# Proving that a matrix is invertible without using determinants

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}}$

$AB=$${C^{-1}} WLOG, B is invertible CAB =C$${C^{-1}}$

$CAB = I$

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

$CA=$${B^{-1}} My question is if it is correct to assume {C^{-1}} does not exist since the proof does not mention anything about {C^{-1}} existing or not. • possible duplicate? If ABC non-singular prove that A, B and C non-singular too. – AccidentalFourierTransform Mar 25 '18 at 14:08 • What do you know about inverses, at that point in the book? For instance, what definition and/or characterizations of inverses and invertible matrices do you have available? – Federico Poloni Mar 25 '18 at 20:05 • @FedericoPoloni I know An n × n matrix A is invertible when there exists an n × n matrix B such that AB = BA = I and if A is an invertible matrix, then the system of linear equations Ax = b has a unique solution x = A^(-1)b. I used the second fact in my proof, where I made vector b a zero vector. So at this point in the book, I know nothing of elementary matrices or determinants. – Jay Mar 25 '18 at 21:10 • @Jay Do you know that checking only one of those two equations (AB=I, for instance) is sufficient for invertibility, or do you have to check both conditions? Most of the answers build on this fact. – Federico Poloni Mar 26 '18 at 6:35 • @FedericoPoloni Yes, there is a proof in the textbook that shows this theorem – Jay Mar 26 '18 at 23:51 ## 4 Answers It can be shown, via elementary methods, that if M and N are square matrices such that MN = I, then NM= I. Thus, if ABC = A(BC) = I, then (BC)A = B(CA) = I, which shows that B is invertible and B^{-1}=CA. A square matrix A is invertible if and only if there is another matrix A^{-1} such that A^{-1}A=I. In the expressABC=I, chose X=AB and we have XC=I. Thus C^{-1}=X. Similarly show A is invertible. Now,$$ABC=IAB=C^{-1}CAB=ICA=B^{-1}$$• The first statement is only true for square matrices, since A^{-1}A=I implies AA^{-1}=I. However, for matrices in general, it should be "A matrix A is invertible iff. there is another matrix A^{-1} which satisfies A^{-1}A=AA^{-1}=I. – Erlend Graff Mar 25 '18 at 12:53 • @ErlendGraff noted and edited! – Prathyush Poduval Mar 25 '18 at 14:15 A square matrix is invertible if and only if its rank is n. Also, we know that \operatorname{rank}(AB) \le \min(\operatorname{rank}(A),\operatorname{rank}(B) ) In this question$$ABC=I$$Hence \operatorname{rank}(ABC)=n$$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$B^{-1}=(A^{-1}C^{-1})^{-1}=CA$• +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. – leftaroundabout Mar 25 '18 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.

• The first statement is not just the definition of invertible matrices. It requires a non-trivial proof. – Adayah Mar 25 '18 at 11:57
• I edited my answer. – grontim Mar 25 '18 at 14:17