Year: 2024
Paper: 3
Question Number: 5
Course: LFM Pure
Section: 3x3 Matrices
No solution available for this problem.
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, $\mathbf{M}$ and $\mathbf{N}$ are non-singular $2 \times 2$ matrices.
The \emph{trace} of the matrix $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is defined as $\operatorname{tr}(\mathbf{M}) = a + d$.
\begin{questionparts}
\item Prove that, for any two matrices $\mathbf{M}$ and $\mathbf{N}$, $\operatorname{tr}(\mathbf{MN}) = \operatorname{tr}(\mathbf{NM})$ and derive an expression for $\operatorname{tr}(\mathbf{M}+\mathbf{N})$ in terms of $\operatorname{tr}(\mathbf{M})$ and $\operatorname{tr}(\mathbf{N})$.
\end{questionparts}
The entries in matrix $\mathbf{M}$ are functions of $t$ and $\dfrac{\mathrm{d}\mathbf{M}}{\mathrm{d}t}$ denotes the matrix whose entries are the derivatives of the corresponding entries in $\mathbf{M}$.
\begin{questionparts}
\setcounter{enumi}{1}
\item Show that
\[\frac{1}{\det \mathbf{M}} \frac{\mathrm{d}}{\mathrm{d}t}(\det \mathbf{M}) = \operatorname{tr}\!\left(\mathbf{M}^{-1} \frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t}\right).\]
\item In this part, matrix $\mathbf{M}$ satisfies the differential equation
\[\frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t} = \mathbf{MN} - \mathbf{NM},\]
where the entries in matrix $\mathbf{N}$ are also functions of $t$.
Show that $\det \mathbf{M}$, $\operatorname{tr}(\mathbf{M})$ and $\operatorname{tr}(\mathbf{M}^2)$ are independent of $t$.
In the case $\mathbf{N} = \begin{pmatrix} t & t \\ 0 & t \end{pmatrix}$, and given that $\mathbf{M} = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ when $t = 0$, find $\mathbf{M}$ as a function of $t$.
\item In this part, matrix $\mathbf{M}$ satisfies the differential equation
\[\frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t} = \mathbf{MN},\]
where the entries in matrix $\mathbf{N}$ are again functions of $t$.
The trace of $\mathbf{M}$ is non-zero and independent of $t$. Is it necessarily true that $\operatorname{tr}(\mathbf{N}) = 0$?
\end{questionparts}
This question was a little less popular than question 4 but was less successful with a mean score of under 8/20. The first part was very well-answered with some efficiently realising that elements not on the leading diagonal did not need calculating. Sadly, some overlooked the second result required. Part (ii) was well-answered too, with the same efficiency as in (i) being employed by some. Part (iii) was less well-answered, with the non-conjugate nature of matrix multiplication often being overlooked, and in the last result treating A, B, C, and D as constants. Applying the scalar version of the chain rule to differentiate M² was not an uncommon error, but those that answered this part successfully usually rewrote tr(M²) in terms of tr(M) and det(M). Part (iv) caused the most difficulty. Only a handful attempted to provide an explicit counterexample to the statement. Some gave a counterexample that did not satisfy all the conditions on M and N, and a larger number of students convinced themselves that there is no good reason for the claim to hold, but did not give a counterexample. Some students attempted to prove the claim was true. Due to this there were many more 17/20 solutions than 18 or 19/20 solutions. Only 6 candidates achieved 20/20.