If $A=\begin{bmatrix}1 & 2 \\3 & 1\end{bmatrix}$, $B_2=\{x\in\Bbb{R}^2|\|x\|_2\le1\}$, parameterize the borders to draw the image of $T(x)=A.x$ for $x\in B_2$.
I only thought that $x_1^2+x_2^2=1$ is the border of $B_2$ and the transformation do $T(x)=\begin{bmatrix}x_1+2.x_2\\3.x_1+x_2\end{bmatrix}$. But what would be the equations of $T(B_2)$?
One way to find equations (or inequalities) for $T(B_2)$ is as follows.
Note that the point $x = (x_1,x_2)$ will be an element of $T(B_2)$ if and only if $T^{-1}(x) \in B_2$. We note that $T^{-1}(x) = A^{-1}x$, and $$ A^{-1} = \frac 15\pmatrix{1&-2\\-3&1}. $$ So, the inequality that defines $T(B_2)$ is $\|A^{-1}x\| \leq 1$; that is $$ \frac {(x_1 - 2x_2)^2}{5^2} + \frac {-3x_1 + x_2}{5^2} \leq 1 \implies (x_1 - 2x_2)^2 + (-3x_1 + x_2)^2 \leq 5^2. $$