TITLE: Sum of elements of the inverse matrix (without deriving the inverse matrix) using elementary methods.
QUESTION [5 upvotes]: I have the matrix 
  $$\begin{pmatrix}
3&2&2&\\
2&3&2\\
2&2&3
\end{pmatrix}.$$ Find the sum of elements of the inverse matrix without computing the inverse. 

I have seen this post, but I need much more elementary method. I have checked that inverse exists, and $\Delta=7$. The answer is $\frac37$.
What I did, is by (not so)simple computation, I worked out that every matrix of the form 
$\begin{pmatrix}
a&b&b&\\
b&a&b\\
b&b&a
\end{pmatrix}$ always achieves an inverse of the form 
$$\frac{a-b}{a+2b}
\begin{pmatrix}
a+b&-b&-b&\\
-b&a+b&-b\\
-b&-b&a+b
\end{pmatrix}.$$
And by this technique, the expected result follows for my case $a=3, b=2$. 
But I don't think this is an elegant method. Because I need to prove this lemma, then have to claim the result. And also, I was told in question not to compute inverse, where I, here, am computing inverse of a general case.

REPLY [6 votes]: The post you link tells you that the sum you want is $(\vec{1})^t\cdot A^{-1}\cdot \vec{1}$ (where $\vec{1}$ is the vector with all entries equal to $1$). Your matrix has the property that all lines have the sum of the elements equal to $7$; this is equivalent to writing
$$A\cdot \vec1= 7\vec1,$$and from that you get that
$$A^{-1}\cdot \vec{1}=\frac{1}{7}\vec1.$$
Thus you have immediately
$$(\vec{1})^t\cdot A^{-1}\cdot \vec{1}=\frac17(\vec{1})^t\cdot \vec1=\frac{3}{7}.$$

REPLY [4 votes]: Notice that $u^T=(1,1,1)$ is an eigenvector with eigenvalue $7$.
So we have $Au=7u$ and hence $A^{-1}u=\frac{1}{7}u$.
But the sum of the elements of $\frac{1}{7}u$ is just the sum of the elements of $A^{-1}$. So the answer is $\frac{3}{7}$.