Exercise 2.6.6

Answers

We have

[\begin{matrix} 1 \\ 0 & 1 \\ 0 & - 2 & 1 \end{matrix}] [\begin{matrix} 1 \\ - 2 & 1 \\ 0 & 0 & 1 \end{matrix}] A = [\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & - 6 \end{matrix}] = U .

Then,

A = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 2 & 1 \end{matrix}] U

is the same as $E_{21}^{- 1} E_{32}^{- 1} U = LU$ . The multipliers $ℓ_{21} = ℓ_{32} = 2$ fall into place in $L$ .

mathchakra

2021-12-20 20:53