GATE CSE 2004 | Question: 27

Question

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

• A.    $D^{-1}C^{-1}A^{-1}$
• B.    $CDA$
• C.    $ADC$
• D.    Does not necessarily exist

Answer 1

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

$B^{-1} = CDA$

Comments

Best answer..👍

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@Gupta731    i think you get your answer much earlier :)

multiplication is not commutative.

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From this:
$B=A^{-1}D^{-1}C^{-1}\\
to \\B^{-1}=(A^{-1}D^{-1}C^{-1})^{-1}$
Please someone tell me how this was done.
It looks like we took the inverse of LHS and RHS. Is that allowed?

Answer 2

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.

Answer 3

$ABCD=I$

=>$(AB)(CD)=I$  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

Comments

we know that $A.A^{-1}=I$

If $A.B=I$ then we can write like $A^{-1}=B$  $($means $B$ is an inverse of $A)$

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My doubt:

For ABCD = I Apart from (AB)(CD) = I There are other choices too like (A)(BCD) = I or (ABC)D = I

Can we conclude?

• From (A)(BCD) = I, that BCD is inverse of A?
• From (ABC)D = I, that D is inverse of ABC?

Answer 4

$ABCD = I$

$(ABCD)^{-1} = I^{-1}$ (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 $

$Answer: Option (B)$