What is the dimension of this linear subspace? 
We are in $\mathbb{R}^{3}$ and we have the vectors $a_{1}=
\begin{pmatrix} 1\\  1\\ 0 \end{pmatrix}, a_{2}=\begin{pmatrix} 1\\ 
0\\  1 \end{pmatrix}$ and the linear subspace $U = \left\{
\begin{pmatrix} x\\  y\\  z \end{pmatrix} \in \mathbb{R}^{3}: x-3y+z =0
\right\}$
What's the dimension of the linear subspace $\text{span} \left\{
a_{1},a_{2} \right \}+U$?

$\text{span} \left\{
a_{1},a_{2} \right \}+U = s\begin{pmatrix}
1\\ 
1\\ 
0
\end{pmatrix}+t\begin{pmatrix}
1\\ 
0\\ 
1
\end{pmatrix}+\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}= \begin{pmatrix}
1 & 1 & x\\ 
1 & 0 & y\\ 
0 & 1 & z
\end{pmatrix} \begin{pmatrix}
s\\ 
t\\ 
1
\end{pmatrix}$
Now I tried to calculate the rank of the $3 \times 3$ matrix by using Gauss:
Take first line and subtract it with second line:
$\begin{pmatrix}
1 & 1 & x\\ 
0 & 1 & x-y\\ 
0 & 1 & z
\end{pmatrix}$
Now take second line and subtract it with third line:
$\begin{pmatrix}
1 & 1 & x\\ 
0 & 1 & x-y\\ 
0 & 0 & x-y-z
\end{pmatrix}$
We see that the rank is $3$ because we didn't get any full zero line, thus the dimension must be $3$.

Please can you tell me if I did it correctly and if not, how is it done easily and correctly? I need to know for my exam, would be very very nice of you! :)
 A: Although you are correct that elements of $\text{span}(a_1, a_2) + U$ are linear combinations of elements of both spaces, you were wrong in assuming that elements of $U$ are vectors $(x,y,z)$, since you did not take into account that the elements of $U$ satisfy $x -3y + z = 0$. 
So how would we solve this question? Well, you already have two vector spanning $\text{span}(a_1, a_2)$, namely $a_1, a_2$ (this you did correct!). Now let us find vectors which span $U$. Since we are working in $\mathbb{R}^3$, you might see that $U$ is a plane, hence two dimensional. Note that two vectors spanning $U$ are $(1,0, -1)$ and $(3, 1, 0)$ (you might find other vectors, but in order to keep things as easy as possible, choose vectors with as many zeros as possible!). Now we apply what you have done: we put the four vectors in a matrix and use Gaussian elimination to find (possible) zerorows. We have the matrix
$$\begin{pmatrix}
1 & 1 & 1 & 3\\
1 & 0 & 0 & 1 \\
0 & 1 & -1 & 0 
\end{pmatrix}$$
and after Gaussian Elimination, we find:
$$\begin{pmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1
\end{pmatrix}$$
so we find that the rank is 3, hence the dimension is 3 (as you concluded, but not in a correct way).
$\textbf{EDIT 1}$: I have made a terrible mistake: $(1,0,0)$ is NOT an element of $U$, I will edit it! This will change the resulting Gaussian Elimination, but (afther I have checked it) not the result about the dimension!
$\textbf{EDIT 2}$: in case you were wondering how we could see that the dimension of $U$ is 2: note that we have that the elements in $U$ satisfy $x - 3y + z = 0$. If you do not know that this represents a plane (through the origin) in $\mathbb{R}^3$, note that we have that 
$$x + z = 3y$$
and equivalently 
$$\frac{x + z}{3} = y.$$
Hence the elements in $U$ are of the form 
$$\begin{pmatrix}
x\\
\frac{x+z}{3}\\
z
\end{pmatrix}.$$
Therefore, we only have two variables. Taking $x = 3, z = 0$ results in $(3, 1, 0)$ (Note that I could have taken $x = 1$, and this would result in the second coordinate being $\frac{1}{3}$, but I do not like fractions in my matrix and scaling a vector does not change anything). Taking $x = 0, z = 3$, we find $(0, 1, 3)$. This are not the vectors I have taken before, but it shows (in my opinion) better that $U$ has dimension 2. This implies that you can pick any two linearly independent vectors as a basis of $U$ (making my choice valid).
$\textbf{EDIT 3}$ (content of my comment on this answer): If you are not used to the notion of 'rank', we could also argue as follows: Clearly $a_1, a_2$ are linearly independent, so the span a space of dimension 2. Since $U$ is also of dimension 2, we have two possibilities: either $U$ is a part of $\text{span}(a_1, a_2)$, in which case the total dimension is 2, or $U$ is not a part of $\text{span}(a_1, a_2)$. You can show that the second case must hold by finding a vector $u$ in $U$ (such as the ones I already described) and show that $a_1, a_2, u$ forms a triple of linearly independent vectors. 
