Why is the product of elementary matrices necessarily invertible? Why is the product of elementary matrices necessarily invertible? I understand that each elementary matrix is invertible, but why is their product also invertible? Is it the indirect result of this theorem?

 A: Each elementary matrix $E_i$ is invertible so $\det(E_i)\ne0$.
As $$\det(E_iE_j)=\det(E_i)\det(E_j)\ne0$$ then $E_iE_j$ is invertible.
A: It has nothing to do with linear algebra.  It is a general fact that in any unital ring, a product of invertible elements: $a_1a_2\dots a_n$, is invertible, and its invertible, and its inverse is the product of the inverses of the factors, in reverse order. In other words:
$$(a_1a_2\dots a_n)^{-1}=a_n^{-1}\dots a_2^{-1} a_1^{-1}. $$
A: That $A$ is invertible means precisely that there is another matrix $A^{-1}$ such that
$$AA^{-1}=I=A^{-1}A.$$
It is easy to show that the inverse, if it exists, must be unique.
Now suppose $A,B$ to be invertible, and denote $C:=AB$. $C$ will be invertible if we can find $C^{-1}$ such that $CC^{-1}=I=C^{-1}C$. Observe that $$C(B^{-1}A^{-1})=(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I.$$
Similarly we get $(B^{-1}A^{-1})C=I$. Therefore, since inverses are unique, by definition we can conclude that $C^{-1}=B^{-1}A^{-1}$.
