Show that the sum of the elements on every line of matrix $A^{-1}$ is $s^{-1}$.

Question

We have $A$ $(n×n)$ matrix with complex entries. We know that $\det(A)≠0$ and the fact that the sum of the elements on every line is s. Show that the sum of the elements on every line of matrix $A^{-1}$ is $s^{-1}$.

Answer 1

Let's take $V = (1,1...,1)$

The sum of the elements on every row is $s$ is equivalent to $AV = sV$

Applying $A^{-1}$ in both sides we get :

$V = sA^{-1}V$ and since $s \neq 0$ (otherwise, $V$ would be in the kernel of $A$) you have $A^{-1}V = s^{-1}V$