
> 
> Given sum of elements of each row of a positive definite square matrix $M$ of order $n$ all of whose entries are non-negative, when is it possible to find the sum of all elements of the matrix $M^{-1}$?
>  Are there any papers or research done in this direction?
> 
> 
> 


To fully explain my question ,


Let $M$ be any matrix of order $n$ which has all its entries non-negative.
Also all eigenvalues of $M$ are positive.
If the sum of each row of $M$ is known, when is it possible to find sum of all the elements of $M^{-1}$?


I will be grateful if someone can say if there exist any conditions on $M$ to answer this question?


Any help will be useful.



When is it possible to find the sum of all elements of inverse of a matrix?
If the condition on $M$ (in order for the sum of all entries of $M^{-1}$ to be uniquely determined) is to be expressed only in terms of the row sums of $M$ (with the additional restrictions that all entries of $M$ be non-negative and all eigenvalues of $M$ be positive), then the answer is the following: 



> 
> The sum of all entries of $M^{-1}$ is uniquely determined if and only if all the row sums are the same. 
> 
> 
> 


Indeed, the "if" part of this claim is proved in the answer by Robert Israel. 


Consider now the "only if" part. We have to show that, if any two of the prescribed row sums of $M$ are different, then the sum of all entries of $M^{-1}$ is not uniquely determined, even with the additional restrictions that all entries of $M$ be non-negative and all eigenvalues of $M$ be positive. By considering block-diagonal matrices and rescaling and rearranging the rows, without loss of generality we may assume that $n=2$ and 
$$M=\left(
\begin{array}{cc}
 a & 1-a \\
 b & s-b \\
\end{array}
\right),$$
where $0\le b<as\le s$ (the latter condition holds iff all entries of $M$ are non-negative and all eigenvalues of $M$ are positive). Here $s\in(0,\infty)\setminus\{1\}$ is given, but (say) $a$ may freely vary in the interval $(b/s,1]$. 
The sum of all entries of $M^{-1}$ here is 
$$S:=\frac{2 a-2 b+s-1}{a s-b},
$$
and the partial derivative of $S$ in $a$ is 
$$\frac{(s-1) (2 b-s)}{(a s-b)^2}\ne0
$$
if, for instance, $b=0$ and $a=1/2$ (recall that $s\ne1$). So, the sum of all entries of $M^{-1}$ is not uniquely determined. 


