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.

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.

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