The name is absent

16

+-•••-?
-  +  ∙∙∙  -  ?

-  -  ∙∙∙  +  ?

о  о  •••  о  -

The submatrix of A, which is obtained by deleting the last column and the last row,
is a productive Leontief matrix. Thus there exists a positive integral combination
of columns one through n — 1 for which the last element is zero, the и-th element
is strictly negative and the other elements are strictly positive. We can therefore
transform A to the matrix B = AV by subtracting a large positive integral multiple
of this combination from the last column.

Next we shall give a procedure to transform the matrix B = (ðʊ) to the standard
form. The procedure is described as follows:

Step (a) If there are indices i and j (г ψ j^) for some 1 < i,j ≤ n with Ьц ≤ ∣‰₇∙∣, we
can find a positive integer c and an integer d ∈ { 0,1,..., ⅛ — 1 } such that
∖bi_j I = сЬц + d, where c is the lower integer part of ()'ɪ, and then add c multiple
of column i to column j.

Step (b) Repeat Step (a) until there are no indices i and j (г ψ j) for 1 ≤ i,j' ≤ n with
Ьц N I b{j i.

It is obvious that the above operation is a unimodular transformation. We still
have to demonstrate that the procedure is feasible. Recall that the origin of Rⁿ is
in the interior of the convex hull of the vectors α₁,...,α_ra+1. It implies that there are
n ⅛ 1 strictly positive convex combination coefficients ʌɪ,..., Λ_ra+ι such that

∏+1

∑a_λ = o.                       (4.1)

г = 1

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