I have the following vectorial subspaces :
U = $\left(\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix} \right)$ and V = $\left(\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} \right)$
To calculate the basis of U + V should have use this method : $$ U + V = \left(\begin{pmatrix} (2+0) \\ (0+2) \\ (1+1) \\ (-2+0) \end{pmatrix} , \begin{pmatrix} (3-1) \\ (6+1) \\ (9+0) \\ (-12+1)\end{pmatrix} \right)$$ and use the Gaussian elimination to calculate $$\left(\begin{array}{cc|c} 2 & 2 & x_1 \\ 2 & 7 & x_2 \\ 2 & 9 & x_3 \\ -2 & -11 & x_4 \end{array}\right) $$
Or I am wrong and it's $\left(\begin{array}{cccc|c} 2 & 3 &0 & -1& 0 \\ 0 & 6 &2&1 & 0 \\ 1 & 9 &1&0&0 \\ -2 & -12&0&1 & 0 \end{array}\right)$ which generate this solution after reduction $\left(\begin{array}{cccc|c} 1 & 0 &0 & -\frac{1}{2}& 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$ $\sim$ $\begin{pmatrix} 1 & 0 & 0 & -\frac{1}{2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} \end{pmatrix}$
Thank you for your responses!
The former is wrong. To illustrate why, consider a simpler example: $$U = \operatorname{span}((1, 0)) \quad V = \operatorname{span}((0, 1)).$$ Then $U + V = \Bbb{R}^2$, as any point $(a, b) \in \Bbb{R}^2$ can be expressed as $$(a, b) = (a, 0) + (0, b)$$ where $(a, 0) = a(1, 0) \in U$ and $(0, b) = b(0, 1) \in V$. However, using your method, we get $$U + V \stackrel{?}{=} \operatorname{span}((1, 0) + (0, 1)) = \operatorname{span}((1,1)),$$ which is clearly false, as $\operatorname{span}((1,1))$ is the line $y = x$, not all of $\Bbb{R}^2$.
Note also that this method of adding subspaces is sensitive to choice of basis. We could just as easily write $V = \operatorname{span}((0, 2))$, but $\operatorname{span}((1, 0) + (0, 2)) \neq \operatorname{span}((1, 0) + (0, 1))$. This is another hint that this method is not sound.
On the other hand, the latter method is sound. In general, if $B_1$ and $B_2$ are sets of vectors, then $$\operatorname{span} B_1 + \operatorname{span} B_2 = \operatorname{span}(B_1 \cup B_2).$$ This follows immediately from working with linear combinations; a linear combination of $B_1$ plus a linear combination of $B_2$ is always a linear combination of $B_1 \cup B_2$. Conversely, one can take a linear combination of $B_1 \cup B_2$, separate out the terms, and write it as a linear combination of $B_1$ plus a linear combination of $B_2$.
So, if you wish to find a basis for $U + V$, union their bases (for a set of four vectors), then row reduce to find a basis for the span. The resulting basis is a basis for $U + V$.