Suppose that (x,y) --- (s,t) is the linear operator on R^2 definted by the equations 2x+y=s 6x+2y=t. Find the image of the line x+y=1 under this operator.

Question

Suppose that (x,y) ---> (s,t) is the linear operator on R^2 definted by the equations 2x+y=s 6x+2y=t. Find the image of the line x+y=1 under this operator.

anonymous · Answer

I don't know how to approach this question

anonymous · Answer

im a little confused myself....but lets see.  If i plug in a vector (x,y), and i know that x + y = 1, then when i plug it in the formula for s, i would get:
$$s = 2x + y \Rightarrow s = x+x+y \Rightarrow s = x+1$$
using the formula for t:
$$t = 6x+2y \Rightarrow t = 4x+2x+2y \Rightarrow t = 4x+2$$
so what does that mean? ima think about this for a sec <.<

anonymous · Answer

i could solve for x in the first equation, which gives:
$$s = x+1 \Rightarrow x = s-1$$
Then i'll plug that in the second equation and get:
$$t = 4x+2 \Rightarrow t = 4(s-1)+2 \Rightarrow t = 4s-2$$
Keep in mind t corresponds with y, and s with x, so its like we have a "new line" caused by the trasnformation:
$$y = -x+1 \rightarrow t = 4s-2$$

anonymous · Answer

Oh clever didn't think to isolate x, makes sense thanks for your help

anonymous · Answer

joemath could you attempt my question?