Finding the inverse of a symmetric $28\times28$ matrix

inverselinear algebramatrices

Let $$A$$ be a $$28 \times 28$$ matrix whose $$(i,j)$$-th entry is $$a_{ij} = \begin{cases} 28 &\text{if } i = j\\ -1 &\text{if } i \neq j \end{cases}$$ What is the inverse of this matrix?

I tried doing it by the adjoint method, but my resolution is wrong. First I tried finding the determinant of $$A$$, which I believe to be $$\det(A) =28^{28}+27-14\cdot28^2-14$$. Then, I tried finding the cofactor of $$a_{ij}$$ where $$i=j$$, which I believe is $$\operatorname{cof}(28) =28^{27}-27-28^3$$ and also the cofactor of $$a_{ij}$$ where $$i\neq j$$ which I calculated to be $$\operatorname{cof}(-1)=28^{26}+27-1+27\cdot28$$. Since the matrix is symmetric, I thought that the cofactor matrix is equal to the adjoint matrix, so the elements of the inverse matrix would be:
$$a_{ii}^{-1}=\frac{\operatorname{cof}(28)}{\det(A)}$$ and
$$a_{ij}^{-1}=\frac{\operatorname{cof}(-1)}{\det(A)}$$
where $$i\neq j$$. But this doesn't give the correct value. I think it might be a simple mistake, but I really can't find it so any help would be appreciated.

Note that $$A=29I - e e^T$$, where $$e$$ is a vector of ones. It is straightforward to check that the eigenvalues are $$1,29$$ hence it is invertible.
To compute $$A^{-1}$$, solve $$Ax=b$$.
This is $$29x-(e^Tx)e = b$$, hence $$29 e^T x -(e^Tx)e^T e = e^T b$$, or $$e^T x = e^T b$$. The first equation can be written as $$29 x = (e^T b)e+b=(I+e e^T)b$$, and so $$x = {1 \over 29} (I+e e^T)b$$, from which we get $$A^{-1} = {1 \over 29} (I+e e^T)$$.