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4 Answers

Best answer
64 votes
64 votes

Given $ABCD = I$

Multiplying both sides 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 $B^{-1}= (A^{-1}D^{-1}C^{-1})^{-1}$

$B^{-1} = CDA$

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26 votes
26 votes

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.

16 votes
16 votes
$ABCD=I$

=>$(AB)(CD)=I$  i.e. CD is the inverse of AB

=>$(AB)^{-1}=CD$

=>$B^{-1}A^{-1}=CD $

=>$ B^{-1}A^{-1}A=CDA$

=>$B^{-1}I=CDA$

= >$ B^{-1}=CDA$

Ans- B
0 votes
0 votes
$ABCD = I$

$(ABCD)^{-1} = I^{-1}$ (Taking inverse on both sides)

$D^{-1}C^{-1}B^{-1}A^{-1} = I$

$DD^{-1}C^{-1}B^{-1}A^{-1} = DI$ (Multiplying on left by D on both sides)

$C^{-1}B^{-1}A^{-1} = D $   $ ( \because DD^{-1} = I) $

$CC^{-1}B^{-1}A^{-1} = CD $

$B^{-1}A^{-1} = CD $

$B^{-1}A^{-1}A = CDA $

$B^{-1} = CDA $

$Answer: Option (B)$
Answer:

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