Trick for Inverse Hollow Matrix Calculation (Self-Answered) Let $A$ be the hollow matrix :
$$
A=\begin{pmatrix}
0&1&1&1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
\end{pmatrix}
$$
Find the inverse matrix $A^{-1}$ without using any elementary row/ column operations.

Notice that 
$$
A=
\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1
\end{pmatrix} - I_{4\times4}
$$
Let 
$$
B=
\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1
\end{pmatrix}
$$
then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding, 
$$
A^2+2A+I = 4(A+I) \implies A^2-2A=3I
$$
Therefore $A^{-1}=\frac{1}{3}(A-2I)$.
More generally, this method gives us an easy way to express the inverse of a hollow matrix $A$ (of the particular stated form) of any order $n$, which is the main reason for posting this question.
