p = (1, 2)T
It was tedious to figure that out (and would be much worse if A were, say, 5 × 5). It would be nice to have a better way. Say that
(1) q = Ap
for column vectors p and q and n×n matrix A . We know A and q and are trying to figure out what p is. Is it possible to find a matrix Bn×n so that
(2) p = Bq
If it is, then using (1) and (2)
(3) q = A(Bq)
(4) q = (AB)q
If (4) is true, then (AB) = I. (Remember that I is unique). B, if it exists, is the inverse of A, written A-1, and the following is true:
AA-1 = A-1A = I