If $rank(A)=n$ then $A$ is an equivalent matrix to $I_{n\times n}$. This means that there are some elementary matrices $E_1,E_2,...,E_s$ such that $E_s...E_2E_1A=I$ and so $$A=E_1^{-1}...E_s^{-1}I$$ But $E_i^{-1}$ are also invertible as $I$ so $A$ is invertible as well.
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
The idea is correct, but not the computations. What happens is that$$\operatorname{rank}T=2\neq\dim\mathbb R^3=3.$$More generally, no linear map $f$ from a vector space with dimension $n$ into a vector space $V$ with dimension $m>n$ can be impossible, since$$\operatorname{rank}f\leqslant n<m=\dim V.$$