What does calculating the inverse of a matrix mean? Assume I have 3 equations, $x+2y+z=2, 3x+8y+z=12, 4y+z=2$ which could be represented in matrix form ($Ax = b$) like this:
$\begin{pmatrix}
1 & 2 & 1\\ 
3 & 8 & 1\\ 
0 & 4 & 1
\end{pmatrix}\bigl .\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} = \begin{pmatrix}
2\\ 
12\\ 
2
\end{pmatrix}$
Then, the inverse of $A$, $A^{-1}$, would be: $\begin{pmatrix}
2/5 & 1/5 & -3/5\\ 
-3/10 & 1/10 & 1/5\\ 
6/5 & -2/5 & 1/5
\end{pmatrix} $.
So, my question is, what does this even mean? We know that $A$ is a coefficients matrix that represents the 3 equations above, so what does $A^{-1}$ mean with respect to these 3 equations? What have I done to the 3 equations is exactly my question. 
Please note that I understand very well how to find the inverse of a matrix, I just don't understand the intuition of what's happening and sort of the meaning of the manipulations I am applying to the equations when they are in matrix form. 

Let us organize our equations in this way: 
$x\begin {bmatrix}
           1 \\
           3 \\
           \vdots \\
           0
         \end{bmatrix}$
+ $y\begin {bmatrix}
           2 \\
           8 \\
           \vdots \\
           4
         \end{bmatrix}$
+ $z\begin {bmatrix}
           1 \\
           1 \\
           \vdots \\
           1
         \end{bmatrix}$
= $\begin {bmatrix}
           2 \\
           12 \\
           \vdots \\
           2
         \end{bmatrix}$
When you have done inverse we got:
$2\begin {bmatrix}
           2/5 \\
           -3/10 \\
           \vdots \\
           6/5
         \end{bmatrix}$
+ $12\begin {bmatrix}
           1/5 \\
           1/10 \\
           \vdots \\
           2/5
         \end{bmatrix}$
+ $2\begin {bmatrix}
           -3/5 \\
           1/5 \\
           \vdots \\
           1/5
         \end{bmatrix}$
= $\begin {bmatrix}
           x \\
           y \\
           \vdots \\
           z
         \end{bmatrix}$
Initially on right side we had $[2 \ 12 \ 2 ]^T$. Using inverse, we want to place $[x \ y \ z ]^T$ on right side.
An another way is to think $[2 \ 12 \ 2 ]^T$ as a point represented using three vectors $[1 \ 3 \ 0 ]^T$, $[2 \ 8 \ 4 ]^T$, $[1 \ 1 \ 1 ]^T$. $x,y,z$ was the scaling factors. Now we transformed same point into $[x \ y \ z ]^T$ using vectors associated with column vectors of $A^{-1}$.
