Unformatted text preview: 35. Chapter 1 ISM: Linear Algebra This leaves the possible solutions :3 = s = [1,5,1ﬂ and 15, and we can compute the
corresponding values of 1'1 = 40+ %s and $2 = ﬁll — 3&5 in each case. 4] 53 66 T9
Soweﬁndthe following solutions: ﬁll , 42 , 24 and ﬁ . ﬂ 5 ll] 15 . This problem is similar in nature to Exercise ﬁll, and we will follow that example, reveal— ‘ 3 2
ing the matrix: [1 1 1 HID] . We reduce this to 1 I] _§ ‘ —1III , which
3 2 % int) n 1 g 3 2m
gag — 1][l
yields solutions of the form —%33 + ﬂﬂl] . Since all the values must be positive [there
3:! are at least one man, one woman and one child}, we see that 36 4:: 1'3 at El], and 33 must
2 5 S 11 14
he even. From this, we use I; to ﬁnd our solutions: 3] , 25 , 2D , 15 , 1D
68 7D 72 7'4 751
17
and 5
’53 Rather than setting up a huge system, here we will reason this out logically. Since there are 311] barrels, each son will get 111] of them. If we use the content of a full barrel as our
unit for wine, we see that each brother will get % = 5 barrel—fulls of wine. Thus, the
ten barrels received by each son will, on average, be half full, meaning that for every full barrel a son receives, he also receives an empty one. Now let 31, 1'2, and 33 be the numbers of half—full barrels received by each of the three
sons. The ﬁrst son, receiving $1 half—full barrels will also gain 1U — 1:1 other barrels, half . . ~ 111—:
of which must be full and half of which must be empty, each equal to the quanth —2'~ Thus, 31 must be even. The same works for $2 and 1'3. Since 1'1 + $2 + 1‘3 = II], we have
boiled down our problem to simply ﬁnding ﬁsts of three non—negative even numbers that
add up to 10. We ﬁnd our solutions by inspection: [Elli]=El*Ell§l=llllElEl=Ell§l*
lllllllll=ElEl*lll*llllElllllllMlll As we stated before, the number of full and empty barrels is dependent on the number
of half—full barrels. Thus, each solution here translates into exactly one solution for the 34 ...
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 Spring '09
 Derksen
 Algebra

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