Show that sum of elements of rows / columns of a matrix is equal to reciprocal of sum of elements of rows/colums of its inverse matrix Suppose $A=(a_{ij})_{n\times n}$ be a non singular matrix. Suppose sum of elements of each row is $k\neq 0$, then the sum of elements of rows of $A^{-1}$ is $\frac{1}{k}$. 
Let    $\,A^{-1}=(b_{ij})_{n\times n}$.
Then 
\begin{align*}
&A\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=k\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &\frac{1}{k}A^{-1}A\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=\frac{1}{k}A^{-1}k\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &\frac{1}{k}I_n\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=A^{-1}\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &\frac{1}{k}I_n\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=(b_{ij})\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies& (b_{ij})\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=\frac{1}{k}I_n\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &(\displaystyle\sum_{r=1}^n b_{1r}, \displaystyle\sum_{r=1}^n b_{2r},\cdots, \displaystyle\sum_{r=1}^n b_{nr})=(\frac{1}{k}, \frac{1}{k}\cdots, \frac{1}{k})\\
\implies & \displaystyle\sum_{r=1}^n b_{ir}=\frac{1}{k}, \forall i=1,2,\cdots,n 
 \end{align*}

apply ${1\over k}A^{-1}$ to
$$
A 
\begin{bmatrix}
1 \\
1 \\
\vdots \\
1 \\
\end{bmatrix}
=
k 
\begin{bmatrix}
1 \\
1 \\
\vdots \\
1 \\
\end{bmatrix}
$$
