How do I prove that $(ABC)^{-1} = C^{-1} B^{-1} A^{-1}$ [closed]

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Please help me answering this problem! thank you :)

Prove that for any nonsingular matrices $A$, $B$, and $C$, the equation $$(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$$ holds. (Hint: Assume $D$ is the inverse of $ABC$, thus $DABC =I$. Post-multiply both sides of the matrix equation by $C^{-1}B^{-1}A^{-1}$ and proceed from there.)

Source.

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2 Answers

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$$(ABC)C^{-1}B^{-1}A^{-1}=AB\underbrace{(CC^{-1})}_{=I}B^{-1}A^{-1}=ABB^{-1}A^{-1}=AA^{-1}=I$$ Therefore, the inverse of $ABC$ is $C^{-1}B^{-1}A^{-1}$. I let you make all the necessary justification (like associativity of matriciel product...)

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1. let A,B,C be matrices and $ C=AB $
2. $ B=A^{-1}AB = A^{-1}C $
3. $ A=ABB^{-1} = CB^{-1} $, then by using $C=AB$ we have
4. $C=AB =[CB^{-1}][A^{-1}C] $
5. $C=AB =[CB^{-1}A^{-1}]C $ where $[CB^{-1}A^{-1}] = I$

Therefore:

$B^{-1}A^{-1} = C^{-1}$

By using this result you can extend it to n matrices.

For instance, for you question let $ABC = F$ and $BC = D$. Then you will have $F^{-1} = D^{-1}A^{-1}$ where $D^{-1} = C^{-1}B^{-1} $. Substituting back $D^{-1}$ gives $F^{-1} = C^{-1}B^{-1}A^{-1}$.

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