crisis of identity (matrix transformations)

Hi folks,

I read the following example in my maths book and feel the authors have not explained what they are doing very clearly. We have to find the equations of any straight lines that are invariant under the following transformation:

$\displaystyle \left( \begin{array}{c} x' \\ y' \end{array} \right) =

\left( \begin{array}{cc} -6 & 3 \\ 4 & -2 \end{array} \right)

\left( \begin{array}{c} x \\ y \end{array} \right) +

\left( \begin{array}{c} 4 \\ -1 \end{array} \right)

$

Where x' and y' are the image points of x and y. Any straight line will be of the form y = mx + c and any point on the line will have coordinates of the form (k, mk + c).

Thus

$\displaystyle \left( \begin{array}{c} x' \\ y' \end{array} \right) =

\left( \begin{array}{cc} -6 & 3 \\ 4 & -2 \end{array} \right)

\left( \begin{array}{c} k \\ mk + c \end{array} \right) +

\left( \begin{array}{c} 4 \\ -1 \end{array} \right) $

=

$\displaystyle \left( \begin{array}{c} -6k + 3mk + 3c + 4 \\

4k - 2mk - 2c - 1 \end{array} \right)

$

Now if y = mx + c is invariant, then this point (x', y') must also lie on y = mx + c

i.e. $\displaystyle 4k - 2mk - 2c - 1 = m(-6k + 3mk + 3c + 4) + c $

giving: $\displaystyle k(4 + 4m - 3m^2) = 3mc + 4m + 3c + 1 $ (equation 1).

This equation must be true for all values of k (i.e. equation 1 is an identity in k). Thus:

$\displaystyle 4 + 4m - 3m^2 = 0 $ and $\displaystyle 3mc + 3c + 1 = 0$

So: for those who have followed so far, I have to ask, how is equation 1 split into two parts? k is a variable and I know the difference between an equation and an identity (one is for all k, the other only particular values of k) but I don't understand the theory behind the way equation 1 suddenly becomes two equations both equal to zero. Can anyone help?

best regards