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Ch.1 Solutions Pg31 - SD 61 ISM Linear Algebra Section 1.2...

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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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