Given the linear transformation matrix:
$$T=\pmatrix{2&-3\\1&1}$$
Find the equation for the curve that $T$ transforms a circle with equation $x^2+y^2=6$ into.
What I know:
My basis is going to be $[1,0]^T$ and $[0,1]^T$ because I'm in $\Bbb{R}^2$.
I have my transformation matrix $T$ (usually called $A$?). $T(x)=Ax$
I have one equation for the circle.
Need to find the equation for the curve. (circle transformed into curve)
I've been working vector/matrix type transformations, so the 'equation' type has me confused.
Any help would be greatly appreciated! Thanks!!
Best Answer
May I suggest a less sophisticated approach for this problem. Let us define $T(x,y)=(u,v)$ hence $$ u = 2x-3y \qquad v = x+y $$ Solve these equations for $x$ and $y$ as functions of $u,v$ and plug-these into $x^2+y^2=6$ and you'll obtain the formula for the image of the circle under the $T$-transformation. Of course, you can understand what happens in terms of deeper theory, but the approach I outline is totally valid for this problem.