# Can someone explain these 2 linear algebra proofs

## Homework Statement

The proofs:

show (A')^-1 = (A^-1)'

and

(AB)^-1 = B^-1A^-1

## The Attempt at a Solution

for the first one:

(A^-1*A) = I
(A^-1*A)' = I' = I
A'(A^-1)' = I

but im not sure how this proves that a transpose inverse = a inverse transpose...

the second i have the same problem. Not sure how this really proves what it's asking.

http://tutorial.math.lamar.edu/Classes/LinAlg/InverseMatrices_files/eq0041M.gif [Broken]

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## Answers and Replies

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## Homework Statement

The proofs:

show (A')^-1 = (A^-1)'

and

(AB)^-1 = B^-1A^-1

## The Attempt at a Solution

for the first one:

(A^-1*A) = I
(A^-1*A)' = I' = I
A'(A^-1)' = I
Both proofs are using the same basic idea: if the product of two (square) matrices is I, then the two matrices are inverses of each other.

In the last line above, you have AT and (A-1)T multiplying to make I. That means that AT is the inverse of (A-1)T.
but im not sure how this proves that a transpose inverse = a inverse transpose...

the second i have the same problem. Not sure how this really proves what it's asking.

http://tutorial.math.lamar.edu/Classes/LinAlg/InverseMatrices_files/eq0041M.gif [Broken]
Same thing in the work above, which says that the inverse of AB is B-1A-1.

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