Find the matrix A that has two rows and two columns and has $$A\pmatrix{1\\ 1}=\pmatrix{2\\ 1}\text{ and }A\pmatrix{-1\\1}=\pmatrix{1\\-1}.$$
Question:
How do I write out the corresponding system of linear equations with $4$ equations and $4$ unknowns so I can determine $A$?
If you want to write a system of linear equations, then you can write $$A=\begin{pmatrix}a & b\\ c&d\end{pmatrix}$$
and then see that $$A\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}a & b\\ c&d\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}a+b \\ c+d\end{pmatrix}$$
so that gives you the first two equations $a+b=2$ and $c+d=1$.
You do similarly for the other vector and you will have four equations.
If you learned linear algebra before, you may also take an alternative approach. Remember that matrices are really just representations of linear operators given a basis. If you do not know what that means, don't worry. You will probably learn that soon.
You can express $\begin{pmatrix}2 \\ 1\end{pmatrix} = \alpha \begin{pmatrix}1 \\ 1\end{pmatrix} + \beta\begin{pmatrix}-1 \\ 1\end{pmatrix}$
and $\begin{pmatrix}1 \\ -1\end{pmatrix} = \gamma \begin{pmatrix}1 \\ 1\end{pmatrix} + \delta \begin{pmatrix}-1 \\ 1\end{pmatrix}$
and then you can conclude that in the basis $$\left\{\begin{pmatrix}1 \\ 1\end{pmatrix},\begin{pmatrix}-1 \\ 1\end{pmatrix}\right\}$$
the linear operator $\mathcal A$, which in the standard basis is represented by $A$, is represented by a matrix $A' = \begin{pmatrix}\alpha&\beta1 \\ \gamma & \delta\end{pmatrix}$
Then, finding $A$ is just a matter of changing the basis for $\mathcal A$, since $A$ is, by definition, the representation of $\mathcal A$ in the standard basis.