Exercise 3.1.3

When $k=1$, we have $M^{-1}={(A-u{v}^T)}^{-1}=A^{-1}+A^{-1}u{(I-v^T{A}^{-1}u)}^{-1}{v}^T{A}^{-1}$

When $A=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$, we have $A^{-1}=\left[\begin{array}{cc}\hfill \frac{1}{3}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{2}\hfill \end{array}\right]$, so $(A-u{v}^T)=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$, then ${(A-u{v}^T)}^{-1}=\left[\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{2}\hfill \end{array}\right]$.

According to the formula, we have

These two results agree with each other.