Linear Algebra: How to interpret vectors and equations of a system of equations? From Strang's 18.06 Linear Algebra OCW Course I realized that there is a row view and column view to interpret a matrix and the solution to Ax = b. Well the view made sense but it was hard understanding the matrix operations from this. So I checked out other lectures on Youtube and found this https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab which explained matrices as Linear Transformations which eased me on the geometric part.
But as I stared to wonder the relationship between the system of equations and the columns of Matrix A in Ax = b. I couldn't find any connection!
Suppose I have this system of equations:
x + 2y = 4
x + 3y = 5
Solution is x = 2 and y = 1 which is essentially the point of intersection of line x = 4 - 2y and x = 5 - 3y.
Now if I go to solve this equation in vector form. I have the matrix A = \begin{bmatrix}1&2\\1&3\end{bmatrix} 
Now I have 2 vectors [1 1] and [2 3] which are the basis for my newly transformed space and I have to find the vector in my original vector space that gives me b = [4 5].
Or by Strang's way I have to find such a linear combination of column vectors of A that gives me b.
So my question is not why but HOW this equations of line is converted into vector and how can we say that they are equivalent? I can't seem to find an explanation of the relationship between the coordinate space and vector space. How does the coefficients of the x and y variable of the equations transform into i and j basis for the vector space?
 A: Not sure I entirely get your question right but in essence
$$\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}4 \\ 5\end{bmatrix}$$
is the same as splitting $A$ into its columns via
$$\begin{bmatrix}1 \\ 1\end{bmatrix}\cdot x + \begin{bmatrix}2 \\ 3\end{bmatrix}\cdot y= \begin{bmatrix}4 \\ 5\end{bmatrix}$$
which is the same as 
$$\begin{bmatrix}x \\ x\end{bmatrix} + \begin{bmatrix}2y \\ 3y\end{bmatrix}= \begin{bmatrix}4 \\ 5\end{bmatrix}$$
which gives you
$$\begin{bmatrix}x+2y=4 \\ x+3y=5\end{bmatrix} $$
which is the description in coordinte space you were referring to. Is it any of these steps that you are struggling with or did I misunderstand your question?
edit: Geometrically, the two interpretations live in different spaces. The second equation lives in the column space of the matrix and asks: how much do I need to go in direction of the first column and second column to reach the vector on the RHS? How can I express the RHS as a linear combination of the two columns?
Here is a picture (solid lines: first and second column vector (up to first dot); dashed lines: space spanned by each column, red cross: target (RHS), reachable by one green and two blue or two blue and one green):

The second interpretation lives in the row space of the matrix and says: each row of the matrix is one line (in general hyperplane) of possible solutions. If I intersect all the lines (hyperplanes) I get the one solution that satisfies all equations simultaneously. 
Here is a picture (notice that here the axis are x and y, not so in interpretation one, where x and y are the "lengths" of the vectors, i.e., how far you go along each):

Geometrically, these two are separate, as they are in different spaces and therefore we cannot directly connect them. Just imagine the matrix would not be square but, say, 3x2. Then columnspace vectors would live in $\mathbb{R}^3$ whereas rowspace vectors would live in $\mathbb{R}^2$. 
