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.
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 \(\bf M\).
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
\[
\Delta\!\! \left( \bf a, \bf b \right) :
\Delta\!\! \left( \bf b, \bf c \right) :
\Delta\!\! \left( \bf c, \bf a \right) =
\Delta\!\! \left( \bf p, \bf q \right) :
\Delta\!\! \left( \bf q, \bf r \right) :
\Delta\!\! \left( \bf r, \bf p \right).
\]
Solution:
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.
\(\,\)
\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.