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

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SUMMARY

The discussion centers on finding the inverse of the product of two squared matrices, specifically (A^2B^2)^-1. The correct formula for the inverse of a product of matrices is established as (AB)^-1 = B^-1A^-1, which is crucial for solving the problem. The user initially misapplied the formula, leading to confusion regarding the cancellation of matrices. The correct approach involves recognizing that (A^2B^2)^-1 can be expressed as (B^2)^-1(A^2)^-1, which simplifies to (B^-1)^2(A^-1)^2.

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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?
 
Last edited by a moderator:
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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