TITLE: Basic Understanding of Linear Combinations Geometrically
QUESTION [5 upvotes]: I am currently working through MIT's Introduction to Linear Algebra by Gilbert Strang, with no previous matrix experience.  In the first lecture, we are given the following linear equations:
$$2x -  y = 0\\
-x + 2y = 3$$
The solution to the system of equations is $(1,2)$.
Following this Professor Strang rewrites the system of linear equations in a column picture:
x [ 2 ] + y [-1] = [ 0 ]
 [-1] +  [2] = [ 3 ]

In the following steps, the vectors $[2, -1]$ and $[-1, 2]$ are plotted to show that the solution $(0, 3)$ can be found by geometrically by multiplying the "$x$-vector" by $1$ and the "$y$-vector" by $2$, and adding the two results.
My confusion is as follows, looking at the column vectors:
In the first vector from the $x$ coefficients we get $[2,-1]$.   What property allows the two $x$ coefficients to be drawn as a vector in $x$ and $y$ on an $xy$ plot?  The result of this is that both coefficients from equation $1$ give magnitude only in the $x$-axis and the coefficients from equation $2$ are on the $y$-axis? Which confuses me if this is an $xy$ plot 
I am new to this so I am not articulating this very well, so apologies and thanks in advance for your time.
Diagrams and formal class notes on P1 & 2:
https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/the-geometry-of-linear-equations/MIT18_06SCF11_Ses1.1sum.pdf

REPLY [3 votes]: I was very lucky; I was searching on YouTube for videos on linear algebra and found some of the best videos ever made (at least in my opinion). The 15 videos, made by 3Blue1Brown (they are also on the Khan Academy site) can be found here. I would highly encourage you to watch these videos. They will help you so much, by providing the geometric intuition necessary for linear algebra to make sense. 
As for the system of equations, here's how it works. You're given the system of equations $$2x-y=0$$ $$-x+2y=3$$ Now, we can take the coefficients of the variables, and put them in a matrix:
$$\begin{bmatrix}2&-1\\-1&2\end{bmatrix}$$ Note that we put the "x" coefficients in a column, and the "y" coefficients in a column. Put them in in the order you have them in the equations, and you'll be good. However, make sure that when you do this, you don't put in a x coefficient and a y coefficient in the same column. Now, we can create a vector with the variables:
$$\begin{bmatrix}x\\y\end{bmatrix}$$ And a vector with the constants, on the other side of the equals sign:
$$\begin{bmatrix}0\\3\end{bmatrix}$$ And finally, put it all together: $$\begin{bmatrix}2&-1\\-1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\3\end{bmatrix}$$ This is our new equation to solve. Let's represent this equation for now by the equation $$Ax=b$$ where $A$ represents our matrix of coefficients, $x$ is our vector of variables, and $b$ is our vector of constants. What we can do is find the inverse of $A$, or $A^{-1}$, and multiply it by both sides, which leaves us with $$x=A^{-1}b$$ 
I won't go into how to find the inverses of matrices here; you can look it up in a YouTube video, but let it suffice to say that the inverse of $A$ in this case is $$\begin{bmatrix}\frac{2}{3}&\frac{1}{3}\\\frac{1}{3}&\frac{2}{3}\end{bmatrix}$$ Multiply by both sides, and we are left with $$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\3\end{bmatrix}\begin{bmatrix}\frac{2}{3}&\frac{1}{3}\\\frac{1}{3}&\frac{2}{3}\end{bmatrix}$$ Reduce the right hand side and we are left with $$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}$$ So the solution to the system is $$x=1, y=2$$
Hope this helps!