Sherman-Morrison Formula to determine the inverse of a $3 \times 3$ matrix Given $$A = \pmatrix{2 & 0 & -1 & \\ -1 & 1 & 1 \\ -1 & 0 & 1}$$ and $$A^{-1} = \pmatrix{1 & 0 & 1 & \\ 0 & 1 & -1 \\ 1 & 0 & 2}$$ I want to use the Sherman-Morrison Formula to determine $A^{-1}$ if we update $A$ so that the $a_{32}$ element is $2$. In other words, now $$A = \pmatrix{2 & 0 & -1 & \\ -1 & 1 & 1 \\ -1 & 2 & 1}$$
I've seen how to do this for $2\times2$ matrices, but I cannot seem to figure out how to do this for $3 \times 3$ matrices.
Note: I believe the answer is the following: $$\pmatrix{1 & 2 & -1 & \\ 0 & -1 & 1 \\ 1 & 4 & -2}$$

you want to find the inverse of $A + 2e_3e_2^T?$ we will use the facts that 
$$(I + ab^T)^{-1} = I - \frac{ab^T}{1 + a^Tb}, (A+ab^T)^{-1} = A^{-1} - \frac{A^{-1}ab^TA^{-1}}{1+b^TA^{-1}a.}$$
we will take $$a = 2e_3, b =e_2, A^{-1}a = 2(1,-1,2)^T, b^TA^{-1} = (0,1,-1), b^TA^{-1}a=-2$$ putting all these together gets you 
$$(A + ab^T)^{-1} = A^{-1} +2 \pmatrix{1\\-1\\2}\pmatrix{0&1&-1} 
=\pmatrix{1&0&1\\0&1&-1\\1&0&2} +2\pmatrix{0&1&-1\\0&-1&1\\0&2&-2} = 
\pmatrix{1&2&-1\\0&-1&1\\1&4&-2} $$
