Inverses of Matrices Homework: Find (A^2B^2)^-1

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To find the inverse of the product of two squared matrices, (A^2B^2)^-1, it is essential to recognize that the inverse of a product of matrices is not simply the product of their inverses in the same order. The correct formula is (A^2B^2)^-1 = (B^2)^-1(A^2)^-1, which can be expressed as (B^-1)^2(A^-1)^2. The discussion highlights the importance of reversing the order when dealing with matrix inverses due to the non-commutative nature of matrix multiplication. The user initially confused the notation but clarified their understanding after receiving guidance. This exchange emphasizes the need for careful manipulation of matrix operations to avoid common mistakes.
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Homework Statement


hi, i have 2 nxn matrices A and B i found that (AB)^-1=A^-1xB^-1 but how do i do
(A^2B^2)^-1=?

please i really need some help here I am kinda lost i have
AA^-1=I
AI=A
(AB)(A^-1xB^-1)=I

anyone have any ideas?
 
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fredrick08 said:

Homework Statement


hi, i have 2 nxn matrices A and B i found that (AB)^-1=A^-1xB^-1 but how do i do
(A^2B^2)^-1=?
Well, it's really unfortunate that you "found" that because it is not true!

please i really need some help here I am kinda lost i have
AA^-1=I
AI=A
(AB)(A^-1xB^-1)=I
No! you can't cancel BA-1! That's why (AB)-1 is NOT A-1B-1.

In order to be able to "cancel" A with A-1 or B with B-1, because multiplication of matrices is not commutative, you have to reverse the order of multiplication. (AB)(B-1A-1)= A(BB-1)A-1= A(I)A-1= AA-1= I. (AB)-1= B-1A-1.

One thing you could do is think of (A2B2)-1 as (UV)-1= V-1U-1 where U= A2 and V= B2- that is (A2B2)-1= (B2)-1(A-1)2= (A-1)2(B-1)2.

Similarly, (AABB)(B-1B-1A-1A-1)= AAB(BB-1)B-1A-1A-1= AA(B-1B-1)A-1A-1= A(AA-1)A-1= AA-1= I


anyone have any ideas?
 
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oh yes... sorry that was a typo it should of been B^-1A^-1 because of socks and shoes formula... sorry, ok yes i understand what u mean, yer i kept thinking of it as A^2 and B^2 instead of AA and BB just got myself confused... thankyou heaps!
 
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