Linear Algebra Solutions 44

# Linear Algebra Solutions 44 - 2-3 a 12-5 b-14 → 1-2 b 3...

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Finally we have to determine the solution for which x 2 + y 2 + z 2 is least. x 2 + y 2 + z 2 = (5 z ) 2 + (1 - 4 z ) 2 + z 2 = 42 z 2 - 8 z + 1 = 42( z 2 - 4 21 z + 1 42 ) = 42 ( µ z - 2 21 2 + 1 42 - µ 2 21 2 ) = 42 ( µ z - 2 21 2 + 13 882 ) . We see that the least value of x 2 + y 2 + z 2 is 42 × 13 882 = 13 21 and this occurs when z = 2 / 21, with corresponding values x = 10 / 21 and y = 1 - 4 × 2 21 = 13 / 21. 9. Let Δ = 1 - 2 b a 0 2 5 2 0 f f f f f f be the coe±cient determinant of the given system. Then expanding along column 2 gives Δ = 2 f f f f a 2 5 0 f f f f - 2 f f f f 1 b a 2 f f f f = - 20 - 2(2 - ab ) = 2 ab - 24 = 2( ab - 12) . Hence Δ = 0 if and only if ab = 12. Hence if ab 6 = 12, the given system has a unique solution. If ab = 12 we must argue with care: AM = 1 - 2 b 3 a 0 2 2 5 2 0 1 1 - 2 b 3 0 2 a 2 - ab
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Unformatted text preview: 2-3 a 12-5 b-14 → 1-2 b 3 1-5 b 12-7 6 2 a 2-ab 2-3 a → 1-2 b 3 1-5 b 12-7 6 12-ab 6 6-2 a 3 = 1-2 b 3 1-5 b 12-7 6 6-2 a 3 = B. Hence if 6-2 a 6 = 0, i.e. a 6 = 3, the system has no solution. If a = 3 (and hence b = 4), then B = 1-2 4 3 1-5 3-7 6 → 1-2 / 3 2 / 3 1-5 3-7 6 . 47...
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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