## Proof of Equation for Inverse of Product of Square Matrices

Theorem
If
$A$
and
$B$
are square invertible matrices then
$(AB)^{-1} = B^{-1} A^{-1}$

Proof
$(AB)^{-1}$
is the inverse of
$AB$
since
$(AB)(AB)^{-1} - (AB)^{-1}AB=I$

The inverse of a matrix is unique since if
$C,\: C'$
are both inverses of
$AB$
then
$C(AB)=C'(AB)=I$
then
$C(AB)(AB)^{-1}=C'(AB)(AB)^{-1}=(AB)^{-1} \rightarrow c=C'$
.
Multiply
$(AB)^{-1}AB=I$
by
$B^{-1}A^{-1}$
.
$(AB)^{-1}ABBY^{-1}A^{-1}=B^{-1}A^{-1}$

$(AB)^{-1}AA^{-1}=B^{-1}A^{-1}$

$(AB)^{-1}AA^{-1}=B^{-1}A^{-1}$

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

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