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