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