# Inverses

• May 1st 2010, 04:10 PM
CSG18
Inverses
Hi,

Why is it that if B = A^-1,

B^-1 = A

I understand that whatever you do to one side, you have to do the same for the other side, but I don't understand how (B^-1)(B^-1) = B

Can anyone please explain?

Thanks.
• May 1st 2010, 05:02 PM
jakncoke
What is this in regards to? What field are we working in? Matricies?
• May 1st 2010, 09:06 PM
mr fantastic
Quote:

Originally Posted by CSG18
Hi,

Why is it that if B = A^-1,

B^-1 = A

I understand that whatever you do to one side, you have to do the same for the other side, but I don't understand how (B^-1)(B^-1) = B

Can anyone please explain?

Thanks.

\$\displaystyle B^{-1} = (A^{-1})^{-1} = A\$.

Alternatively, note that \$\displaystyle B B^{-1} = I\$. But if \$\displaystyle B = A^{-1}\$ then \$\displaystyle A^{-1} B^{-1} = I \Rightarrow A A^{-1} B^{-1} = A \Rightarrow I B^{-1} = A ....\$
• May 2nd 2010, 04:34 AM
HallsofIvy
The definition of \$\displaystyle X^{-1}\$, in general, (matrices, elements of a group or field, etc.) is that \$\displaystyle Y= X^{-1}\$ if and only if \$\displaystyle XY= I\$ and \$\displaystyle YX= I\$ where I is the "multiplicative identity" for the algebraic structure. T

If you know that \$\displaystyle B= A^{-1}\$ then you know that \$\displaystyle AB= I\$ and [mathy]BA= I[/tex], by replacing X above with A and Y with B. But if we were to replace X with B and Y with A, we wold get exactly the same equations!

Therefore, \$\displaystyle A= B^{-1}\$.