Year: 2021
Paper: 3
Question Number: 2
Course: LFM Pure
Section: 3x3 Matrices
The total entry was a marginal increase from that of 2019, that of 2020 having been artificially reduced. Comfortably more than 90% attempted one of the questions, four others were very popular, and a sixth was attempted by 70%. Every question was attempted by at least 10% of the candidature. 85% of candidates attempted no more than 7 questions, though very nearly all the candidates made genuine attempts on at most six questions (the extra attempts being at times no more than labelling a page or writing only the first line or two). Generally, candidates should be aware that when asked to "Show that" they must provide enough working to fully substantiate their working, and that they should follow the instructions in a question, so if it says "Hence", they should be using the previous work in the question in order to complete the next part. Likewise, candidates should be careful when dividing or multiplying, that things are positive, or at other times non-zero.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Let
\[
x = \frac{a}{b - c}, \qquad y = \frac{b}{c - a} \qquad \text{and} \qquad z = \frac{c}{a - b},
\]
where $a$, $b$ and $c$ are distinct real numbers.
Show that
\[
\begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix}
\begin{pmatrix} a \\ b \\ c \end{pmatrix}
=
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\]
and use this result to deduce that $yz + zx + xy = -1$.
Hence show that
\[
\frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} \geqslant 2.
\]
\item Let
\[
x = \frac{2a}{b+c}, \qquad y = \frac{2b}{c+a} \qquad \text{and} \qquad z = \frac{2c}{a+b},
\]
where $a$, $b$ and $c$ are positive real numbers.
Using a suitable matrix, show that $xyz + yz + zx + xy = 4$.
Hence show that
\[
(2a + b + c)(a + 2b + c)(a + b + 2c) > 5(b+c)(c+a)(a+b).
\]
Show further that
\[
(2a + b + c)(a + 2b + c)(a + b + 2c) > 7(b+c)(c+a)(a+b).
\]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix}
\begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} a-xb+xc \\ ay+b-yc \\ -za+zb+c \end{pmatrix} \\
&&&= \begin{pmatrix} a-x(b-c) \\ b-y(c-a) \\ c-z(a-b) \end{pmatrix} \\
&&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{align*}
Notice since $a,b$ and $c$ are distinct real numbers the vector $\langle a,b,c \rangle$ cannot be the zero vector, so the determinant of the matrix is zero, ie $0= 1(1+yz)+x(y-yz)+x(yz+z) = 1 +yz+yx+zx$.
Notice also then that
\begin{align*}
&& \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} &= x^2+y^2+z^2 \\
&&&= (x+y+z)^2 - 2(xy+yz+zx) \\
&&&= 2 + (x+y+z)^2 \geq 2
\end{align*}
\item $\,$ \begin{align*}
&& \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix}
\begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} 2a-xb-xc \\ -ay+2b-yc \\ -za-zb+2c \end{pmatrix} \\
&&&= \begin{pmatrix} 2a-x(b+c) \\ 2b-y(c+a) \\ 2c-z(a+b) \end{pmatrix} \\
&&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\
\Rightarrow && 0 &= \det \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \\
&&&= 2(4 -yz)+x(-2y-yz)-x(yz+2z) \\
&&&= 8 - 2yz-2yx-2xyz-2zx\\
\Rightarrow && 4 &= xyz+xy+yz+zx
\end{align*}
\begin{align*}
&& (2a + b + c)(a + 2b + c)(a + b + 2c) &> 5(b+c)(c+a)(a+b) \\
\Leftrightarrow && \left ( \frac{2a}{b+c}+1 \right)\left ( \frac{2b}{c+a}+1 \right)\left ( \frac{2c}{a+b}+1 \right) &> 5 \\
\Leftrightarrow && \left ( x+1 \right)\left ( y+1 \right)\left ( x+1 \right) &> 5 \\
\Leftrightarrow && xyz+xy+yz+zx+x+y+z+1 &> 5 \\
\Leftrightarrow && 5+x+y+z&> 5 \\
\end{align*}
Which is clearly true since if $a,b,c$ are positve real numbers so are $x,y,z$.
This final inequality is equivalent to showing $x+y+z > 2$ ie
\begin{align*}
&& x+y+z &> 2 \\
\Leftrightarrow && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} &> 1 \\
\\
&& \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} & > \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = 1
\end{align*}
So we're done.
\end{questionparts}
This was the fourth most popular question being attempted by very nearly four fifths of the candidates. It was the third most successful with a mean mark of just over 9/20, though very few achieved full marks. With four "Show that"s, marks were frequently lost for lack of proper justification, and with inequalities to demonstrate involving fractional quantities, positivity was often not considered, let alone proved, or stated as relevant. Even if candidates stated det(M)=0, which they sometimes didn't, only a minority of candidates realised that they had to justify using det(M)=0, and of these only some could do so convincingly; there were a number of incorrect arguments used. Some candidates sacrificed marks by, for example, attempting to show that a²/(b−c)² + b²/(c−a)² + c²/(a−b)² ≥ 2 via a purely algebraic approach rather than using the result just found (i.e. ignoring the "hence"). For the very last part of the question, candidates used a variety of methods in order to explain why x + y + z > 2. The most common method was to express the sum in terms of a, b and c and then show that this was greater than 2, but approaches using the AM-GM inequality or by splitting into different cases were sometimes used successfully.