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+20 votes

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
asked in Linear Algebra by Veteran (59.9k points)
edited by | 2.6k views

3 Answers

+6 votes
Best answer

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


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

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


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

Ans- B
answered by (303 points)
selected by
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 $A)$
+40 votes


$ABCD = I$

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

$B^{-1} = CDA$

answered by Active (2.3k points)
edited by
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

someone please explain?


After step  BCD=A-1  

we can multiply B-1 on both sides to get

CD=B-1A-1  then multiply A on both sides , CDA=B-1A-1A we get



Why can't I directly multiply B-1  on both sides and get the answer as B-1=ACD.

B and B-1 gets cancelled in LHS

Best answer..👍

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

multiplication is not commutative.

+17 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.

answered by Boss (42.6k points)

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