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's look at your original three equations.  
$$x+2y+z=2$$
$$3x+8y+z=12$$
$$4y+z=2$$

Now let's multiply by $\frac{2}{5}$, $\frac{1}{5}$, and $\frac{-3}{5}$ respectively.  We get 
$$\frac{2x}{5}+\frac{4y}{5}+\frac{2z}{5}=\frac{4}{5}$$
$$\frac{3x}{5}+\frac{8y}{5}+\frac{z}{5}=\frac{12}{5}$$
$$\frac{-12y}{5}+\frac{-3z}{5}=\frac{-6}{5}$$
Now add the three equations together.  We get 
$$x + 0y + 0z = 2$$
or 
$$x=2$$

Now multiply the same three equations by $\frac{-3}{10}$, $\frac{1}{10}$, and $\frac{1}{5}$ respectively.  We get 
$$\frac{-3x}{10}+\frac{-6y}{10}+\frac{-3z}{10}=\frac{-6}{10}$$
$$\frac{3x}{10}+\frac{8y}{10}+\frac{z}{10}=\frac{12}{10}$$
$$\frac{4y}{5}+\frac{z}{5}=\frac{2}{5}$$
Summing 
$$0x + y + 0z = \frac{10}{10}$$
or 
$$y = 1$$

Now multiply the same three equations by $\frac{6}{5}$, $\frac{-2}{5}$, and $\frac{1}{5}$ respectively.  We get 
$$\frac{6x}{5}+\frac{12y}{5}+\frac{6z}{5}=\frac{12}{5}$$
$$\frac{-6x}{5}+\frac{-16y}{5}+\frac{-2z}{5}=\frac{-24}{5}$$
$$\frac{4y}{5}+\frac{z}{5}=\frac{2}{5}$$
Summing 
$$0x + 0y + z = \frac{-10}{5}$$
or 
$$z = -2$$

And if you look at the numbers by which we multiplied, they are from 
$$\begin{pmatrix}
2/5 & 1/5 & -3/5\\ 
-3/10 & 1/10 & 1/5\\ 
6/5 & -2/5 & 1/5
\end{pmatrix} = A^{-1}$$
We essentially did the matrix multiplication of $A \cdot A^{-1}$ to get $I$ manually when we could have just done 
$$\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} = \begin{pmatrix}
2/5 & 1/5 & -3/5\\ 
-3/10 & 1/10 & 1/5\\ 
6/5 & -2/5 & 1/5
\end{pmatrix}\begin{pmatrix}
2\\ 
12\\ 
2
\end{pmatrix}=\begin{pmatrix}
2\\
1\\
-2
\end{pmatrix}$$
and gotten the same answer.  $A^{-1}$ is essentially the numbers by which we multiply the equations so we can add them together and get the solutions.  Solving for the inverse is determining those numbers.  Of course, if you're working with the equations, it would be easier to substitute in than to come up with all nine numbers.  The convenient thing here is that we don't have to multiply $A\cdot A^{-1}$, as we already know the result.  We can just do the right side multiplication.  
