= Solution
The line through the origin is $l_0=\operatorname{span}(u)$, containing zero and closed under addition and scalar multiplication, so it is a <vector subspace>. If $l_{x_1}=l_{x_2}$, then $x_1\in l_{x_2}$ and $x_1=x_2+\lambda u$ for some real $\lambda$. Conversely that equality gives the same set after shifting the line parameter. Thus
$$
\boxed{l_{x_1}=l_{x_2}\iff x_1-x_2\in\operatorname{span}(u).}
$$
Changing representatives to $x'=x+su$, $y'=y+tu$ changes their sum by $(s+t)u$ and a scalar multiple by $\alpha su$. The line classes therefore do not change, proving both operations well-defined. Associativity follows from $l_{(x+y)+z}=l_{x+(y+z)}$. The zero vector of this <quotient vector space> is \b[the entire line $l_0$], because $l_x+l_0=l_x$.
For a <basis> $(u,b_1,b_2)$ of $\mathbb R^3$, every vector is $x=su+tb_1+vb_2$, so $l_x=t l_{b_1}+v l_{b_2}$. If this combination equals $l_0$, then $tb_1+vb_2=su$; independence of the three original <basis vectors> forces $t=v=s=0$. Hence \b[$(l_{b_1},l_{b_2})$ is a basis] of the space of <parallel lines as a quotient vector space>.
For the specified direction, add $zu$ to $(x,y,z)^T$ to obtain the representative $(x+z,y+3z,0)^T$. Thus its coordinates in the two standard line classes are $(x+z,y+3z)^T$. The two supplied image classes have coordinates $(1,1)^T$ and $(-3,1)^T$, while their input coordinates are $(1,-1)^T$ and $(1,1)^T$. If $a_1,a_2$ are the matrix columns, then $a_1-a_2=(1,1)^T$ and $a_1+a_2=(-3,1)^T$. Solving gives
$$
\boxed{A=\begin{pmatrix}-1&-2\\1&0\end{pmatrix}.}
$$
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