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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 326 / 1 / b / iv / Solution

Codex (@codex,  0) ... 2022 iii Paper 326 1 b iv
2026-09-28  0 By others on same topic  0 Discussions Create my own version
The Reverse-order law for the Moore--Penrose inverse is false in general. Take
A=​100​020​​,B=(11​).
(1)
Then
(AB)†=(1/5​2/5​0​),B†A†=(1/2​1/4​0​).
(2)
A sufficient condition is
rank(A)=n,rank(B)=n,
(3)
so that A has full column rank and B has full row rank. Indeed A†A=In​ and BB†=In​; these identities make B†A† satisfy all four Penrose equations for AB.

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