Where is the vector $\vec{x_{1}} = (1,0)$ being sent under $T$? Where is $\vec{x_{2}} = (0,1)$ being sent?
Do you know why the answers to the above questions basically gives you the answer to the problem?
Update
Recall that every linear transformation $T: \Bbb R^{n} \to \Bbb R^{m}$ has an associated matrix $A_{m \times n}$ such that for any $\vec{x}$, $T(\vec{x}) = A\vec{x}$. So, we can characterize any linear transformation $T$ by a matrix $M$ such that, to figure out what happens to $\vec{x}$ under $T$, we could just multiply out $A \vec{x}$.
Now, also recall that if $A$ is a $2 \times 2$ matrix, which it is in your example, then multiplying $A$ with the vector $(1,0)$ gives you the first column of $A$. Multiplying $A$ by $(0,1)$ gives you the second column of $A$. So, to figure out the two columns of $A$, since $T(1,0)$ is the same thing as $A$ times $(1,0)$, then computing $T(1,0)$ should give you the first column of $A$. Similarly, $T(0,1)$ should give you the second column of $A$.
Where is the vector $\vec{x_{1}} = (1,0)$ being sent under $T$? Where is $\vec{x_{2}} = (0,1)$ being sent?
Do you know why the answers to the above questions basically gives you the answer to the problem?
Update
Recall that every linear transformation $T: \Bbb R^{n} \to \Bbb R^{m}$ has an associated matrix $A_{m \times n}$ such that for any $\vec{x}$, $T(\vec{x}) = A\vec{x}$. So, we can characterize any linear transformation $T$ by a matrix $M$ such that, to figure out what happens to $\vec{x}$ under $T$, we could just multiply out $A \vec{x}$.
Now, also recall that if $A$ is a $2 \times 2$ matrix, which it is in your example, then multiplying $A$ with the vector $(1,0)$ gives you the first column of $A$. Multiplying $A$ by $(0,1)$ gives you the second column of $A$. So, to figure out the two columns of $A$, since $T(1,0)$ is the same thing as $A$ times $(1,0)$, then computing $T(1,0)$ should give you the first column of $A$. Similarly, $T(0,1)$ should give you the second column of $A$.