In $\mathbb{R}^3$, let $B = \{ a(3, 2, 0), b(1, -1, 0) \}$. I have to prove that B is linearly independent. So I have that, $\exists \alpha, \beta \in \mathbb{R}$:
$$ \alpha\left(3,2,0\right)+\beta\left(1,-1,0\right)=0 \\ \left(3\alpha+\beta,2\alpha-\beta,0\right)=0 $$
from which I get the equation system:
$$ 3\alpha+\beta=0 \\ 2\alpha-\beta=0 $$
By adding both equations:
$$ 3\alpha+\beta+2\alpha-\beta=0 \\ 5\alpha=0 \\ \alpha=0 \\ $$
And then:
$$ 3\alpha+\beta=0 \\ 3\left(0\right)+\beta=0 \\ \beta=0 $$
Therefore, B is linearly independent. Am I correct, so far?
Then I have to describe the space generated by B. So I don't get what the book means by "describing". As I understand it, the space generated by B is all the possible linear combinations of B, so the following combinations are in Gen{B}:
$$ 1\left(a\right)+3\left(b\right)=\left(3+3\left(1\right),2+3\left(-1\right),0+3\left(0\right)\right)=\left(6,-1,0\right) \\ 2\left(a\right)+1\left(b\right)=\left(2\left(3\right)+1\left(1\right),2\left(2\right)+2\left(-1\right),2\left(0\right)+2\left(0\right)\right)=\left(7,2,0\right) \\ \alpha\left(a\right)+\beta\left(b\right)=\left(3\alpha+\beta,2\alpha-\beta,0\alpha+0\beta\right)=\left(3\alpha+\beta,2\alpha-\beta,0\right) $$
Would this be enough to "describe" Gen{B}? Do I need to add something else?
I think you meant $B=\{ (3,2, 0), (1,-1,0)\}$.
To show that $B$ is linearly independent, show that $$\alpha(3,2,0)+\beta(1,-1,0)=0 \implies \alpha = \beta = 0$$
To describe the space generated by $B$, it is a $2$-dimensional plane that passes through the origin. You can find the equation explicitly.
$$\begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix}\times \begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ -5\end{bmatrix}$$
Hence the plane is $z=0$.