When the inverse is known at order n , compute it at order n+1 . Non-symmetric matrix.
Let A be a generic, possibly non symmetric, invertible matrix of order n and A -1 its known inverse. Let b and d be vectors of length n and c a scalar. Compute the inverse of the block matrix:
[ A b d T c ]
The inverse has the form:
[ X y w T z ]
with X matrix of order n , y and w vectors of length n and z a scalar, all to be determined so that:
[ A b d T c ][ X y w T z ]=[ I 0 0 T 1 ]
where I is the identity matrix of order n and 0 a vector of length n with all components 0 .
By applying the row-by-column product to the block matrices one obtains the following relations:
A X + b w T = I
A y + b z= 0
d T X +c w T = 0 T
d T y +cz=1
Apply A -1 to the first two relations:
X + A -1 b w T = A -1
y + A -1 b z= 0
Now use the auxiliary variable p (a vector of length n ):
p A -1 b
p is known because A -1 is known. Then:
y =-z p
Substitute in the last equation:
- d T p z+cz=1
and solve for z :
z= 1 c- d T p
Of course this can be computed if c d T p , which is the invertibility condition.
y is then obtained from: y =-z p
Now multiply the first equation by d T on the left:
d T X + d T p w T = d T A -1
and use the second equation:
d T X =-c w T
to obtain, after collecting w T :
( -c+ d T p ) w T = d T A -1
Then use the expression of z to obtain: w T =-z d T A -1
Remark that by taking the transpose, since A is not assumed to be symmetric, this is written as: w =-z ( A -1 ) T d
Finally substitute w T in the first equation to obtain:
X = A -1 - p w T
It is also possible to compute an inverse matrix of order n from a known matrix of order n+1. The problem is still stated as:
[ A b d T c ][ X y w T z ]=[ I 0 0 T 1 ]
However in this case X , y , w and z are known, and A -1 has to be computed. Use: A -1 b =- 1 z y
Substitute in: X +( A -1 b ) w T = A -1
So that:
A -1 = X - 1 z y w T
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.
Licenza Creative Commons Francesco Uboldi 2014,2015,2016,2017