Let B, A, and e be square matrices with e small, and define
![\begin{displaymath}
{\hbox{\sf B}}\equiv{\hbox{\sf A}}({\hbox{\sf I}}+{\hbox{\sf e}}),
\end{displaymath}](i_468.gif) |
(1) |
where I is the Identity Matrix. Then the inverse of
is approximately
![\begin{displaymath}
{\hbox{\sf B}}^{-1}=({\hbox{\sf I}}-{\hbox{\sf e}}){\hbox{\sf A}}^{-1}.
\end{displaymath}](i_470.gif) |
(2) |
This can be seen by multiplying
Note that if we instead let
, and look for an inverse of the form
, we obtain
In order to eliminate the
term, we require
. However, then
, so
so there can be no inverse of this form.
The exact inverse of
can be found as follows.
![\begin{displaymath}
{\hbox{\sf B}}'={\hbox{\sf A}}+{\hbox{\sf e}}={\hbox{\sf A}}({\hbox{\sf I}}+{\hbox{\sf A}}^{-1}{\hbox{\sf e}}),
\end{displaymath}](i_485.gif) |
(5) |
so
![\begin{displaymath}
{\hbox{\sf B}}'^{-1}=[{\hbox{\sf A}}({\hbox{\sf I}}+{\hbox{\sf A}}^{-1}{\hbox{\sf e}})]^{-1}.
\end{displaymath}](i_486.gif) |
(6) |
Using a general Matrix Inverse identity then gives
![\begin{displaymath}
{\hbox{\sf B}}'^{-1}=({\hbox{\sf I}}+{\hbox{\sf A}}^{-1}{\hbox{\sf e}})^{-1}{\hbox{\sf A}}^{-1}.
\end{displaymath}](i_487.gif) |
(7) |
© 1996-9 Eric W. Weisstein
1999-05-26