Exercise 2.3.4

Elimination on column 4: $b=\left[\begin{array}{c}1\hfill \\ 0\hfill \\ 0\hfill \end{array}\right]\stackrel{E_{21}}{\text{→}}\left[\begin{array}{c}\hfill 1\\ \hfill -4\\ \hfill 0\end{array}\right]\stackrel{E_{31}}{\text{→}}\left[\begin{array}{c}\hfill 1\\ \hfill -4\\ \hfill 2\end{array}\right]\stackrel{E_{32}}{\text{→}}\left[\begin{array}{c}\hfill 1\\ \hfill -4\\ \hfill 10\end{array}\right].$ The original $Ax=b$ has become $Ux=c=(1,-4,10)$. Then back substitution gives $z=-5,y=\frac{1}{2},x=\frac{1}{2}$. This solves $Ax=(1,0,0)$.