$$ A= \pmatrix{ 1 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1} $$
How would we compute the inverse of A using the Sherman-Morrison Formula?
I believe the answer will be below, but I like to know the steps, please. $$ A^{-1}=\pmatrix{ 1 & -2 & 4 & -8 \\ 0 & 1 & -2 & 4 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 1 } $$
Sherman-Morrison formula is about rank-one updates. Your $A$ is not a rank-one update of a matrix whose determinant and inverse are easy to compute (unless you are familiar with circulant matrices). It is a bad idea to apply Sherman-Morrison formula here.
A simpler idea is to exploit the fact that $A=I+2J$ where $J$ is a nilpotent matrix. Since $J^4=0$. We have $I-(2J)^4=I$. So, from the factorisation $1-x^4=(1+x)(1-x+x^2-x^3)$, we obtain $$ (I+2J)\left[I-(2J)+(2J)^2-(2J)^3\right]=I-(2J)^4=I. $$ Therefore $A^{-1}=(I+2J)^{-1}=I-(2J)+(2J)^2-(2J)^3$.