Exercise 2.6.5

Answers

We have

EA = [\begin{matrix} 1 \\ 0 & 1 \\ - 3 & 0 & 1 \end{matrix}] [\begin{matrix} 2 & 1 & 0 \\ 0 & 4 & 2 \\ 6 & 3 & 5 \end{matrix}] = [\begin{matrix} 2 & 1 & 0 \\ 0 & 4 & 2 \\ 0 & 0 & 5 \end{matrix}] = U .

With $E^{- 1}$ as $L$ ,

A = LU = [\begin{matrix} 1 \\ 0 & 1 \\ 3 & 0 & 1 \end{matrix}] [\begin{matrix} 2 & 1 & 0 \\ 0 & 4 & 2 \\ 0 & 0 & 5 \end{matrix}] = [\begin{matrix} 2 & 1 & 0 \\ 0 & 4 & 2 \\ 6 & 3 & 5 \end{matrix}] .

mathchakra

2021-12-20 20:52