Example: Sherman-Morrison Formula

Consider the following equation \(Ax=b\),\[\begin{bmatrix}
 1 &  2 &  3 \\
 1 &  4 &  9 \\
 4 &  5 &  7 \\
\end{bmatrix}x = \begin{bmatrix}
 2 \\
 4 \\
 6 \\
\end{bmatrix}.\] The solution of this equation is,\[x = \begin{bmatrix}
 0.75 \\
 0.25 \\
 0.25 \\
\end{bmatrix}.\] Now let us suppose we wanted to solve a similar problem ( \(A'x' = b\) )with a slightly perturbed matrix,\[A' =  \begin{bmatrix}
 1 &  2 &  3 \\
 1 &  6 &  9 \\
 4 &  5 &  7 \\
\end{bmatrix}.\] Notice that this matrix differs from \(A\) in the (2,2) element (6 instead of 4).

The Sherman-Morrison formula tells us to form \(u = [0~1~0]^T, v = [0~~2~~0]^T\), so that \[A' = A +  u v^T.\] Hence \[y = x = A^{-1}b = \begin{bmatrix}
 0.75 \\
 0.25 \\
 0.25 \\
\end{bmatrix},\text{ and } z = A^{-1} u = \begin{bmatrix}
 0.125 \\
 -0.625 \\
 0.375 \\
\end{bmatrix}.\] Thus,\[ x' = y - \left({v^T y \over 1 + v^T z}\right) z\]  This implies \[x' = \begin{bmatrix}
 0.75 \\
 0.25 \\
 0.25 \\
\end{bmatrix} \frac{0.25}{1-0.625} \begin{bmatrix}
 0.125 \\
 -0.625 \\
 0.375 \\
\end{bmatrix} =\begin{bmatrix}
   1 \\
  -1 \\
   1 \\
\end{bmatrix},\] which is indeed the correct solution.