Unformatted text preview: SD. 61. ISM: Linear Algebra Section 1.2 111 c o [1 _%§. D
[1111 o e _%E D
tDI'EVEEI [l U 1'] [l _%E D.ThﬂS,A=%§W,B=%W,C—E=%W,D=
u a 11111 —%E D
u a u o 1 —£5 D %WandE=%W. Ifwe choose 721 to he the depth of the well, then A = 265, s = 191, c = 1413,19 = 129
and s = are. We let 31,32 and 1:3 be the numbers of roosters, hens and chicks respectively. Then,
since we buy,r a total of a hundred birds, and spend a hundred (boins on them, we ﬁnd the equations 1'1 + $2 + $3 = 10'] and 5:31 + 3172 + %1.'3 = lﬂl]. We ﬁt these into our matrix, [1 1 1 1GB] 1
5 3 % ' lﬂﬂ
_£ : _
which redqu to 1 I] 3 ' lﬂl] _
I] 1 g 5 2m
30* $1 _ i333 = ‘10”: and 1‘2 + git'3 = EDD. Now, we can write our solution vectors in 1‘1 %1‘3 — ill]
terms of 2:3: $2 = —%3:3 + 2111] . Since all of our values must be non—negative, 31
5'33 1‘3 must be greater than or equal to zero, or %2:3 — 1m 3 I], which means that I; 3 75. Also, 33 must he greater than or equal to zero, meaning that —%$3+ EDD 3 ID or 1'3 5 %.
Since 1'3 must he an integer, this forces 33 E 85. Thus, we are looking for solutions where 75 E 33 E 85. We notice, however, that $1 and 32 are only integers when I; is a multiple
of 3. Thus, the possible values for era are ?5, YB, 31 and 34. roosters [l 4 S 12
Now the possible solutions for hens are 25 , IS , 11 , and 4 ,
chicks 75 TS Si 84 We let 31,32,113 and It; be the numbers of pigeons, sarasabirds, swans and peacocks
respectively. We ﬁrst determine the cost of each bird. Each pigeon costs g panas, each
sarasabird costs % panas, the swans cost % panas apiece and each peacock costs 3 panas. We use these numbers to set up our system, but we must remember to make sure we are 31 ...
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 Spring '09
 Derksen
 Algebra

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