Let $A$ be a $n\times n$ square matrix such that $$ a_{ij}= \begin{cases} 2 &\text{if}\, i=j \\ -1 &\text{if}\, i\neq j, \end{cases} $$ where $a_{ij}$ represents the entry in the $i$th row and $j$th column of $A$. I would like to get a generalized form of the inverse of this matrix given $n>1$.
I have noticed that the inverse exists for all $n$ except when $n=3$. I have computed the inverse until $n=8$ and observed that the inverse matrix $A^{-1}$ is given by \begin{align} a^{-1}_{ij}= \begin{cases} \frac{n-4}{3(n-3)} &\text{if}\, i=j \\ \frac{-1}{3(n-3)} &\text{if}\, i\ne j. \end{cases} \end{align} For instance, for $n=2$, we have $$ A^{-1} = \left( \begin{matrix} 2/3 & 1/3\\ 1/3 & 2/3 \end{matrix} \right) $$ and for $n=4$, we have $$ A^{-1} = \left( \begin{matrix} 0 & -1/3 & -1/3 & -1/3\\ -1/3 & 0 & -1/3 & -1/3\\ -1/3 & -1/3 & 0 & -1/3\\ -1/3 & -1/3 & -1/3 &0 \end{matrix} \right) . $$ As mentioned above, I have observed the inverse until $n=8$ and this seems to be the trend for all $n$, except $n=3.$ However, I couldn't find a formal way to show this mathematically. Can someone offer me guidance on how to show that the inverse exists if $n\ne 3$ and that $A^{-1}$ is indeed given by $$ a^{-1}_{ij}= \begin{cases} \frac{n-4}{3(n-3)} &\text{if}\, i=j \\ \frac{-1}{3(n-3)} &\text{if}\, i\ne j \end{cases} $$ if this is true?
