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$A$, $B$ and $C$ are square matrices with $ABC=I$. I need to show that $B$ is invertible and $B^{-1} = CA$.

I have proved it using the fact stated here.

Since we only need to prove invertiblity of $B$ there may be a simpler proof. Any ideas?

Here is my proof.

$(AB)C=I \implies C(AB) = I \implies (CA)B=I \implies B^{-1}=CA$

Here I have used the fact in linked question. I am seeking an elementary proof but seems like there isn't one.

Community
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Ashot
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  • Do you mean "invertible"? –  Jun 28 '14 at 17:47
  • The inverse of AB is $(AB)^{-1}=B^{-1}A^{-1}$ (All matrices are invertible because of 1=det(I)=det(ABC)=det(A)det(B)det(C) ). Now u can easy show the statement. – Marm Jun 28 '14 at 17:47
  • @Balla: I was about to write this, or will you do it yourself? – user99680 Jun 28 '14 at 17:50
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    I don't think you're going to get away with not using the fact in the linked question, or, as some people have done below, properties of determinants. It's not true in general for rings in which right inverses need not be left inverses. – Dustan Levenstein Jun 28 '14 at 19:20

5 Answers5

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All matrices are invertible. The we obtain:

$ABC=I \iff AB=C^{-1} \iff A=C^{-1}B^{-1} \iff CA=B^{-1}$

Marm
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Using the fact mentioned in your question we see that $A$ and $C$ are invertible and so $B$ is also invertible and that $$B=A^{-1}C^{-1}$$ Can you take it from here?

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First of all, it's clear that $A$ is invertible, because it has the inverse $BC$.

If you want to know that $B^{-1}=CA$, try writing their product and showing that it must be $I$:

$B(CA) = IBCA = A^{-1}ABCA=A^{-1}(ABC)A = A^{-1}IA = I$

G Tony Jacobs
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As stated by Balla, by properties of the determinant , $$Det(ABC)=DetADetBDetC=1 $$, so each of DetA, DetB, DetC must be nonzero, so invertible. You can then check that $ C^{-1}B^{-1}A^{-1}$ is the inverse of $ABC$, since : $$(C^{-1}B^{-1}A^{-1})(ABC)=C^{-1}B^{-1}(AA^{-1})BC=....=I$$, so that $$B^{-1}C^{-1}A^{-1}=I $$ Now you can multiply both sides by $AC$, to find the inverse of $B$.

user99680
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As many answers mention, it is fairly easy to show that $A$ (or $C$, if you prefer) must be invertible.

  • We can use the properties of determinants: $\det(A) \det(B) \det(C) = \det(ABC) = \det(I) = 1$, so $A$ must be invertible.
  • We can use the rank inequality $\text{rank}(PQ) \leq \min\{\text{rank}(P),\text{rank}(Q)\}$, setting $A = P$ and $BC = Q$, noting $PQ = I$ has full rank

With that in mind, we have the following proof of a slightly different flavor. $$ B(CA) = A^{-1}(ABC)A = A^{-1} I A = I $$ So that $CA = B^{-1}.$

Ben Grossmann
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