I need to set up a proof for this problem:
Given that $A$ and $B$ are both $n\times n$ matrices. $A$ is invertible, and $AB=BA$.
Prove that $A^{-1}B=BA^{-1}$.
I'm just unsure how to go about this particular proof.
I need to set up a proof for this problem:
Given that $A$ and $B$ are both $n\times n$ matrices. $A$ is invertible, and $AB=BA$.
Prove that $A^{-1}B=BA^{-1}$.
I'm just unsure how to go about this particular proof.
Hint: Multiply both sides by $A^{-1}$ on the left, then on right:
$A^{-1} AB A^{-1} = A^{-1} BA A^{-1} \\ BA^{-1} = A^{-1}B$
You could try to start with the fact that $AB=BA$ and multiply one side by $A^{-1}$. After that, do the same with the other side. You should find what you wanted to prove.
Question for you: what does $AA^{-1}$ equals?
Since A is invertible we have that $AA^{-1}=I$.
We are told$ AB = BA$ Then if you multiply both sides by $A^{-1}$we have:
$A^{-1}AB = A^{-1}BA$ $IB = A^{-1}BA B=A^{-1}BA, BA^{-1}=A^{-1}BAA^{-1}, BA^{-1}=A^{-1}B(AA^{-1}),BA^{-1}=A^{-1}BI$ $BA^{-1}=A^{-1}B$