How would I find the inverse of this matrix?
I know we can pull the a out so that:
\begin{equation} aI - \frac{a}{b}J = a\left[I - \frac{1}{b}J\right] \end{equation}
and then we can focus on the inverse of the matrix in the brackets, but this seems to be outside the scope of how I know how to invert a matrix.
Observe that $J^2 = nJ$, when $J$ is of order $n \times n$.
Then $(I - rJ)(I - sJ) = I - (r + s)J + rs J^2 = I + (nrs - r - s)J$.
Now, for any given $r$, we can find $s$ such that the above becomes $I$ by solving $nrs - r - s = 0$. That is, $s = \frac{r}{nr - 1}$. This does not work if $r = \frac 1 n$. But when $r = \frac 1 n$, we have $I - rJ = I - \frac 1 n J$, which is non-invertible — there are multiple ways to see this — observe that it is idempotent (but is not the identity matrix, and therefore must be singular), or observe that $0$ is an eigenvalue, since $n$ is an eigenvalue of $J$.
Thus, for any $r \ne \frac 1 n$, $(I - rJ)^{-1} = I - \frac{r}{nr - 1}J$.
Therefore, $$\left(a - \dfrac a b J\right)^{-1} = \dfrac 1 a I - \dfrac{1}{a(n - b)} J$$ for any $a \ne 0, b \ne n$.