As soon as you have the matrix of a linear map $f$, you have a system of generators of the image of $f$.
Indeed the column-vectors of the matrix are the images of the vectors of the basis in the source-space, say $v_1,v_2,v_3$. Hence for any vector $v$ in $\mathbf R^3$, $v=\lambda_1 v_1+\lambda_2v_2+\lambda_3v_3$, we have:
$$f(v)=\lambda_1 f(v_1)+\lambda_2f(v_2)+\lambda_3f(v_3).$$
Naturally, this system of generators is not minimal, i. e. it is not a basis of the image in general. But from this system, you can deduce a basis by column reduction.
It will not be necessary here, because the rank-nullity theorem ensures $\dim\operatorname{Im}f=2$ since you've found that $\dim\ker f=1$. Thus, $f$ is surjective, i. e. the image of $f$ is the whole of $\mathbf R^2$.
Finding the kernel first is a good idea. (though actually what we need is just to find the nullity)
Construct the augmented matrix (note that the last column is actually not needed):
$$\left[\begin{array}{ccc|c}2 & -1 & 0 & 0\\ 1 & 1 & -3 & 0 \\ -1 & 0 & 1 & 0\end{array}\right]$$
And find its corresponding RREF:
$$\left[\begin{array}{ccc|c}1 & 0 & -1 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$$
See that the third column is not a pivot column, let $c=t$. from the second equation, $b-2c=0$, hence $b=2t$, from the first equation, $a-c=0$, hence $a=t$.
Hence solution to the system $Tx=0$ is $(a,b,c)=t(1,2,1)$.
A basis of the kernel would be $\{ 1+2x+x^2\}$.
As mentioned at the start, the goal is to find the nullity. From the RREF, we can see that $rank(T)=2$ and the the number of columns is $3$, hence $nullity(T)=1 > 0$, hence it is not injective.
Your codomain is $\mathbb{R}^3$, however as you can see from the RREF, $rank(T)=2 < 3$, it does not have enough vectors to span $\mathbb{R}^3$.
Best Answer
Hint: If $u$ and $v$ are vectors with the same image, what can you say about the image of $u-v$?