Basic Understanding of Linear Combinations Geometrically 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
 A: 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!
A: You can ignore the $x,y$ variables in the system of equations, since they have nothing to do with the plot. Think about it as, you want to find two numbers $a,b$ such that number $a$ times the vector $(2,-1)$ plus number $b$ times the vector $(-1,2)$ equals another vector $(0,3)$. Now because $2$ is the first coordinate in the first term, you plot it as $x$-coordinate. $-1$ is the second coordinate in the first term, you plot it as $y$-coordinate. Same idea for the second term, and the answer. 
A: You're missing your $x$ and $y$ in the second equation and your brackets seem to be confusing you.  The system of equations can be reinterpreted as saying that we have $$x\begin{bmatrix}2\\-1\end{bmatrix}+y\begin{bmatrix}-1\\2\end{bmatrix}=\begin{bmatrix}0\\3\end{bmatrix}.$$
If you plot the two vectors $\begin{bmatrix}2\\-1\end{bmatrix}$ and $\begin{bmatrix}-1\\2\end{bmatrix}$, you can indeed verify by drawing (and by arithmetic) that $x=1$, $y=2$ works.
