# Proof for inverse matrices

• Jul 11th 2010, 05:32 PM
arze
Proof for inverse matrices
Prove that for three non-singular matrices A, B, C that $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$

I know that for two square matrices, C, D, say, $(CD)^{-1}=D^{-1}C^{-1}$. I tried using this result to prove $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$, by considering AB as one square matrix, and multiplying by C, and here is my problem.
$(AB)^{-1}=B^{-1}A^{-1}$, I don't know how to continue after this. Or am I off the right track?

Thanks!
• Jul 11th 2010, 06:31 PM
MattMan
You're on the right track. Since we know that for all nonsingular matrices $A,B$, that $(AB)^{-1}=B^{-1}A^{-1}$
So for three nonsingular matrices $(ABC)^{-1}=((AB)C)^{-1}=C^{-1}(AB)^{-1}=C^{-1}B^{-1}A^{-1}$
• Jul 11th 2010, 06:43 PM
roninpro
You may also verify the rule directly.

Recall that two matrices $A,B$ are inverses if $AB=BA=I$. We claim that $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$. We can check it by computing $ABC(C^{-1}B^{-1}A^{-1})$ and $(C^{-1}B^{-1}A^{-1})ABC$. Everything should work out nicely.
• Jul 11th 2010, 07:12 PM
arze
Quote:

Originally Posted by MattMan
You're on the right track. Since we know that for all nonsingular matrices $A,B$, that $(AB)^{-1}=B^{-1}A^{-1}$
So for three nonsingular matrices $(ABC)^{-1}=((AB)C)^{-1}=C^{-1}(AB)^{-1}=C^{-1}B^{-1}A^{-1}$

thank you very much, I wasn't sure how exactly to work it out from where I left.