Exercise 1.4.4

• 1.

  $E=E_2{E}_1=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -a\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill \mathit{ac}-b\hfill & \hfill -c\hfill & \hfill 1\hfill \end{array}\right]$, and we have $\mathit{EA}=I$.

• 1.

  $E_1^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill a\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill b\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ and $E_2^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill c\hfill & \hfill 1\hfill \end{array}\right]$

So $L=E_1^{-1}{E}_2^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill a\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill 1\hfill \end{array}\right]=A$

We notice that the multipliers $a$, $b$, $c$ are mixed up in $E=L^{-1}$ but they are perfect in $L$.