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 May 21st, 2009, 05:15 AM #1 Newbie   Joined: May 2009 Posts: 15 Thanks: 0 Linear Algebra - Singularity problem Hi all, I'm having a little trouble with a simple question: I have to proof that if A is singular, and B is not, then AB is singular
 May 21st, 2009, 07:20 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Linear Algebra - Singularity problem Hint: A matrix $X$ is singular iff $\det(X)=0.$
 May 21st, 2009, 10:55 AM #3 Newbie   Joined: May 2009 Posts: 15 Thanks: 0 Re: Linear Algebra - Singularity problem Actually I'm trying to prove that det(AB) = det(A)*det(B), so I can't use this equation and, naturally, have to start by the case where A or B or both are singular... still looking for an answer, or a hint
 May 21st, 2009, 11:03 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Linear Algebra - Singularity problem If $AB$ is nonsingular, then there exists $X$ such that $ABX=I.$ What can you then say about $BX?$
 May 21st, 2009, 11:16 AM #5 Newbie   Joined: May 2009 Posts: 15 Thanks: 0 Re: Linear Algebra - Singularity problem hmmm then (BX) would be equal to A^(-1), but wait a minute, A was not supposed to have an inverse, since its singular! mattpi, you're a genious!!!! sorry for bothering you with this trivial question. =D see you soon, probably I'll be back with more trivial questions!

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