Is (B^-1)^T = (B^T)^-1 for symmetric invertible matrices?

[pyroknife]

the matrices A and B are
invertible symmetric matrices and AB = BA.
Show that A*B^-1 is symmetric


(A*B^-1)^T
=A^T * (B^-1)^T
=A^T * (B^T)^-1

Since A and B are symmetric
=A*B^-1


Is this right? Is (B^-1)^T = (B^T)^-1?

 

[HallsofIvy]

the matrices A and B are
invertible symmetric matrices and AB = BA.
Show that A*B^-1 is symmetric


(A*B^-1)^T
=A^T * (B^-1)^T
=A^T * (B^T)^-1

Since A and B are symmetric
=A*B^-1


Is this right? Is (B^-1)^T = (B^T)^-1?

Yes, that is true, but "(A*B^-1)^T= A^T*(B^-1)^T" isn't.
Rather, (A*B^{-1})^T= (B^{-1})^T*B^T.

 

[pyroknife]

Yes, that is true, but "(A*B^-1)^T= A^T*(B^-1)^T" isn't.
Rather, (A*B^{-1})^T= (B^{-1})^T*B^T.

Oh I memorized the identity wrong. (AB^T)=B^t*A^T

 

[pyroknife]

The solution in the book first proves
IF AB=BA, then B^-1 * A *B=A, so B^-1*A=AB^-1

For the last type, the "B^-1*A=AB^-1", part how did they go from B^-1 * A *B=A to
B^-1*A=AB^-1.

Did they divide both sides by B?

 

[Dick]

The solution in the book first proves
IF AB=BA, then B^-1 * A *B=A, so B^-1*A=AB^-1

For the last type, the "B^-1*A=AB^-1", part how did they go from B^-1 * A *B=A to
B^-1*A=AB^-1.

Did they divide both sides by B?

They multiplied both sides on the right by B^(-1). Talking about 'dividing' matrices by B is ambiguous. You can 'divide' on the left or the right.