Given matrices a, b, c and d, we wish to determine under what conditions there exists matrices x, y, z and w such that the block two by two matrices
are inverses of each other. Also, we wish to find formulas for x, y, z and w.
This problem was solved in a paper by W.W. Barrett, C.R. Johnson, M. E. Lundquist and H. Woerderman [BJLW] where they showed it splits into several cases depending upon which of a, b, c and d are invertible. In our next example, we assume that a, b, c and d are invertible and derive the result which they obtain. If one runs NCProcess1 on the polynomial equations which state that a, b, c and d are invertible together with the eight polynomial equations which come from the two matrices above being inverses of each other, one gets the spreadsheet:
THE ORDER IS NOW THE FOLLOWING: a < a < b < b < c < c < d < d « z « x « y « w
YOUR SESSION HAS DIGESTED  

THE FOLLOWING RELATIONS  
w a d z b d
x d  d z b
y
c

b
z
c
a a 1
a a 1
b b 1
b b 1
c c 1
c c 1
d d 1
d
d
1
 

USER CREATIONS APPEAR BELOW 
 
 

SOME RELATIONS WHICH APPEAR BELOW 
 
 MAY BE UNDIGESTED 
 
z
b
z
z
+
d
a
c
This spreadsheet shows that, if a, b, c and d are invertible, then one can find x, y, z and w such that the matrices in (§) are inverses of each other if and only if z b z = z + d a c. The spreadsheet also gives formulas for x, y and w in terms of z.
In [BJLW] they also solve the problem in the case that a is not invertible  the answer is more complicated and involves conditions on ranks of certain matrices. It is not clear whether or not these can be derived in a purely algebraic fashion.