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 \[B^{-1}A^{-1}\]
.\[(AB)^{-1}ABBY^{-1}A^{-1}=B^{-1}A^{-1} \]
\[(AB)^{-1}AA^{-1}=B^{-1}A^{-1} \]
\[(AB)^{-1}AA^{-1}=B^{-1}A^{-1} \]
\[(AB)^{-1}I=B^{-1}A^{-1} \]
\[(AB)^{-1}=B^{-1}A^{-1} \]