Prove if $A$, $B$, and $C$ are square matrices and $ABC = I$, then $B$ is invertible and $B^{-1}= CA$.

I know that this proof can be done by taking the determinant of $ABC=I$ and showing that $A$, $B$, and $C$ are invertible and then finding the inverse of $B$. However, in this chapter of the book, we have not yet learned determinants so I would like to solve the problem without determinants. My proof method involves using a contradiction and is as follows:

Assume ${C^{-1}}$ does not exist, then $\exists$ $x$ $\neq$ $0$ such that $Cx = 0$.
$ABCx =Ix$

$AB0 = x$

$0=x$, which is a contradiction since we know that $x$ $\neq$ $0$, and therefore ${C^{-1}}$ exists.

$AB$${C^{-1}}$ $=I$${C^{-1}}$


WLOG, B is invertible

$CAB =C$${C^{-1}}$

$CAB = I$

$CAB$${B^{-1}}$ $=I$${B^{-1}}$


My question is if it is correct to assume ${C^{-1}}$ does not exist since the proof does not mention anything about ${C^{-1}}$ existing or not.

  • $\begingroup$ possible duplicate? If ABC non-singular prove that A, B and C non-singular too. $\endgroup$ Mar 25, 2018 at 14:08
  • 1
    $\begingroup$ What do you know about inverses, at that point in the book? For instance, what definition and/or characterizations of inverses and invertible matrices do you have available? $\endgroup$ Mar 25, 2018 at 20:05
  • $\begingroup$ @FedericoPoloni I know An n × n matrix A is invertible when there exists an n × n matrix B such that AB = BA = I and if A is an invertible matrix, then the system of linear equations Ax = b has a unique solution x = A^(-1)b. I used the second fact in my proof, where I made vector b a zero vector. So at this point in the book, I know nothing of elementary matrices or determinants. $\endgroup$
    – Jay
    Mar 25, 2018 at 21:10
  • $\begingroup$ @Jay Do you know that checking only one of those two equations ($AB=I$, for instance) is sufficient for invertibility, or do you have to check both conditions? Most of the answers build on this fact. $\endgroup$ Mar 26, 2018 at 6:35
  • $\begingroup$ @FedericoPoloni Yes, there is a proof in the textbook that shows this theorem $\endgroup$
    – Jay
    Mar 26, 2018 at 23:51

4 Answers 4


It can be shown, via elementary means, that if $M$ and $N$ are square matrices such that $MN = I$, then $NM= I$.

Thus, if $ABC = A(BC) = I$, then $(BC)A = B(CA) = I$, which shows that $B$ is invertible and $B^{-1}=CA$.


A square matrix $A$ is invertible if and only if there is another matrix $A^{-1}$ such that $A^{-1}A=I$. In the express$ABC=I$, chose $X=AB$ and we have $XC=I$. Thus $C^{-1}=X$. Similarly show $A$ is invertible. Now, $$ABC=I$$ $$AB=C^{-1}$$ $$CAB=I$$ $$CA=B^{-1}$$

  • $\begingroup$ The first statement is only true for square matrices, since $A^{-1}A=I$ implies $AA^{-1}=I$. However, for matrices in general, it should be "A matrix $A$ is invertible iff. there is another matrix $A^{-1}$ which satisfies $A^{-1}A=AA^{-1}=I$. $\endgroup$ Mar 25, 2018 at 12:53
  • $\begingroup$ @ErlendGraff noted and edited! $\endgroup$ Mar 25, 2018 at 14:15

A square matrix is invertible if and only if its rank is $n$.

Also, we know that $\operatorname{rank}(AB) \le \min(\operatorname{rank}(A),\operatorname{rank}(B) )$

In this question


Hence $\operatorname{rank}(ABC)=n$

$$n \le \min(\operatorname{rank}(A),\operatorname{rank}(B), \operatorname{rank}(C) )$$

Hence $\operatorname{rank}(A)=\operatorname{rank}(B) =\operatorname{rank}(C)=n$ and they are all invertible.

Hence $B=A^{-1}C^{-1}$ and $B^{-1}=(A^{-1}C^{-1})^{-1}=CA$

  • 2
    $\begingroup$ +1 for no-nonsense proof that they must all be invertible. (Another way to phrase this: if one of the matrices were noninvertible, then there would have to be some subspace which is at some point in the composition chain mapped to zero, but then $ABC$ couldn't be the identity.) — However the question did also ask about $B^{-1} = CA$, which can't be shown that easily. $\endgroup$ Mar 25, 2018 at 15:51

Since, $ABC = I$, then $A$ is invertible (see below). Same with $C$ since $(AB)C=I$. The inverse of $C$ is $AB$. So, $B = A^{-1}C^{-1}$ and $BCA= I$. So, the inverse of $B$ is $CA$.

Proof-sketch of AB=I $\implies$ A and B are invertible and they are inverses of each other

Consider the linear transformations induced by matrices $A$ and $B$ on $\mathbb R^n$. For, convenience I will represent the corresponding linear transformations by $A$ and $B$ as well. Since $AB=I$, it implies the composition of linear transformations $A$ and $B$ is the identity linear transformation. So, $B$ is an injective linear transformation and $A$ is a surjective linear transformation. Since, we are in a finite-dimensional vector space this implies both $A$ and $B$ are invertible linear transformation (by rank-nullity theorem). So, $A$ and $B$ are invertible as matrices. Also not that since $AB=I$, we get $BABA=BA$. Now by multiplying inverse of $A$ and $B$ from right on both sides, we get $BA=I$ as well. Now, from the definition inverse of matrices, we get $A$ and $B$ are inverses of each other.

  • $\begingroup$ The first statement is not just the definition of invertible matrices. It requires a non-trivial proof. $\endgroup$
    – Adayah
    Mar 25, 2018 at 11:57
  • $\begingroup$ I edited my answer. $\endgroup$
    – grontim
    Mar 25, 2018 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.