If $ABC$ is invertible, then show matrices $A$, $B$, and $C$ are all invertible. In Linear Algebra, I am working with inverting matrices and doing proofs. The exact question is:
Show that if $ABC$ is invertible, then $A, B, C$ are all invertible.
Each matrix is a $n\times n$ matrix, but I do not understand how to show that each individual matrix is invertible.
 A: Hints:
$M$ is invertible iff $\det(M)\ne0$.
$\det(ABC)=\det(A)\det(B)\det(C).$
A: You only really need to show that, if $AB$ is invertible, then $A$ and $B$ are invertible. If you then have $ABC = (AB)C$ is invertible, then $AB$ and $C$ are invertible (by the simple result), and hence $A, B, C$ are all invertible (by the simple result again).
Another thing to think about is that this is a property that is going to rely on finite-dimensionality in some way. The result is not true for infinite matrices (or, more generally, operators on infinite-dimensional spaces). So, expect finite dimensions to come in some way. There's a few ways to attack this problem, but all of them will make use of the finite-dimensionality of the problem.
Here's one way. Recall that, in finite dimensions, if $MN = I$, where $M$ and $N$ are square matrices, then $NM = I$ too, i.e. $M = N^{-1}$. Then, if $D = (AB)^{-1}$, then
$$I = (AB)D = A(BD).$$
Therefore, $A$ is invertible, and $BD = A^{-1}$. Similarly,
$$I = D(AB) = (DA)B,$$
so $B$ is invertible with inverse $DA$.
A: It is enough, by induction, to show that if $AB$ is an invertible matrix, then both $A$ and $B$ are invertible. By definition, we have a matrix $C$ such that $(AB)C=I$ (where $I$ is the identity matrix.) Now, obviously, by the associative property, we see that $A(BC)=I,$ so $BC$ is an inverse of $A.$ In other words, $A$ is invertible.
The same argument applies for $B,$ since $C(AB)=I,$ so $CA$ is an inverse of $B.$
(Note: this argument applies for any ring.)
A: If exists $X$ such that $(ABC)X=\mathbb1$ then $X$ has inverse too, and $BCX$ is the inverse of $A$. Then $BC=A^{-1}X^{-1}$. But also $(XAB)C=\mathbb1$, hence $C$ is invertible. So $B=(CXA)^{-1}$ then $B(CXA)=\mathbb1$, that is $B$ is also invertible.
