Why do you use the inverse matrix to find the image of a curve in the plane? Given a $2\times2$ linear transformation matrix and the equation of a curve in the plane (e.g. $x^2+y^2=1$), why does one use the inverse matrix to find the equation of the image of the equation after the transformation?
I recently watched a video that explained how $2\times2$ matrices represent linear transformations in the plane.  For example, if I wanted to find the image of the point $(1,0)$ after the transformation given by the matrix
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right],
$$
I would simply multiply:
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]\left[\begin{matrix}1\\0\\\end{matrix}\right]=\left[\begin{matrix}2\\-1\\\end{matrix}\right]\rightarrow(2,-1).
$$
To extend this idea, I wanted to see the image of the circle $x^2+y^2=1$ after the transformation given by the same matrix.  I figured that since
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}2x+y\\-x\\\end{matrix}\right],
$$
the equation of the transformed circle would be $(2x+y)^2+(-x)^2=1$.  But after testing a few points on the circle, I quickly realized that it was not the correct equation.  Instead, the correct equation turned out to be
$$
(-y)^2+(x+2y)^2=1,
$$
which is obtained by multiplying $x$ and $y$ by the inverse matrix:
$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]^{-1}\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}0&-1\\1&2\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}-y\\x+2y\\\end{matrix}\right]
$$
So why do we use the original matrix to find the coordinates of the image of individual points (i.e. $(1,0)\rightarrow(2,-1)$), but the inverse matrix to find the equation of the image of the equation of a curve?  Is there a geometric explanation for this (or how do I explain this to my students)?
 A: $(x_1,y_1)$ gets map to $(x_2, y_2)$.
We know that $x_1^2+y_1^2=1$, but after mapping, we can't promise that $x_2^2+y_2^2=1$ still holds. 
what we are sure is it's preimage obey the rules, hence we apply the inverse.
A: For a point $(x,y)$ on the circle we have $x^2+y^2=1$ and the transformed point is $(X,Y)$ such that
$$\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}X\\Y\\\end{matrix}\right]\implies
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]^{-1}\left[\begin{matrix}X\\Y\\\end{matrix}\right]=\left[\begin{matrix}0&-1\\1&2\\\end{matrix}\right]\left[\begin{matrix}X\\Y\\\end{matrix}\right]=\left[\begin{matrix}-Y\\X+2Y\\\end{matrix}\right]=\left[\begin{matrix}x\\y\\\end{matrix}\right]$$
then
$$(-Y)^2+(X+2Y)^2=1$$
A: We can be even more general here. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be any transformation of the plane (e.g. your matrix). Now suppose you have a subset $C$ of the plane defined by an equation $f(x,y) = 0$ (so for the circle, $f(x,y) = x^2 + y^2 - 1$.) Then if a point $p$ is on the circle $C$, $T(p)$ satisfies
$$f \circ T^{-1}(T(p)) = f(p) = 0.$$
Thus the transformed circle $T(C)$ is defined by the equation $f \circ T^{-1}$, not by the equation $f \circ T$.
A: A nice way to think about it is to consider the transformed curve as a geometric locus. Fist of all let the starting curve be $S$ and the transformed curve be $C$ and $\alpha$ the transformation itself. What we need to obtain is a function $f(X,Y)=0$ defined as the set $C$ of points $(X,Y) \implies (x,y) \in S$, namely:  $(X,Y) \in (C) \iff (x,y) \in (S)$.
Generally, when we calculate a geometric locus, we choose an equation that implies that a point is on the curve, and then we write the variables in that equation in terms of $x$ and $y$.
For example: the condition for a point to be on the unit circle is $d=1$, where $d$ is the distance from the origin. At this point we start writing the distance from the origin in terms of $(x,y)$.
For the transformed curve it is exactly the same. The condition for a point to be on the transformed curve is: $(x,y) \in S$, which is the equation of the original curve. Now we just need to write $(x,y)$ in terms of $(X,Y)$, and the job is done :)
