Exercise 2.4.29

Answers

Matrices

E_{21} = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]

and

E_{31} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 4 & 0 & 1 \end{matrix}]

produce zeros in the $2, 1$ and $3, 1$ entries. Multiply $E$ ’s to get

E = E_{31} E_{21} = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ - 4 & 0 & 1 \end{matrix}] .

Then

EA = [\begin{matrix} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 3 \end{matrix}]

is the result of both $E$ ’s since

(E_{31} E_{21}) A = E_{31} (E_{21} A) .

mathchakra

2021-12-06 18:07