Given linear trans. $T(x) = Ax$, $A = \begin{pmatrix}1 && - 2 \\3 &&1 \end{pmatrix}$, find trans of $x + y=2$ Given linear trans. $T(x) = Ax$, $A = \begin{pmatrix}1 && - 2 \\3 &&1
\end{pmatrix}$.
Find to what line x + y = 2 is transformed to.
Well, I've written the transfromation as $T(x) = (x - 2y, 3x + y)$. and I don't know how to continue from here.
The answer is: $-2x + 3y = 14$
I believe a hint will be sufficient, it is really easy but somehow I couldn't find any info. about how to approach this kind of exercise.
 A: Hint:
Write the line as $r(t)=(0,2)+t(2,-2)$ where $t$ is a real number and consider $T(r(t))$ using linearity.
A: The line is $$[1,1]\left[\begin{array}{c}x\\y\end{array}\right]=2\\
[1,1]A^{-1}A\left[\begin{array}{c}x\\y\end{array}\right]=2\\
\left([1,1]A^{-1}\right)T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=2$$
A: Part of the problem is that you're equivocating on what $x$ means. In some cases, you're using it as a vector; in others, as the vector's first component. Sometimes you use both in the same equation! No wonder you're confused!
As you've noticed, $x+y=2$ is equivalent to $y=2-x.$ Hence, letting $\vec v$ be an arbitrary vector on the line, we have $\vec v=\begin{bmatrix}t\\2-t\end{bmatrix}$ for some constant $t.$
Since $T(\vec v)=A\vec v=\begin{bmatrix}3t-4\\2t+2\end{bmatrix},$ as you can (and should) verify, put $\begin{bmatrix}x\\y\end{bmatrix}=A\vec v,$ then find constants $a,b,c$ such that $ax+by=c.$
A: Write your initial vector $x$ as $x=(t,2-t)$ then apply the transformation on it and see what you obtain. :)
