ADJUSTMENT BY LEAST SQUARES 171 
Multiplying by 52, 15, 17, 15, 8 and 8 respectively and adding, we 
get 
161M — 17M — 15Mo3 — 8(me20 + m3 + M22) = 161Dao, (13) 
where 
161Do = 52d + 17d + 15(doo + doz) + 8(dio + diz) 
and moi’ = Du gives a first approximation to mo. It should be noted 
that all the rectangles whose correlates are thus eliminated are those 
which are laterally and diagonally once-removed from the key rec- 
tangle o1, whilst the residual correlates mz0, moi, me2, mos and mig are 
the ones adjacent externally to those eliminated. 
Similarly for o2 we should get 
TOLMo2 — 17M — 15(moo + mos) — 8( M10 + mar + Moz + ms) 
= 52d2 + 17di2 + 15(do + dos) + 8(diu + dis), 
and so on for. any other boundary rectangle not a corner. 
The drawback to the above series for moi, mos, - - - is that the 
coefficients involved are not so simple as in the case of equations (11) 
and (12). There is some advantage in using for these rectangles the 
same multipliers, namely 7, 2 and 1 for the key equation, for all first- 
remove equations and for common second-remove equations, as were 
used in equation (12). 
These give the equation 
22m. — my — 2(ma + mos) — (Meo + mis + mez) 
= 7d + 2(doo + du + doz) + dio + diz = 22D’, (14) 
in which my, is le{t as an additional residual to the previous series; but 
the series of residuals left for m’’, after we have written 
vee ’ ” ! 
mi = mi + mai + --- and mo = Da’, 
is very easy to use for further approximation. 
For the inner rectangles, such as 11, 12, 21, 22, - - - , which have 
four first-remove and four common second-remove rectangles, to- 
gether forming a rectangle of three by three cells, we use multipliers 
6, 2 and 1 respectively for each order of remoteness and obtain 
20M, — 2(mi3 + ma) — (mos + M23 + ms. + mao) 
= 6dut2(da + diz + da + dio) + (doo + doz + doz + deo), 
20M. — 2(m10 + ms2 + mus) — (moo + mos + mau + m3 + Mar + Mao) 
= 6dy + 2(du + doe + dis + doz) + (dor + dig + doz + dai), 
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