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.
Best Answer
Matrix multiplication corresponds to substituting new variables for the given ones in the system of linear equations. In more detail, for a system of $n$ equations in $n$ unknowns $X_1,\dots ,X_n $, suppose that $A$ represents the system of equations. Suppose now that you introduce new variables $Y_1,\dots ,Y_n$ and you express each $X_i$ as a linear combination of the new variables. If you write $B$ for the matrix of coefficients of the $X_i$ represented as combinations of the $Y_i$, then the matrix $AB$ corresponds to the matrix of coefficients of the original system of equations after substituting the new variables in. If you work this out for the case $n=2$ it's easy to see what is going on. This in fact is one way to motivate the definition of matrix multiplication (in general, not just for square matrices).
Now, what all this tells you is that if you have $A$ and you found that $B=A^{-1}$ is its inverse, then if you introduce new variables $Y_1,\dots , Y_n$ and express the $X_i$ in terms of those by reading the coefficient in the inverse matrix $B$, then substituting these variables into the original system will result in a very very simple system. Namely, the coefficient after substituting will be the coefficients in $AB=I$. This is the simplest system in the world. So, find the inverse of a matrix is equivalent to finding a change of coordinates, from the $X_i$'s to the $Y_i$'s, which make the system of equations particularly nice.
Again, this holds true for all systems, not just $n\times n$.