# Sum of row elements in an invertible matrix

Let $A_{nxn}$ be an invertible matrix such as that the sum of the elements of each row is equal to $c$.

Prove that the sum of the elements of every row in $A^{-1}$ is equal to $d$ and write $d$ using $c$.

Clue: you can use the column vector $v=(1,1,1,...,1)^t$.

This is the question I got and I have no idea how to solve it.

Start with $$A\begin{bmatrix}1\\1\\...\\1\end{bmatrix}=\begin{bmatrix}c\\c\\...\\c\end{bmatrix}$$

This is true by assumption. Now A is invertible, so multiply with the inverse from the left and see what happens:

$$\underbrace{A^{-1}A}_{I}\begin{bmatrix}1\\1\\...\\1\end{bmatrix}=A^{-1}\begin{bmatrix}c\\c\\...\\c\end{bmatrix}$$

So $$\begin{bmatrix}1\\1\\...\\1\end{bmatrix}=A^{-1}\begin{bmatrix}c\\c\\...\\c\end{bmatrix}$$

What does that tell you?

• If I got you right, it means that the sum of every row in $A^{-1}$ is actually $d = 1 - c$ ? – Dan Revah Nov 29 '15 at 12:36
• or it is actually $d = 1/c$ ? – Dan Revah Nov 29 '15 at 12:46
• Maybe you can factor out c on the right hand side and then multiply by 1/c. :-) – Denis Düsseldorf Nov 29 '15 at 13:14
• So what does $d$ equals to? I'm not entirly sure – Dan Revah Nov 29 '15 at 13:17
• If you factor c out you get the RHS $cA^{-1}\begin{bmatrix}1\\...\\1\end{bmatrix}$ and then you divide by c (assuming it is not 0) and then you get 1/c in each component on the LHS. As the multiplication with the vector $(1,1,1,...,1)^T$ gives you the sum of each row, the answer is d=1/c :-) – Denis Düsseldorf Nov 29 '15 at 13:24