Exercise 3.3.5

Answers

The system

[\begin{matrix} 1 & 2 & - 2 & b_{1} \\ 2 & 5 & - 4 & b_{2} \\ 4 & 9 & - 8 & b_{3} \end{matrix}] \to [\begin{matrix} 1 & 2 & - 2 & b_{1} \\ 0 & 1 & 0 & b_{2} - 2 b_{1} \\ 0 & 0 & 0 & b_{3} - 2 b_{1} - b_{2} \end{matrix}]

is solvable if $b_{3} - 2 b_{1} - b_{2} = 0$ . Back-substitution gives the particular solution to $Ax = b$ and the special solution to $A x = 0$ :

x = [\begin{matrix} 5 b_{1} - 2 b_{2} \\ b_{2} - 2 b_{1} \\ 0 \end{matrix}] + x_{3} [\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}] .

mathchakra

2022-01-23 14:30