How can I prove that $\text{adj}(AB)=\text{adj}(B)\:\text{adj}(A)$, if $A$ and $B$ are any two $n\times n$ singular matrices Here, $\text{adj}(A)$ means the adjugate of the matrix $A$.
I was wondering if someone could help me?
How to prove $\text{adj}(AB)=\text{adj}(B)adj(A)$?
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matrices
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0What have you tried? Also, the title and body of your post ask different questions - which one is it? – 2017-02-06
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2Possible duplicate of [$\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A$](http://math.stackexchange.com/questions/799284/operatornameadjab-operatornameadj-b-operatornameadj-a) – 2017-02-06
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0@Callus my question is different. I am asking here for singular matrices. – 2017-02-06
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1The question linked to is also about singular matrices, and the top voted solution works for singular matrices. IN fact, in the question linked, there is this line "I know how to prove it for non singular matrices, but I have no idea what to do in this case." – 2017-02-06
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0No, it has not been solved yet for singular case. – 2017-02-06
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0Yes it has. I think what might be confusing is that the solution uses $X^{-1}$, but the inversion is over the field of rational functions in indeterminates of the entries of the two matrices. This inverse always exists. IN other words, the matrices are formally inverted, which is fine. – 2017-02-06
1 Answers
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$Aadj(A)=det(A)I$, $ABadj(B)adj(A)=A(BAdj(B))adj(A)=det(B)AAdj(A)=det(B)det(A)I=det(AB)I.$
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0I don't see how this proves the claim. – 2017-02-06
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0I proved what OP asked in the question, not in the title – 2017-02-06
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0Sorry, I'm sure this is me missing something obvious, but I don't see how what you've written proves $adj(AB)=adj(B)adj(A)$. – 2017-02-06
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0Maybe this shows $AB\text{adj}(B)\text{adj}(A) = AB\text{adj}(AB)$, but you have to clear the $AB$ somehow. On the other hand, OP accepted this answer, so probably it's me that is missing something. – 2017-02-06