Hi,
This is probably a simple problem, but I just fell in love with that
Prove that for any $\displaystyle n\in\mathbb{N}$, the square matrix consisting in only 1's in the diagonal and n everywhere else is always invertible.
Algebra is great!
Here's a similar problem, perhaps a little harder.
If $\displaystyle n$ is an even positive integer, then the $\displaystyle n\times n$ matrix $\displaystyle A$ given by $\displaystyle A_{ij}=1 \ (i \neq j), \: \: A_{ii}=0$, is invertible.
The inverse is :
$\displaystyle -\frac{1}{n-1} I + \frac{n}{(n-1)(nm-n+1) } A $
where $\displaystyle [A]_{ij} = 1 ~~ 1 \leq i,j \leq m $ and $\displaystyle m $ is the size of the matrix .
To obtain the determinant , not only the residue ...
Consider the system of the linear equations :
$\displaystyle [ nA - (n-1)I ]x = y ~~ , ~ x = (x_1 ,x_2 ,...,x_m )^T ~,~ y = (0,0,0,...,0,1)^T $ and note that $\displaystyle nA - (n-1)I $ is your matrix , let it be $\displaystyle B_m$ ( If i read the problem correctly )
Let $\displaystyle S $ be the quantity $\displaystyle \sum x_i $ then we have
$\displaystyle nS - (n-1) x_i = 0 ~,~ 1 \leq i \leq m-1 ~~,~~ nS - (n-1)x_m = 1 $
$\displaystyle x_i = \frac{nS}{n-1} ~,~ 1 \leq i \leq m-1 ~~,~~ x_m = \frac{nS - 1}{n-1}$
$\displaystyle \frac{n(m-1)S}{n-1} + \frac{nS - 1}{n-1} = S $
$\displaystyle S = \frac{1}{nm - n + 1 } $
So $\displaystyle x_m = - \frac{n(m-1)-n+1}{(n-1)(nm-n+1) } = \frac{det(B_{m-1})}{det(B_m)} $
I believe it is true that $\displaystyle det(B_m) = (1-n)^{m-1}(nm-n+1)$