# Why is that if every row of a matrix sums to 1, then the rows of the inverse matrix sums to 1 too?

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?

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.)
• To extend the assertion we can say that property of row sum equal 1 holds also for any matrix $A^k$ ($A$ invertible) when $k$ is an integer $(...,-2,-1,0,1,2, ..)$ – Widawensen Jan 25 '18 at 10:33