The equation of C is $x^2 + y^2 =1 $
How do I find the equation of the curve $C'=f(C)$
This is the image of $C$ under the linear transformation $f$ represented by the matrix $A=\begin{bmatrix}2 & 1 \\ 1 & 2 \end{bmatrix}$
The inverse image of the curve $C'$ is the circle $C$
a)What is the equation for $C'$ ?
b)I'm also supposed to calculate the area enclosed by $C'$
I'm not sure what I'm supposed to do to obtain the equation. I have no idea how to start with (a) but I suppose that (b) can be calculated with the determinant of $C'$? Can someone confirm this too?
This is a neat question. I assume that you're working in the real plane, $\mathbb{R}^2$. It may be easier to think about the inverse of $f$ (which exists, since $A$ is invertible). Notice that $C'$ is the set of points $(x,y)$ in $\mathbb{R}^2$ such that $f^{-1}(x,y)$ lies on the unit circle. Now $$f^{-1} \left( \begin{bmatrix} x \\ y \end{bmatrix} \right) = \frac{1}{3}\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 2x - y \\ 2y - x \end{bmatrix} $$
so $C'$ consists of the points $(x,y) \in \mathbb{R}^2$ such that $(\frac{1}{3}(2x - y))^2 + (\frac{1}{3}(2y - x))^2 = 1$, i.e. $5x^2 - 8xy + 5y^2 = 9$.
For question (b), you are correct in that the area enclosed changes by a factor of (the absolute value of) the determinant of $A$. In this case $\det A = 3$, so the area enclosed by $C'$ is $3$ times the area enclosed by $C$, i.e. $3\pi$.