Year: 2020
Paper: 2
Question Number: 6
Course: LFM Pure
Section: Matrices
No solution available for this problem.
There were just over 800 entries for this paper, and good solutions were seen to all of the questions. Candidates should be aware of the need to provide clear explanations of their reasoning throughout the paper (and particularly in questions where the result to be shown is given in the question). Short explanatory comments at key points in solutions can be very helpful in this regard, as can clearly drawn diagrams of the situation described in the question. The paper included a few questions where a statement of the form "A if and only if B" needed to be proven – candidates should be aware of the meaning of such statements and make sure that both directions of the implication are covered clearly. In general, candidates who performed better on the questions in this paper recognised the relationships between the different parts of each question and were able to adapt methods used in earlier parts when working on the later sections of the question.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A $2 \times 2$ matrix $\mathbf{M}$ is real if it can be written as $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $a$, $b$, $c$ and $d$ are real. In this case, the \emph{trace} of matrix $\mathbf{M}$ is defined to be $\mathrm{tr}(\mathbf{M}) = a + d$ and $\det(\mathbf{M})$ is the determinant of matrix $\mathbf{M}$. In this question, $\mathbf{M}$ is a real $2 \times 2$ matrix.
\begin{questionparts}
\item Prove that
\[\mathrm{tr}(\mathbf{M}^2) = \mathrm{tr}(\mathbf{M})^2 - 2\det(\mathbf{M}).\]
\item Prove that
\[\mathbf{M}^2 = \mathbf{I} \text{ but } \mathbf{M} \neq \pm\mathbf{I} \iff \mathrm{tr}(\mathbf{M}) = 0 \text{ and } \det(\mathbf{M}) = -1,\]
and that
\[\mathbf{M}^2 = -\mathbf{I} \iff \mathrm{tr}(\mathbf{M}) = 0 \text{ and } \det(\mathbf{M}) = 1.\]
\item Use part \textbf{(ii)} to prove that
\[\mathbf{M}^4 = \mathbf{I} \iff \mathbf{M}^2 = \pm\mathbf{I}.\]
Find a necessary and sufficient condition on $\det(\mathbf{M})$ and $\mathrm{tr}(\mathbf{M})$ so that $\mathbf{M}^4 = -\mathbf{I}$.
\item Give an example of a matrix $\mathbf{M}$ for which $\mathbf{M}^8 = \mathbf{I}$, but which does not represent a rotation or reflection. [Note that the matrices $\pm\mathbf{I}$ are both rotations.]
\end{questionparts}
This was the second most popular question on the paper. In general, candidates need to be careful when proving statements of the form "A if and only if B" and should be aware that in some cases it may not be possible to prove both directions in one go. Candidates should also be aware that, in some cases, the algebra is not sufficient on its own to demonstrate the reasoning and explanations of the steps are often helpful. Part (i) of the question was generally completed well. In part (ii) many largely successful attempts were seen, but few candidates picked up all of the marks for this section. The main errors arose from not adequately considering cases and so dividing by 0, and from not noticing that a² = d² = 1 could result in a = −d = ±1. The most successful attempts in this part were the ones that separated the two directions of the implications. Many candidates misused the condition M ≠ ±I in trying to prove the implication in one direction or did not check this condition when proving the implication in the opposite direction. Few attempts at part (iii) were seen and a common mistake was to do the component-wise algebra to find M⁴ instead of using the results from previous parts. In general, those who had understood the previous parts and attempted part (iv) were able to solve the final part of the question.