Numerical Linear Algebra with Applications

Example 11.11

Let $A=[\begin{array}{cccc}-1& -1& 0& 1\\ -1& 1& 1& 0\\ 1& 1& 1& 1\\ 2& 0& 1& 0\end{array}]$. This example illustrates the process just described that derives GEPP.

Start: $L=[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}],P=[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}]$ i = 1: Exchange rows 1 and 4. Eliminate entries at indices (2, 1) − (4, 1):

$\begin{array}{l}{E}_1=E_{41}\left(-\frac{1}{2}\right)E_{31}\left(\frac{1}{2}\right)E_{21}\left(-\frac{1}{2}\right)=[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0.5& 0& 0& 1\end{array}][\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -0.5& 0& 1& 0\\ 0& 0& 0& 1\end{array}][\begin{array}{cccc}1& 0& 0& 0\\ 0.5& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}]=[\begin{array}{cccc}1& 0& 0& 0\\ 0.5& 1& 0& 0\\ -0.5& 0& 1& 0\\ 0.5& 0& 0& 1\end{array}]\\ P_1=[\begin{array}{cccc}0& 0& 0& 1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 1& 0& 0& 0\end{array}]\\ E_1{P}_{1}A=[\begin{array}{cccc}2& 0& 1& 0\\ 0& 1& 1.5& 0\\ 0& 1& 0.5& 1\\ 0& -1& 0.5& 1\end{array}]\end{array}$

i = 2: A row exchange is not ...