$ABCD = I$ then $B^{-1} =?$ I got this question in a practice book.
A,B,C and D are $n\times n$ matrices with non-zero determinant.
$ABCD = I$ , then $B^{-1}$ = ?
The answer to this was $B^{-1}= CDA$. 
How was that answer arrived at ?
 A: Since A, B, C, D all have non zero determinant this implies that their respective inverses exist.
It's really an exercise in matrix algebra and in particular matrix multiplication.
\begin{align}
ABCD &=I\\
AB &=D^{-1}C^{-1}\\
B &=A^{-1}D^{-1}C^{-1}\\
B^{-1} &=CDA
\end{align}
A: If $ABCD = I$, then $BCD = A^{-1}$, $BC = A^{-1}D^{-1}$, $B = A^{-1}D^{-1}C^{-1}$,
from which we get $B^{-1} = (A^{-1}D^{-1}C^{-1})^{-1} = CDA$.
A: If $ABCD = I$, then we could find a somewhat cyclic relation between these matrices:
$$
ABCDA = A \Rightarrow BCDA = A^{-1}A=I
$$
by exploiting multiplication from left or right, in addition to that matrix and the inverse of a matrix can commute in a multiplication.
Doing the same trick leads us to: $CDAB = I, DABC = I$ also. 
$BCDA = B(CDA)=I$ and $CDAB = (CDA)B = I$ give you the answer.
A: $ABCD=I$, $BCD=A^{-1}$, $CD=B^{-1}A^{-1}$, $CDA=B^{-1}$. 
A: $(AB)(CD)=I \Rightarrow (CD)(AB)=I \Rightarrow (CDA)(B)=I \Rightarrow CDA=B^{-1}$.
Edit: Alternatively,
$(A)(BCD)=I \Rightarrow (BCD)(A)=I \Rightarrow (B)(CDA)=I \Rightarrow CDA=B^{-1}$.
A: Suppose $A$ is a $n \times n$ matrix with  non-zero determinant and $AB=I$. Then we can say that $B$ is the inverse of $A$, and we can say that $B^{-1}=A$?
Using the same rules for $ABCD=I$, partion it in $(ACD)B=I$ and let $ACD=X$ so $XB=I$. Thus we can say that $B^{-1}=X$. Thus $B^{-1}=ACD$.
A: If $ A, B, C, D $ are $n \times n$ matrices with non-zero determinants, it means inverse exists for these matrices.
It's given- $$ABCD=I$$
Take inverse on both sides of the equation. This will result in-
$$ => (ABCD)^{-1} = I^{-1} $$
$$ => D^{-1}C^{1}B^{-1}A^{-1} = I $$
(The order reverses on taking the inverse.)
Since, matrix multiplication is not commutative, we need to maintain the order of matrix multiplication.
$\therefore$ Multiply $D$ from the left.
$$ DD^{-1}C^{-1}B^{-1}A^{-1}=ID $$
Since, we know, for a $n \times n$ matrix A, $AI=A$
And, if A is non-singular matrix, then-  $A.A^{-1}=I$
$$=> C^{-1}B^{-1}A^{-1}=D $$
Multiply $C$ from the left.
$$=> CC^{-1}B^{-1}A^{-1}=CD $$
$$=> B^{-1}A^{-1}=CD $$
Multiply $A$ from the right.
$$=> B^{-1}A^{-1}A=CDA $$
$$=> B^{-1}=CDA $$
