Given the vectors $x_1 = \pmatrix{2\\3} , x_2 = \pmatrix{-1\\1}, y_1 = \pmatrix{1\\1\\1}, y_2 = \pmatrix{0\\1\\-1}$ and $y_3 = \pmatrix{3\\6\\0} $
$1)$ Show that there exists exactly one linear mapping $f :\mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that $f(x_1) = y_1$ and $f(x_2) = y_2$
$2)$ For which $i \in \mathbb{R}$ exists a mapping $g:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ with $g(y_1) = x_1$, $g(y_2)= x_2$ and $g(y_3) = \pmatrix{i\\0}$? Is g unique?
My answers:
$1)$ Because $x_1$ and $x_2$ form a Basis of $\mathbb{R}^2$ (they are linearly independent and their dimension is 2) then the mapping is unique. Proving that it's linear is quite simple (all I have to do is show that the properties of a linear map apply here).
$2)$ I tried to solve this as a set of linear equations and found that $i = 3$, but didn't find a proper transformation matrix that represents this mapping. Does this mean that it's not unique?
I'd really appreciate it if someone could show me how to approach question 2 and confirm if my answer for 1 is right or not.
Your answer to question 1 seems fine to me.
For question 2, notice that $y_3=3y_1+3y_2$. Therefore, as you said, $i$ has to be $3$ since it is uniquely determined by the images of $y_1$ and $y_2$. Moreover, such a map is not unique, since it is not defined on a basis of $\mathbb{R}^3$ (for example, you can map the vector $(0,0,1)$ wherever you want).
EDIT (some two years later): as pointed out by @rax_adaam (see comments below), if we look at the second component of $g(y_3)=g(3y_1+3y_2)$, we notice that its value is $12$, and not $0$ as required. Hence, there is no $i$ doing the job.