Can someone explain these 2 linear algebra proofs

[Kuma]

Homework Statement

The proofs:

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

and

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

Homework Equations

The Attempt at a Solution

for the first one:

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

but I am 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

 

[Mark44]

Homework Statement

The proofs:

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

and

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

Homework Equations

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 A^T and (A^-1)^T multiplying to make I. That means that A^T is the inverse of (A^-1)^T.

but I am 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

Same thing in the work above, which says that the inverse of AB is B^-1A^-1.