Problem source

Given two non-zero vectors  $\mathbf{a}=\begin{pmatrix}a_{1}\\
a_{2}
\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}b_{1}\\
b_{2}
\end{pmatrix}$ define  $\Delta\!\! \left( \bf a, \bf b \right)$ by  $\Delta\!\! \left( \bf a, \bf b \right) = a_1 b_2 - a_2 b_1$.
 
 
Let $A$, $B$ and $C$ be  points with position vectors $\bf a$, $\bf b$ and $\bf c$, respectively, no two of which are parallel. Let   $P$, $Q$ and $R$ be points with position vectors $\bf p$, $\bf q$ and $\bf r$, respectively, none of which are parallel. 
 
\begin{questionparts} 
\item
Show that there exists a $2 \times 2$ matrix $\bf M$ such that 
$P$ and  $Q$ are the images of $A$ and  $B$ under the transformation represented by $\bf M$. 
  
\item
Show that $ \Delta\!\! \left( \bf a, \bf b \right) \bf c  + \Delta\!\! \left( \bf c, \bf a \right) \bf b  + \Delta\!\! \left( \bf b, \bf c \right) \bf a = 0.  $ 
 
Hence, or otherwise, prove that a necessary and sufficient condition for the points  $P$, $Q$, and $R$ to be the images of points $A$, $B$ and  $C$ under the transformation represented by some $2 \times 2$ matrix $\bf M$ is that 
\[ 
\Delta\!\! \left( \bf a, \bf b \right) :  
\Delta\!\! \left( \bf b, \bf c \right) :  
\Delta\!\! \left( \bf c, \bf a \right) =  
\Delta\!\! \left( \bf p, \bf q \right) :  
\Delta\!\! \left( \bf q, \bf r \right) :   
\Delta\!\! \left( \bf r, \bf p \right). 
\] 
\end{questionparts}

Solution source

\begin{questionparts}
\item First notice that there is a matrix taking $(1,0)$ and $(0,1)$ to $P$ and $Q$.

Notice there is also a matrix taking $(1,0)$ and $(0,1)$ to $A$ and $B$. Since $A$ and $B$ are not parallel, this map is invertible. Then we must be able to compose this inverse with the second map to obtain a matrix $\mathbf{M}$ satisfying our conditions.

\item $\,$

\begin{align*}
&& LHS &= \Delta\!\! \left( \bf a, \bf b \right) \bf c  + \Delta\!\! \left( \bf c, \bf a \right) \bf b  + \Delta\!\! \left( \bf b, \bf c \right) \bf a  \\
&&&= (a_1b_2-a_2b_1) \binom{c_1}{c_2} + (c_1a_2-c_2a_1)\binom{b_1}{b_2} + (b_1c_2-b_2c_1)\binom{a_1}{a_2} \\
&&&= \binom{a_1b_2c_1-a_2b_1c_1+c_1a_2b_1-c_2a_1b_1+b_1c_2a_1-b_2c_1a_1}{a_1b_2c_2-a_2b_1c_2+c_1a_2b_2-c_2a_1b_2+b_1c_1a_2-b_2c_1a_2} \\
&&&= \binom{0}{0} \\
&&&= \mathbf{0}
\end{align*}

First note that the matrix taking $P$, $Q$ to $A$, $B$ is unique. 

($\Rightarrow$) Suppose $\mathbf{Ma} = \mathbf{p}$ and $\mathbf{Mb} = \mathbf{q}$ and $\mathbf{Mc} = \mathbf{r}$. Then notice that

\begin{align*}
&& \mathbf{0} &= \mathbf{M0} \\
&&&= \mathbf{M}\left ( \Delta\!\! \left( \bf a, \bf b \right) \bf c  + \Delta\!\! \left( \bf c, \bf a \right) \bf b  + \Delta\!\! \left( \bf b, \bf c \right) \bf a\right) \\
&&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{M} \bf c  + \Delta\!\! \left( \bf c, \bf a \right) \mathbf{M}\bf b  + \Delta\!\! \left( \bf b, \bf c \right) \mathbf{M}\bf a\\
&&&= \Delta\!\! \left( \bf a, \bf b \right)\bf r  + \Delta\!\! \left( \bf c, \bf a \right)\bf q  + \Delta\!\! \left( \bf b, \bf c \right) \bf p\\
\end{align*}

However, since $\mathbf{p}, \mathbf{q}, \mathbf{r}$ are not parallel, then these coefficients must be a scalar multiples of $\Delta(\mathbf{p}, \mathbf{q}), \cdots$ as required.

$(\Leftarrow)$ Suppose we have this relationship, and $\mathbf{Ma} = \mathbf{p}$ and $\mathbf{Mb} = \mathbf{q}$, then

\begin{align*}
&& \mathbf{0} &= \mathbf{M0} \\
&&&= \mathbf{M}\left ( \Delta\!\! \left( \bf a, \bf b \right) \bf c  + \Delta\!\! \left( \bf c, \bf a \right) \bf b  + \Delta\!\! \left( \bf b, \bf c \right) \bf a\right) \\
&&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{M} \bf c  + \Delta\!\! \left( \bf c, \bf a \right) \mathbf{M}\bf b  + \Delta\!\! \left( \bf b, \bf c \right) \mathbf{M}\bf a\\
&&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{Mc} + \Delta\!\! \left( \bf c, \bf a \right)\bf q  + \Delta\!\! \left( \bf b, \bf c \right) \bf p\\
\end{align*}

Since these are scalar multiples of $\Delta(\mathbf{p}, \mathbf{q}), \cdots$ and we write this as

\begin{align*}
&& \mathbf{0} &= \Delta(\mathbf{p}, \mathbf{q})\mathbf{Mc} + \Delta(\mathbf{r}, \mathbf{p})\mathbf{q} + \Delta (\mathbf{q}, \mathbf{r})\mathbf{p}
\end{align*}

But since $\mathbf{p}, \mathbf{q}, \mathbf{r}$ are not parallel, this means that $\mathbf{Mc}$ is uniquely defined to be $\mathbf{r}$ as required.
\end{questionparts}