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Exercise 2.5.9

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Yes, $B$ is invertible ( $A$ was just multiplied by a permutation matrix $P$ ). If you exchange rows 1 and 2 of $A$ to reach $B$ , you exchange columns 1 and 2 of $A^{- 1}$ to reach $B^{- 1}$ . In matrix notation, $B = PA$ has $B^{- 1} = A^{- 1} P^{- 1} = A^{- 1} P$ for this $P$ .

mathchakra

2021-12-19 17:01

Exercise 2.5.9

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