$$X= \begin{pmatrix} 2-n & 1 & 1 & 1 & \ldots & 1 & 1 \\ 1 & 2-n & 1 & 1 & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & 1 & 1 & \ldots & 2-n & 1 \\ 1 & 1 & 1 & 1 & \ldots & 1 & 2-n\end{pmatrix}_{n\times n} $$
Which means that the matrix with the size of $n\times n$ have $n-2$ along the diagonal and $1$ everywhere else. Help please, I am stuck with this problem.
Hint: Use the Sherman–Morrison formula on $$X=\text{diag}(1-n)_{n\times n}+\begin{pmatrix} 1\\ \vdots \\1\end{pmatrix}_{n\times 1}\begin{pmatrix}1 & \ldots &1 \end{pmatrix}_{1\times n}.$$
HINT: Let $B$ be the $n\times n$ matrix of ones. Then $XB=B$, so
$$X(B-I)=XB-X=B-X=(n-1)I\;,$$
where $I$ is the $n\times n$ identity matrix.
By the way, manually calculating the inverses for $n=2$ and $n=3$ was enough to suggest what the answer ought to be, and discovering the argument suggested above was then quite easy.
If you see in terms of systems rather than matrices, you must solve
$X(x_1,x_2,\ldots,x_n)=(y_1,y_2,\ldots,y_n)$ where the $y_i$ are given
and the $x_i$ are the unknowns.
Adding everything up, you obtain
$$ x_1+x_2+\ldots +x_n=y_1+y_2+\ldots y_n (1) $$
Then, comparing (1) with the $i$-th equation gives you the value of $x_i$.
