A = $(a_{ij})$ where $a_{ij}$ = 1. find c such that $(I-A)^{-1} = I-cA$

Question

let A be a matrix of order n such that A = $(a_{ij})$ where $a_{ij}$ = 1 for all i and j. find c such that $(I-A)^{-1} = I-cA$.

I know that there is a trick to solve this. I've tried to calculate the inverse of (I-A) but couldn't. what are some other options?

Answer 1

If A is an $nxn$ matrix, then $$(I-A)(I-cA)$$ $$=I - (c+1)A + cA^2$$ $$=I-(c+1)A+cnA$$ So set $$cn=c+1$$ $$c = \frac{1}{n-1}$$

Answer 2

You have to solve the following equation for $c$:

$$(I-A)(I-cA) = I$$

You'll need the size of A to get $c$.