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Let $A, B, C, D$ be $n \times n$ matrices, each with non-zero determinant. If $ABCD = I$, then $B^{-1}$ is

1.    $D^{-1}C^{-1}A^{-1}$
2.    $CDA$
3.    $ADC$
4.    Does not necessarily exist
edited | 1.7k views

Given

$ABCD = I$

multiply LHS,RHS by $A^{-1}$

$A^{-1}ABCD = A^{-1}I$ (position of $A^{-1}$ on both sides should be left)

$\Rightarrow\qquad BCD = A^{-1}$

$\Rightarrow\, BCDD^{-1}= A^{-1}D^{-1}$

$\Rightarrow\qquad \,\;\;BC= A^{-1}D^{-1}$

$\Rightarrow\;\;\; BCC^{-1}= A^{-1}D^{-1}C^{-1}$

$\Rightarrow\qquad\quad \;\,B=A^{-1}D^{-1}C^{-1}$

Now $B^{-1}= (A^{-1}D^{-1}C^{-1})^{-1}$

$B^{-1} = CDA$

edited
why we have multiplied by A inverse first we can multiply by  B inverse or C inv or D inverse first
Is there a rule behind taking inverse of A first over C?
@kaluti @habi

not necessary to take A

Try with D also

but i am not sure if B or C can be taken as they are inbetween

from this

B−1  = (A−1D−1C−1)−1

to this

B−1 = CDA

There is one shortcut to solve this kind of questions:

If ABCD=I and $B^{-1}$ is asked then interchange the B's left side matrices with its right side matrices. Interchange CD with A so it will be CDA, hence, the answer is $B^{-1}$ = CDA

For example if there are 8 nxn matrices given as follows
ABCDEFGH = I and $F^{-1}$ is asked then interchange GH with ABCDE, therefore

$F^{-1}$ will be GHABCDE.

Remark: This method doesn't work for extreme right or left matrix, for example we can't computer $A^{-1}$ or $H^{-1}$ using this method.