TITLE: Why is that if every row of a matrix sums to 1, then the rows of the inverse matrix sums to 1 too?
QUESTION [47 upvotes]: Why is that if every row of a matrix sums to $1$ then the rows of its inverse matrix sum to $1$ too?

For example, consider 
$$A=\begin{pmatrix}
1/3 & 2/3 \\
3/4 & 1/4
\end{pmatrix}$$
then its inverse is
$$A^{-1}=\begin{pmatrix}
-3/5 & 8/5 \\
9/5 & -4/5
\end{pmatrix},$$
which satisfies the condition. Is it true for every such matrix?

REPLY [106 votes]: Let $v = (1, 1, \ldots , 1)'$ be a column vector of all $1$s. Then the rows of $A$ adding to $1$ is equivalent to saying $Av = v$. So when $A$ is invertible, we will have $$A^{-1}v = A^{-1}Av = v$$ Thus $A^{-1}$ has rows summing to $1$ as well. (Note that $A$ will not always be invertible.)