Lets say I have an equation of a plane, $$x-3y+2z=0 $$
and I get matrix to transform it with say a 3x3 matrix with just a-i as place holders for the values in the matrix.
How would I find what the image of the plane is under the matrix transformation, also If there would be a different method for a function of one variable please let me know, I really dont seem to understand how to apply matrix transformations to equations.
We can write the plane $x - 3 y + 2 z = 0$ as
$$\pmatrix{1 & -3 & 2} \pmatrix{x\\y\\z} = 0,$$
that is, as the kernel of the covector $\omega \in (\Bbb R^3)^*$ given in the standard basis by $\pmatrix{1 & -3 & 2}$. An (invertible) matrix $A \in GL(3, \Bbb R)$ acts on covectors by $A \cdot \omega := \omega \circ A^{-1}$ (i.e., via the contragredient representation), so the transformed plane is $$\pmatrix{1 & -3 & 2} A^{-1} \pmatrix{x\\y\\z} = 0.$$ If one writes $A$ in terms of its components $a_{ij}$ and expands, one recovers an equation of the form $a x + b y + c z = 0$. Since we may scale both sides of the equation without changing the plane, and multiplying both sides by $\det A$ replaces the above equation with $$\pmatrix{1 & -3 & 2} \operatorname{adj}(A) \pmatrix{x\\y\\z} = 0 ,$$ where $\operatorname{adj}(A)$ denotes the adjugate matrix of $A$, which is typically nicer than the inverse.