1. ## matrices problem

Question:

Under a certain transformation the image of the point (x,y) is (X,Y) where $\displaystyle \begin{pmatrix} X\\ Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\ 2 & 3 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}$. This transformation maps any point on the line $\displaystyle y = mx$ onto another point on the line $\displaystyle y=mx$. Find the two possible values of m.

My attempt:
I tried to look at the the effect of the transformation on the unit square $\displaystyle \begin{pmatrix} X\\ Y \end{pmatrix} = \begin{pmatrix} 1 & 4\\ 2 & 3 \end{pmatrix} \begin{pmatrix} 0 &1 &1 &0 \\ 0 &0 & 1 & 1 \end{pmatrix}$
$\displaystyle = \begin{pmatrix} 0 &1 &5 &4 \\ 0 &2 & 5 & 3 \end{pmatrix}$

Since it says y= mx
i thought it had something to do with reflection in the line y=mx where $\displaystyle m = tan \alpha$
I have tried to calculate the angle $\displaystyle \alpha$ between i.e ( 1,0) and its image (1,2) or (0,1) (4,3) but it could not figure it out.

I would like to know if my approach is wrong or right?
Please can i have help on finding the two values of m

2. ## Re: matrices problem

I don't know why you would "look at the the effect of the transformation on the unit square" when the problem says nothing about the unit square.

The vector associated with the point (x, y) on the line y= mx is $\displaystyle \begin{pmatrix}x \\ mx\end{pmatrix}$. Applying the transformation $\displaystyle \begin{pmatrix}1 & 4 \\ 2 & 3\end{pmatrix}$ to that gives $\displaystyle \begin{pmatrix}1 & 4 \\ 2 & 3\end{pmatrix}\begin{pmatrix}x \\ mx\end{pmatrix}= \begin{pmatrix}x+ 4mx \\ 2x+ 3mx\end{pmatrix}$.

We want that to be again of the form "y= mx" or, since I am already using "x" and "y", "v= mu" where u= (1+ 4m)x and v= (2+ 3m)x. So (2+ 3m)x= m(1+ 4m)x or $\displaystyle 2+ 3m= m+ 4m^2$. Solve the quadratic equation $\displaystyle 4m^2- 2m- 2= 0$.