TITLE: What does calculating the inverse of a matrix mean?
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: There are two possible views on what you are doing when computing the inverse of a matrix.
One point of view is the one that edo already mentioned in his answer: The matrix can be imagined as transforming a point in 3D space. For example, you could fill the matrix with the values that describe a rotation about one axis in 3D, like the rotation about $45°$ around the x-axis. Computing the inverse of such a matrix then just describes the inverse transformation. In this case, this would be the rotation about $-45°$ around the x-axis. The inverse lets you "undo" the operation that was done with the original matrix.
Now, this somehow obfuscates the point of the underlying equations. So the other point of view is to describe what you're actually doing with this matrix when it describes a system of equations:
When you have a system of equations expressed as $Ax=b$, you're looking for the vector $x$ that "makes this expression true". You could now use different approaches to find this vector. For example, some sort of iterative approximation, elimination, or even trial-and-error - it does not matter. 
The point is: 

For one vector $b = (4,3,9)$, you could find the vector $x = (3.2, 4.5, 2.3)$ 
For one vector $b = (1,8,4)$, you could find the vector $x = (2.1, 7.5, 1.5)$ 
For one vector $b = (6,1,5)$, you could find the vector $x = (1.6, 6.2, 5.3)$ 
For one vector $b = (4,6,1)$, you could find the vector $x = (5.2, 4.6, 5.7)$ 
...

(The numbers are made up, only for illustration)
What you are doing when computing the inverse of the matrix is:
You are solving all these problems simultaneously. You are deriving a rule that tells you how to obtain the matching $x$ for any $b$, without any real computation, just by "looking up" the value of $A^{-1}b$. So instead of solving one problem, you are solving infinitely many problems. That's a nice thing, and pretty useful.