Hint: how would one write the the polynomial of degree $n<1$ in your space?
The polynomial of degree $0$ is of the form $c \ x^0$. Meaning it's just a constant.
This is how your matrix should look like initially. From top to bottom coefficients for $x^3,x^2,x^1,x^0$. Your vectors should have 4 components not 3.
\begin{bmatrix}
2&0 &1 \\
0&0 &-1 \\
1&1&0 \\
1&-2 &0
\end{bmatrix}
Oh answering your question about the polynomials or the matrices. We use the matrices because it's easier to show linear independence with them. Showing independence is equivalent to showing $A\vec x=\vec 0$ $\Rightarrow \vec x=\vec 0$. It would be hard to use polynomials themselves to show that.
The columns of the following matrix would form a basis since they are linearly independent.
\begin{bmatrix}
2&0 &1 &1 \\
0&0 &-1& 0\\
1&1&0 & 0\\
1&-2 &0 &0
\end{bmatrix}
Let's talk about vectors of real numbers with two elements in them. The set of these vectors is named $\mathbb{R}^2$.
Consider
$$\begin{bmatrix}1\\ 0\end{bmatrix} \text{and} \begin{bmatrix}0\\ 1\end{bmatrix}.$$
Can any other two element vector be written as a combination of these two vectors? Yes, it can! Consider any vector $\begin{bmatrix}a\\ b\end{bmatrix}$ for any values $a$ and $b$. $\begin{bmatrix}a\\ b\end{bmatrix}=a\begin{bmatrix}1\\ 0\end{bmatrix} + b\begin{bmatrix}0\\ 1\end{bmatrix}.$ Since any two element vector can be written as a linear combination of these two vectors, and we need both of these vectors, then these vectors form a basis.
Formally, we say that $\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$ is a basis of $\mathbb{R}^2$. We could also say that $\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$ generates $\mathbb{R}^2$.
What about the vectors $$\begin{bmatrix}1\\ 0\end{bmatrix}, \begin{bmatrix}0\\ 1\end{bmatrix} \text{and} \begin{bmatrix}1\\ 1\end{bmatrix}?$$
We could also write any vector $\begin{bmatrix}a\\ b\end{bmatrix}$ as a linear combination of these three vectors. However, we don't need all three of them. We could just as easily work with just the first two vectors. So the set of these vectors is not a basis of $\mathbb{R}^2$. However, it still generates $\mathbb{R}^2$ because $\text{span}\left(\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix},\begin{bmatrix}1\\ 1\end{bmatrix}\right\}\right)=\mathbb{R}^2$. Therefore, the dimension of $\mathbb{R}^2$ is two, because any basis set of $\mathbb{R}^2$ has two vectors in it.
Can we have a basis other than $\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$? Yes, we can. Consider
$$\begin{bmatrix}1\\ 1\end{bmatrix} \text{and} \begin{bmatrix}0\\ 1\end{bmatrix}.$$
Any vector $\begin{bmatrix}a\\ b\end{bmatrix}$ can be written as a linear combination of these two vectors:
$$ a\begin{bmatrix}1\\ 1\end{bmatrix} + (b-a)\begin{bmatrix}0\\ 1\end{bmatrix}=\begin{bmatrix}a\\ b\end{bmatrix}.$$
So $\left\{\begin{bmatrix}1\\ 1\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$ is also a basis of $\mathbb{R}^2$.
Imagine trying to send a vector space through a communication channel. If we had to send every vector then this would be an impossible thing to do because we would need an infinite number of vectors. What could we do instead? If we had a basis for our vector space, we could just send the basis and tell the person on the other end that the vectors we sent are a basis for our vector space. And that's the intuition. A basis is a small set of vectors that completely represents our vector space. In fact, there's no smaller set of vectors that completely represents our vector space.
Best Answer
$-2(2,-3)=(-4,6)$ so it is not a base.