Yes, that is true, but "(A*B^-1)^T= A^T*(B^-1)^T" isn't.pyroknife said: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 said: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 said: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?
To prove the symmetry of A*B^-1 for invertible matrices A and B, we must show that AB = BA. This can be done by multiplying both sides of the equation by B^-1 and simplifying to show that A*B^-1 = B^-1*A. This demonstrates that the product of A and B^-1 is commutative, thus proving symmetry.
Invertible matrices are matrices that have a unique inverse, meaning that when multiplied by its inverse, the result is the identity matrix. This is significant in proving symmetry because it allows us to manipulate the equation AB = BA to show that A*B^-1 = B^-1*A, which is the definition of symmetry.
No, the symmetry of A*B^-1 can only be proven for invertible matrices. Non-invertible matrices do not have a unique inverse and therefore cannot be used to show the commutativity of the product AB.
Yes, there are other ways to prove the symmetry of A*B^-1, such as using the properties of matrix multiplication. For example, we can show that (AB)^-1 = B^-1*A^-1 and (BA)^-1 = A^-1*B^-1, which are equivalent by the commutativity of matrix multiplication.
The symmetry of A*B^-1 has various real-world applications, such as in cryptography and coding theory, where invertible matrices are used to encode and decode information. It is also important in linear algebra and other mathematical fields, as it helps in understanding the properties and relationships between matrices and their products.