Year: 2013
Paper: 1
Question Number: 3
Course: LFM Pure
Section: Matrices
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were good with candidates finding much to occupy them profitably during the three hours of the examination. In hindsight, two or three of the questions lacked sufficient 'punch' in their later parts, but at least most candidates showed sufficient skill to identify them and work on them as part of their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6 questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-attempts, most or all of which were fantastically accomplished mathematically, and such excellence is very heart-warming.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
For any two points $X$ and $Y$, with position vectors $\bf x$ and $\bf y$ respectively, $X*Y$ is defined to be the point with position vector $\lambda {\bf x}+ (1-\lambda){\bf y}$, where $\lambda$ is a fixed number.
\begin{questionparts}
\item
If $X$ and $Y$ are distinct, show that $X*Y$ and $Y*X$ are distinct unless $\lambda$ takes a certain value (which you should state).
\item Under what conditions are $(X*Y)*Z$ and $X*(Y*Z)\,$ distinct?
\item
Show that, for any points $X$, $Y$ and $Z$,
\[
(X*Y)*Z = (X*Z)*(Y*Z)\,
\]
and obtain the corresponding result for $X*(Y*Z)$.
\item The points $P_1$, $P_2$, $\ldots$ are defined by $ P_1 = X*Y$ and, for $n \ge2$, $P_n= P_{n-1}*Y\,.$
Given that $X$ and $Y$ are distinct and that
$0<\lambda<1$, find the ratio in which $P_n$ divides the line segment $XY$.
\end{questionparts}\begin{questionparts}
\item Suppose $X*Y = Y*X$, then
\begin{align*}
&& X * Y &= \lambda \mathbf{x} + (1-\lambda) \mathbf{y} \\
&& Y * X &= \lambda \mathbf{y} + (1-\lambda) \mathbf{x}\\
\Rightarrow && 0 &= (2\lambda - 1)(\mathbf{x} -\mathbf{y})
\end{align*}
Therefore, either $\mathbf{x} = \mathbf{y}$ or $\lambda = \frac12$. Since we assumed $X,Y$ were distinct, $\mathbf{x} \neq \mathbf{y}$ and so $X*Y$ and $Y*X$ are distinct unless $\lambda = \frac12$
\item Suppose $(X*Y)*Z = X*(Y*Z)$
\begin{align*}
&&(X*Y)*Z &= (\lambda \mathbf{x} + (1-\lambda) \mathbf{y}) * \mathbf{z} \\
&&&= (\lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)\mathbf{z}\\
&&X*(Y*Z) &=\mathbf{x}* (\lambda \mathbf{y} + (1-\lambda) \mathbf{z}) \\
&&&= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z}\\
\Rightarrow && 0 &= (\lambda^2 - \lambda)\mathbf{x} + ((1-\lambda) - (1-\lambda)^2)\mathbf{z} \\
&&&=(1-\lambda)(-\lambda \mathbf{x} +\lambda \mathbf{z}) \\
&&&= \lambda(1-\lambda)(\mathbf{z}-\mathbf{x})
\end{align*}
Therefore they are distinct unless $\lambda = 1, 0$ or $\mathbf{x} = \mathbf{z}$.
\item Claim: $(X*Y)*Z = (X*Z)*(Y*Z)$
Proof:
\begin{align*}
&& (X*Y)*Z &= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\
&& (X*Z)*(Y*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) * (\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\
&&&= \lambda(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) + (1-\lambda)(\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\
&&&= \lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda) \mathbf{z}
\end{align*}
Claim: $X*(Y*Z) = (X*Y)*(X*Z)$
Proof:
\begin{align*}
X*(Y*Z) &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\
(X*Y)*(X*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y})*(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\
&= \lambda (\lambda \mathbf{x} + (1-\lambda)\mathbf{y}) + (1-\lambda)(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\
&= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z}
\end{align*}
\item $P_1 = X*Y$ divides the line segment into the ratio $\lambda:(1-\lambda)$. $P_n$ divides the line segment $P_{n-1}Y$ into the ratio $\lambda:(1-\lambda)$, therefore it divides the line segment $XY$ in the ratio $\lambda^n : 1- \lambda^n$
Alternatively,
\begin{align*}
P_1 &= \lambda \mathbf{x} + (1-\lambda)\mathbf{y} \\
P_2 &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y} )*\mathbf{y} \\
&= \lambda^2 \mathbf{x} + (1-\lambda^2) \mathbf{y}
\end{align*}
Suppose $P_k = \lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}$ then
\begin{align*}P_{k+1} &= (\lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}) * \mathbf{y} \\
&= \lambda^{k+1}\mathbf{x} + \lambda(1-\lambda^k)\mathbf{y} + (1-\lambda)\mathbf{y}\\
& = \lambda^{k+1}\mathbf{x} + (1-\lambda^{k+1})\mathbf{y}\end{align*}
\end{questionparts}
This vector question was actually very straightforward, though its unfamiliar appearance clearly put most candidates off, with only around 350 of them making an attempt at it. There were nine marks available for the first two parts, which were technically undemanding, and it is no coincidence that the mean score on the question was around 9½/20. I suspect that, for the most part, this was considered by candidates to be one of those questions that are done towards the end of the examination in order to bump up their paper total by getting the easier marks at the beginning of the question, with no real intention of making a complete attempt. Candidates usually gave up part-way through (iii) where a stab at the "corresponding result for X * (Y * Z)" was required of them, which was actually just (X * Y) * (X * Z). I imagine this highlights the lack of students' familiarity with such properties as distributivity when considering binary operations.