Question

If $A$ and $B$ are invertible matrices of the same order, then prove that
$(AB{)}^{-1}=B^{-1}{A}^{-1}$.

Answer 1

If $X,Y$ are two matrices

$XY=YX=I$

Then $X=Y^{-1}$

Given $B^{-1}{A}^{-1}={\left(AB\right)}^{-1}$

Multiply with $AB$ to $B^{-1}{A}^{-1}$

$AB{B}^{-1}{A}^{-1}=A\left(B{B}^{-1}\right)A^{-1}=A{A}^{-1}=I$

Multiply with $B^{-1}{A}^{-1}$ to $AB$

$B^{-1}{A}^{-1}AB=B^{-1}IB=B^{-1}B=I$

$AB\left(B^{-1}{A}^{-1}\right)=\left(B^{-1}{A}^{-1}\right)AB=I$

${\left(AB\right)}^{-1}=B^{-1}{A}^{-1}$