Question

We are given with two invertible matrices ( mathrm{A} ) and ( mathrm{B} ), how to prove that ( (A B)^{-1}=B^{-1} A^{-1} ? )
We know that if A.B ( =I ) then it means ( mathrm{B} ) is inverse of matrix ( mathrm{A} )
where ( I ) is an identity matrix.
If, we can prove that ( (A B) cdot B^{-1} A^{-1}=I ) then it means that ( B^{-1} A^{-1} ) is inverse of ( A B . )
In other words proving ( (A B) cdot B^{-1} A^{-1}=I )
( Rightarrow(A B)^{-1}=B^{-1} A^{-1} )
Lets simplify ( A B . B^{-1} A^{-1} )
( Rightarrow A I A^{-1}=A A^{-1}=I quad{A I=A ) and
( left.A A^{-1}=Iright} )
Therefore, from above equation, we can say that ( B^{-1} A^{-1} ) is inverse
of ( A B )
( Rightarrow(A B)^{-1}=B^{-1} A^{-1} )
which is the desired equation.

# Jr.Inter !! 12 Let A and B be invertiable matrices then May '03 show that (AB)-1 = B-1 A-1. Sol. A is invertible matrix then A-l exists and AA-1-A1A=I

Solution