Exercise 3.3.28

Answers

[\begin{matrix} 1 & 2 & 3 & 0 \\ 0 & 0 & 4 & 0 \end{matrix}] \to [\begin{matrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix}];

x_{n} = [\begin{matrix} - 2 \\ 1 \\ 0 \end{matrix}];

[\begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 4 & 8 \end{matrix}] \to [\begin{matrix} 1 & 2 & 0 & - 1 \\ 0 & 0 & 1 & 2 \end{matrix}]

Free $x_{2} = 0$ gives $x_{p} = (- 1, 0, 2)$ because the pivot columns contain $I$ .

mathchakra

2022-01-23 15:09