Is my proof correct? Can you find any errors?
Prove that if matrix $ABC$ invertible($A$,$B$,$C$ all $n$ x $n$), then matrix $B$ is invertible.
Part One:
Let $X$ and $Y$ be two $n$ x $n$ matrices such that $XY$=$I$, if we right multiply boths sides by b $∈ ℝ ^n$ we obtain $X$($Y$b)=b. Thus the corresponding linear transformation $T(z)$=$X$z is onto and this is true IFF there is a pivot in every row which is true for an $n$ x $n$ matrix $X$ IFF $X$ is row equivalent to $I$ which is true IFF $X$ is invertible.
Part Two:
If $ABC$ is invertible then there exist an $n$ x $n$ matrix $D$ such that $ABCD=I$.
$ABCD=A(BCD)=I$
So, A is invertible and its inverse is $A^{-1}=BCD$ and so $BCD$ is invertible.
If $BCD$ is invertible then there exist an $n$ x $n$ matrix $E$ such that $BCDE=I$.
$BCDE=B(CDE)=I$ and so $B$ is invertible.
Best Answer
Matrix $ABC$ is invertible iff $\det(ABC) \neq 0$.
Therefore $\det(A)\det(B)\det(C) \ne 0$.
As a consequence $\det(B) \ne 0$
which is equivalent to say that $B$ is invertible ($A$ and $C$ as well...).