Year: 2022
Paper: 2
Question Number: 8
Course: LFM Pure
Section: Invariant lines and eigenvalues and vectors
No solution available for this problem.
Candidates appeared to be generally well prepared for most topics within the examination, but there were a few situations in questions where some did not appear to be as proficient in standard techniques as needed. In particular, the method for finding invariant lines required in question 8 and the manipulation of trigonometric functions that were needed in question 10 caused considerable difficulties for some candidates. An additional issue that occurred at numerous points in the paper relates to the direction in which a deduction is required. It is important that candidates make sure that they know which statement is the one that they should start from as they deduce the other and that it is clear in their solution that the logic has gone in the correct direction. Clarity of solution is also an issue that candidates should be aware of, especially in the situations where the result to be reached has been given. It is important to check that there are no special cases that need to be considered separately, and when dividing by a function it is necessary to confirm that the function cannot be equal to 0 (and in the case of inequalities that the function always has the same sign). When drawing diagrams and sketching graphs it is useful if significant points that need to be clear are not drawn over the lines on the page as these can be difficult to interpret during the marking process.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ be a real matrix with $a \neq d$. The transformation represented by $\mathbf{M}$ has exactly two distinct invariant lines through the origin.
\begin{questionparts}
\item Show that, if neither invariant line is the $y$-axis, then the gradients of the invariant lines are the roots of the equation
\[bm^2 + (a-d)m - c = 0.\]
If one invariant line is the $y$-axis, what is the gradient of the other?
\item Show that, if the angle between the two invariant lines is $45^\circ$, then
\[(a-d)^2 = (b-c)^2 - 4bc.\]
\item Find a necessary and sufficient condition, on some or all of $a$, $b$, $c$ and $d$, for the two invariant lines to make equal angles with the line $y = x$.
\item Give an example of a matrix which satisfies both the conditions in parts (ii) and (iii).
\end{questionparts}
There were a large number of attempts at this question, but high scores were rare. One of the major reasons for solutions losing marks was a lack of understanding that the case where one invariant line is the y-axis needs to be addressed separately (as is highlighted in part (i) of the question). Many candidates were, however, able to demonstrate excellent understanding of the geometry involved. In part (i) many candidates set up a matrix equation to find invariant points rather than invariant lines. Additionally, a significant number of candidates ignored the fact that the invariant lines must go through the origin, meaning that the algebra was more complicated. The special case of one invariant line being the y-axis was often dealt with incorrectly, with many candidates believing that this meant that the other invariant line must be the x-axis. There were many possible approaches to parts (ii) and (iii), the simplest of which involved the tangent addition formulae for part (ii) and the realisation that the two lines were reflections of one another in the line y = x for part (iii). However, many candidates did not appreciate that the angle between the invariant lines is not necessarily the same as the angle between two vectors representing their directions. Additionally, many solutions did not check carefully that the quantities being used were well defined. Those candidates who had scored well on parts (ii) and (iii) were often able to link them together and find a suitable matrix. A very small number of candidates produced the relevant calculations but failed to write down a matrix.