GATE 2004 question on Matrices

Let A, B, C, D be n x n matrices, each with non-zero determinant. If ABCD = I , then B-1 is:

(A) D-1C-1A
(B) CDA
(C) ADC
(D) Does not necessarily exist

Answer

Things you need to know

For a given Matrix, X 

XX-1 = I

Now, A/Q

     ABCD = I

⇒ A-1ABCD = A-1     ------------------------------------------> Multiplying both sides by A-1 at Leftmost  

⇒ IBCD = A-1           (∵ AA-1 = I, where I: Identity Matrix)

⇒  BCDD-1  =  A-1D-1  ----------------------------------------> Multiplying both sides by D-1 at Rightmost

⇒  BCC-1 = A-1D-1C-1 -----------------------------------------> Multiplying both sides by C-1 at Rightmost

⇒  B = A-1D-1C-1

⇒  ∴ B-1 = CDA

Thus , Ans (B)

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