Exercise 2.5.26

Answers

We have

E_{21} A = [\begin{matrix} 1 & 0 \\ - 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 6 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 2 \end{matrix}]

and

E_{12} E_{21} A = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 2 & 1 \end{matrix}] A = [\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix}] .

Multiply by $D = [\begin{matrix} 1 & 0 \\ 0 & 1 ∕ 2 \end{matrix}]$ to reach

D E_{12} E_{21} A = I .

Then,

A^{- 1} = D E_{12} E_{21} = \frac{1}{2} [\begin{matrix} 6 & - 2 \\ - 2 & 1 \end{matrix}] .

mathchakra

2021-12-19 17:34