Exercise 1.4.6

To make zero in the second pivot position, $c=2$, then $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 1\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]$, we exchange row 2 and row 3 to get an upper triangular matrix $U$.

However, when $c=1$, then $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 1\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$. It’s clear that no row exchange can generate an upper triangular matrix with non-zero pivot elements.