last modified **Jul 2, 1996**

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

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. In the Mathematica input file
MatrixInverseInput.m
we run **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.
What we get is the following spreadsheet:

**THE ORDER IS NOW THE FOLLOWING:**

YOUR SESSION HAS DIGESTED | ||
---|---|---|

THE FOLLOWING RELATIONS | ||

USER CREATIONS APPEAR BELOW | ||
---|---|---|

SOME RELATIONS WHICH APPEAR BELOW | ||||
---|---|---|---|---|

MAY BE UNDIGESTED | ||||

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
above 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.