Pure 2022 Q2 [5]}}