Inverse Question for Matrices: AB vs B A

[eyehategod]

For matrices:
is (AB)$$^{-1}$$=
A$$^{-1}$$B$$^{-1}$$
or
B$$^{-1}$$A$$^{-1}$$

 

[daveb]

Are you talking about linear operators, matrices, members of a group, or what?

 

[robphy]

For matrices:
is (AB)$$^{-1}$$=
A$$^{-1}$$B$$^{-1}$$
or
B$$^{-1}$$A$$^{-1}$$

Since $$I=(AB)^{-1}(AB)=(AB)^{-1}AB$$
.. you can finish this off.

 

[HallsofIvy]

In other words do it! What is $(A^{-1}B^{-1})(AB)$? What is $(B^{-1}A^{-1})(AB)$?

 

[eyehategod]

so the answer is B$$^{-1}$$A$$^{-1}$$

 

[eyehategod]

what I am trying to get to is this:
is there a general property.
for example:
is(BA)$$^{2}$$
equal to:
B$$^{2}$$A$$^{2}$$
or
A$$^{2}$$B$$^{2}$$

 

[morphism]

In general, no.

 

[eyehategod]

so what would the the answer for (BA)^2

 

[morphism]

(BA)^2 = BABA

So if A and B are invertible...

 

[eyehategod]

so its B^2A^2

 

[morphism]

What is B^2A^2?

 

[rock.freak667]

so its B^2A^2

well normally to find a general way(an easy) way to find say A^99194

you can just represent A in the form PDP$^{-1}$ where D is the diagonalizable matrix. But I do not think you have reached that far in your course yet. If you have done eigenvalues and eigenvectors then you will understand.