Exercise 2.5.2

Answers

For the first, a simple row exchange has $P^{2} = I$ so $P^{- 1} = P$ . For the second,

P^{- 1} = [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}] .

Always $P^{- 1} =$ "transpose" of $P$ , coming in Section 2.7.

mathchakra

2021-12-19 16:55