So...I have to either prove or deny that there exists a unique linear map $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ where:
$$\ f(1,1)=(2,6)$$ $$\ f(-1,1)=(2,1)$$ $$\ f(2,7)=(5,3)$$
I know that if, for example, we remove the third condition it can easily be proven that $$\ (1,1),(-1,1)$$ are linearly independent and so they form a basis on $\mathbb{R}^2$, making such a linear map unique.
But now I have three conditions, where $$\ (1,1),(-1,1),(2,7)$$ are, as far as I can see, also linearly independent by checking their determinants in pairs. But how come there can be a basis on $\mathbb{R}^2$ with three vectors? Is this possible or is it a "trick" question?
You cannot check linear independence of a set of three (or more) vectors by checking them pairwise. That is like saying $6$ and $10$ share a factor, and $10$ and $15$ share a factor, and $6$ and $15$ share a factor, so $6$, $10$, and $15$ all share a common factor (they don't). And in fact, no three vectors in a two-dimensional vector space such as $\mathbb{R}^2$ can be linearly independent. So it must be that $(1,1)$, $(-1,1)$, and $(2,7)$ are linearly dependent. Now you need to find a way to write one of them as a linear combination of the others, and see if the three conditions you were given are consistent or not.