0
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ If I got you right, it means that the sum of every row in $A^{-1}$ is actually $d = 1 - c$ ? $\endgroup$ – Dan Revah Nov 29 '15 at 12:36
  • 1
    $\begingroup$ or it is actually $d = 1/c$ ? $\endgroup$ – Dan Revah Nov 29 '15 at 12:46
  • $\begingroup$ Maybe you can factor out c on the right hand side and then multiply by 1/c. :-) $\endgroup$ – Denis Düsseldorf Nov 29 '15 at 13:14
  • $\begingroup$ So what does $d$ equals to? I'm not entirly sure $\endgroup$ – Dan Revah Nov 29 '15 at 13:17
  • 1
    $\begingroup$ 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 :-) $\endgroup$ – Denis Düsseldorf Nov 29 '15 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.