By Felix Y. (ed.)

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**Example text**

Thus n = 6 and k = 3 here. The matrix U is 1 2 . 1 1 Now 5 = 2 × 2 + 1, so q = 2 and r = 1. Assemble the matrix A as indicated above, to get 1 2 0 0 1 1 1 0 0 0 0 0 1 2 0 . A= 0 0 0 1 1 ———————— 0 0 0 0 1 Notice that A has 2 positive entries in each column with other entries 0 and that det(A) = 1, so (a) and (b) hold. 3. Properties of the Balance Module negative column sums of A, which yields 1 2 0 0 1 1 0 0 0 1 0 0 1 2 0 A∗ = 0 1 1 0 0 0 0 0 0 1 −2 −3 −2 −3 −2 49 .

U1 + u2 + · · · + un−1 , 0. Then we have u1 0 u1 + u 2 u1 u1 + u 2 + u 3 u1 + u 2 . θπ (v) = − , . u + u + · · · + u u + u + · · · + u 1 2 n−1 1 2 n−2 0 u1 + u2 + · · · + un−1 which equals u1 u2 u3 . = u. . un−1 −u1 − u2 − · · · − un−1 Our conclusion is that when π is an n-cycle, Im(θπ ) equals Baln (R). The statement in (3) now follows on applying this result to each of the cyclic permutations σi .

The first concern is to provide a means of describing the state of an accounting system at any instant. Now this is most naturally accomplished by listing the “values” of the various accounts in the system. So the first step must be to identify an algebraic structure to which the account values will belong. The structure chosen must be sufficiently rich to accommodate the operations that one would expect to apply to the accounting system. It emerges from the discussion below that ordered integral domains are the natural candidates.