A question about matrices such that the elements in each row add up to $1$. 
Let $A$ be an invertible $10\times 10$ matrix with real entries such that the sum of each row is $1$. Then is the sum of the entries of each row of the inverse of $A$ also $1$?

I created some examples, and found the proposition to be true. I also proved that if two matrices with the property that the sum of the elements in each row is $1$ are multiplied, then the product also has the same property. Clearly, $I$ has this property. I think I have a proof running along the following lines: $$A^{-1}A=I$$ where $A$ and $I$ satisfy the aforementioned property. Also, if $A^{-1}$ did not satisfy this property, then neither would the product of $A$ and $A^{-1}$, which is a contradiction.
Is the proposition true, and if so, is my proof correct?

Any square matrix $\;n\times n\;$ has rows sum equal to  $\;1\;$ iff $\;u:=(1,1,\ldots,1)^t\;$ is an eigenvector with eigenvalue $\;1\;$, but then
$$Au=u\implies A^{-1}\left(Au\right)=A^{-1}u\iff u= A^{-1}u$$
so the answer is yes.
