2.6k views

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
edited | 2.6k views

$ABCD=I$

=>$(AB)(CD)=I$  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
selected
0
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 $A)$

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

$B^{-1} = CDA$

edited
0
why we have multiplied by A inverse first we can multiply by  B inverse or C inv or D inverse first
0
Is there a rule behind taking inverse of A first over C?
0
@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
0

from this

B−1  = (A−1D−1C−1)−1

to this

B−1 = CDA

+4

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

CDA=B-1

0

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

0
0

multiplication is not commutative.

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.

1
2