Thread: Determining matrix relationship of linear transformations

P' is the point of intersection of the line y=x and the line of slope 2 that passes through P.

[Sol] Letting (a,b) and (x',y') be the coordinates of P and P', respectively,

Calculating the equation of the line of slope 2 that passes through P.

y-b=2(x-a)
y=2x-2a+b

Since the lines y=x and y=2x-2a+b intersect at P'(x',y'),

x'=2a-b
y'=2a-b (That's my problem, how do you get x' and y'?)

Well you are solving the system of equations
y=2x-2a+b
x=y

so just substitute to get
$\displaystyle x=2x-2a+b \Rightarrow x=2a+b$
but then you have x=y, so this is also the y coordinate

This gives you the system you have.

I didn't read the title of this, if you want them to be solved as like a matrix equation, you need to first put the system in standard form.

so you have
y=2x-2a+b
x=y

This yields
-2x+y=-2a+b
x-y =0

Which is the same as
$\displaystyle \begin{pmatrix}-2 & 1 \\ 1 &-1 \end{pmatrix} \begin{pmatrix}x\\y \end{pmatrix}=\begin{pmatrix}-2a+b\\0 \end{pmatrix}$

So then you can solve this by row reduction, or finding the inverse of this matrix, or whatever method you want, you will get the same solution as above.