# $ABCD = I$ 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 ?

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

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Wow if you put it that way it makes a lot of sense , Thanks – Gautam Mar 7 '13 at 1:01

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

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

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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}

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Aah, an important point to solving objective questions, "Since A, B, C, D all have non zero determinant this implies that their respective inverses exist." . Thanks . – Gautam Mar 7 '13 at 1:09

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

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This a bit more complicated, but still good to know . Thanks. – Gautam Mar 7 '13 at 1:04

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.

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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 $B^{-1}=ACD$.

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