# 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 \$\displaystyle (ABC)^{-1}=C^{-1}B^{-1}A^{-1}\$

I know that for two square matrices, C, D, say, \$\displaystyle (CD)^{-1}=D^{-1}C^{-1}\$. I tried using this result to prove \$\displaystyle (ABC)^{-1}=C^{-1}B^{-1}A^{-1}\$, by considering AB as one square matrix, and multiplying by C, and here is my problem.
\$\displaystyle (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 \$\displaystyle A,B\$, that \$\displaystyle (AB)^{-1}=B^{-1}A^{-1}\$
So for three nonsingular matrices \$\displaystyle (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 \$\displaystyle A,B\$ are inverses if \$\displaystyle AB=BA=I\$. We claim that \$\displaystyle (ABC)^{-1}=C^{-1}B^{-1}A^{-1}\$. We can check it by computing \$\displaystyle ABC(C^{-1}B^{-1}A^{-1})\$ and \$\displaystyle (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 \$\displaystyle A,B\$, that \$\displaystyle (AB)^{-1}=B^{-1}A^{-1}\$
So for three nonsingular matrices \$\displaystyle (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.