$ABCD = I$ then $B^{-1} =?$

Question

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 ?

Answer 1

$ABCD=I$, $BCD=A^{-1}$, $CD=B^{-1}A^{-1}$, $CDA=B^{-1}$.

Answer 2

$(AB)(CD)=I \Rightarrow (CD)(AB)=I \Rightarrow (CDA)(B)=I \Rightarrow CDA=B^{-1}$.

Edit: Alternatively, $(A)(BCD)=I \Rightarrow (BCD)(A)=I \Rightarrow (B)(CDA)=I \Rightarrow CDA=B^{-1}$.

Answer 3

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}

Answer 4

If $ABCD = I$, then $BCD = A^{-1}$, $BC = A^{-1}D^{-1}$, $B = A^{-1}D^{-1}C^{-1}$, from which we get $B^{-1} = (A^{-1}D^{-1}C^{-1})^{-1} = CDA$.

Answer 5

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.