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Have I got these properties of matrices correct?

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Maybe retry C. Just multiply your result by AB. –  AQP Mar 4 '12 at 21:35
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five of your six answers (counting unchecked entries as answers) are correct. One is wrong. –  Arturo Magidin Mar 4 '12 at 21:37
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$(AB)^{-1}=B^{-1}A^{-1}$ –  Salech Alhasov Mar 4 '12 at 21:38

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Your answers are correct, except that C is false. The right side is the inverse of $BA$, because $$(A^{-1}B^{-1})(BA)=A^{-1}(B^{-1}B)A=A^{-1}IA=A^{-1}A=I.$$ But must the inverse of $AB$ be the inverse of $BA$? See if you can find an explicit counterexample.

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Actually, if the entries of A,B are in a ring, we may only have 1-sided inverses, I think. –  AQP Mar 4 '12 at 21:41
    
Sorry, what I meant is that if the entries of A,B are in some ring R, and AB=I , I think we cannot conclude that BA=I. –  AQP Mar 4 '12 at 21:42
    
I think you're right if $R$ is non-commutative, but for any commutative ring $R$, the collection of invertible matrices, i.e. $$\operatorname{GL}_n(R)=\{A\in M_n(R)\mid \det(A)\in R^\times\},$$ is a group, so there are always two-sided inverses. –  Zev Chonoles Mar 4 '12 at 21:43
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@Zev Chonoles: So I cant distribute the $^{-1}$ across A & B....because that is what I did for A..I thought that was a valid operation. Why is my answer for A correct? –  Jim_CS Mar 4 '12 at 21:45
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@John: Your answer for A is correct because $$(ABA^{-1})^3=(ABA^{-1})(ABA^{-1})(ABA^{-1})=$$ $$AB(A^{-1}A)B(A^{-1}A)BA^{-1}=ABIBIBA^{-1}=AB^3A^{-1}$$ You can't distribute a ${}^{-1}$ in general for matrices, unlike the case for numbers (this is ultimately because matrix multiplication is not always commutative). I'm not sure how distributing a ${}^{-1}$ would have come up in problem A though. –  Zev Chonoles Mar 4 '12 at 21:47

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