Exercise 2.5.17

=$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \\ \hfill -1\hfill & \hfill \hfill & 1\end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill \hfill \\ \hfill -1\hfill & \hfill 1\hfill & \\ \hfill \hfill & \hfill \hfill & 1\end{array}\right]=\left[\begin{array}{cc}\hfill 1& \hfill \\ \hfill -1& \hfill 1& 0\\ \hfill 0& \hfill -1& 1\end{array}\right]$

=E.

-1$\mathit{to}$+1$\mathit{togetinverses}:$$E_{21}^{-1}{E}_{31}^{-1}{E}_{32}^{-1}=\left[\begin{array}{ccc}1\hfill & \hfill & \hfill \\ 1\hfill & 1\hfill & \hfill \\ 1\hfill & 1\hfill & 1\hfill \end{array}\right]$

=L=E^-1.

Notice that the $1$’s are unchanged by multiplying inverses in this order.