Exercise 3.1.9

$\frac{\mathit{dA}(t)}{\mathit{dt}}=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2t\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ $A^{-1}(t)=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -t^2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$, by taking derivative directly on $A^{-1}$, we have

$\frac{d{A}^{-1}(t)}{\mathit{dt}}=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -2t\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

Using formula (12), we have

$\frac{d{A}^{-1}(t)}{\mathit{dt}}=-A^{-1}\frac{\mathit{dA}(t)}{\mathit{dt}}{A}^{-1}=-\left[\begin{array}{cc}\hfill 1\hfill & \hfill -t^2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2t\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill -t^2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -2t\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$