When the inverse is known at order , compute it at order . Non-symmetric matrix.
Let
be a generic, possibly non symmetric, invertible matrix of order
and
its known inverse. Let
and
be vectors of length
and
a scalar. Compute the inverse of the block matrix:
The inverse has the form:
with
matrix of order
,
and
vectors of length
and
a scalar, all to be determined so that:
where
is the identity matrix of order
and
a vector of length
with all components
.
By applying the row-by-column product to the block matrices one obtains the following relations:
Apply
to the first two relations:
Now use the auxiliary variable
(a vector of length
):
is known because
is known. Then:
Substitute in the last equation:
Of course this can be computed if
, which is the invertibility condition.
is then obtained from:
Now multiply the first equation by
on the left:
and use the second equation:
to obtain, after collecting
:
Then use the expression of
to obtain:
Remark that by taking the transpose, since
is not assumed to be symmetric, this is written as:
Finally substitute
in the first equation to obtain:
It is also possible to compute an inverse matrix of order
from a known matrix of order n+1. The problem is still stated as:
However in this case
,
,
and
are known, and
has to be computed. Use:
Substitute in:
Finally, it is important to remark that here the last row and column are attached or eliminated, but this is only for convenience of presentation. The inverse matrix can be updated in this way when any row (and column) is eliminated. When a row (and column) is attached in the last position, row and columns can be sorted afterwards as desired, in both matrices accordingly.
Francesco Uboldi 2014,2015,2016,2017