Fraction-free — the series
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Every intermediate is a minor
Fraction-free elimination divides by the previous pivot at every step and the division is always exact. Not usually, not for these entries — always, because the number being divided is a determinant with that pivot as a factor, which is a theorem and is checked here against the minors themselves.
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A bound on every intermediate at once
Fraction-free elimination's intermediates are minors of the original, which is a theorem about exactness. It is also a bound: Hadamard's inequality applies to every minor, so one inequality bounds the whole run before it starts. The bound on the k-th step is the one on (k+1)×(k+1) minors, not the one on the whole matrix — and on a 10×10 with entries in ±6 the difference is ten bits, with the run reaching 2.7 bits a step against the bound's 3.4.
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Three orders and one last entry
Over the integers there is no stability to pivot for, so a fraction-free elimination swaps rows only when the pivot is zero. Choosing a pivot for length instead does change the sequence of minors — the smallest-nonzero rule makes seven exchanges where the natural order makes none and keeps the profile two bits lower through the middle. It cannot change the peak. The last entry of the elimination is the determinant, and the determinant does not know what order it was computed in.
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The pivot is in every product
A fraction-free elimination's intermediates are minors, and Hadamard bounds a minor by the lengths of the rows it is made of — so the rule that picks the smallest pivot entry looked like a proxy for a rule that picks the shortest row. Measured, the two keep the bit-length profile equally low: 0.969 and 0.966 of the natural order's area. They part on what the arithmetic costs. Counting every multiplication and division at the product of its operands' lengths, the smallest-pivot rule costs 0.70 of the natural order and the shortest-row rule 0.84, because the pivot multiplies every entry of the step and divides every entry of the next. Three per cent of area is thirty per cent of arithmetic.
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What reading the next pivot buys
In a fraction-free elimination every pivot is paid for twice — it multiplies every entry of its own step and divides every entry of the next — and the rule that picks the smallest pivot entry left a median 20 per cent above the least arithmetic any row order reaches. A rule that charges two steps ahead brings the median matrix to within 2.5 per cent, and on one matrix of twenty-four costs 1.87 times the least, worse than the smallest pivot ever does. Charging three steps ahead finds the least of all 40,320 orders on thirteen matrices and is never more than 27 per cent above it. And none of it pays: choosing that way costs forty to two hundred and sixty times the elimination it chooses.
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An order found once knows what the rows know
Searching for a fraction-free elimination's least-cost row order costs forty to two hundred and sixty times the elimination, so it can only pay if one order serves a whole family. The prediction was that it cannot: the best order depends on which rows happen to have short entries, so a reused order should do no better than the natural one. On independent entries and on a shared zero pattern that holds — another member's best order is an ordinary order, at the middle of the 40,320. On a family that shares its row scales it fails completely: another member's order costs a median 1.13 of the least where the natural order costs 1.88. But on every family, including that one, the rule that just takes the smallest pivot entry is cheaper than any reused or trained order — 1.24, 1.08 and 1.01. What an order carries from one matrix to the next is what the family shares — and only when that is a particular matrix, not a visible structure, does an order trained on the family beat the rule.