Exercise 2.6.22

Answers

Eliminating upwards,

[\begin{matrix} 5 & 3 & 1 \\ 3 & 3 & 1 \\ 1 & 1 & 1 \end{matrix}] \to [\begin{matrix} 4 & 2 & 0 \\ 2 & 2 & 0 \\ 1 & 1 & 1 \end{matrix}] \to [\begin{matrix} 2 & 0 & 0 \\ 2 & 2 & 0 \\ 1 & 1 & 1 \end{matrix}] = L .

We reach a lower triangular $L$ , and the multipliers are in an upper triangular $U . A = U L$ with

U = [\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] .

mathchakra

2021-12-20 21:14