solq1 - f(1 = a b c = 2 f(2 = a 2 b 4 c =-1 f(3 = a 3 b 9 c...

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Solutions to quiz # 1 (September 9) 1. Eliminating variables, we obtain x + y + z = 1 x + 2 y + z = 2 3 x + 4 y + kz = 4 = x + y + z = 1 y = 1 y + ( k - 3) z = 1 = x + y + z = 1 y = 1 ( k - 3) z = 0 . If k 6 = 3 then we must have z = 0, y = 1 and x = 0, so the system has one solution. If k = 3 then z can be any number, y = 1 and x = - z , so the system has infinitely many solutions. Answer: The system has a unique solution if k 6 = 3 and infinitely many solutions if k = 3. 2. Substituting t = 1 , 2 , 3, we obtain the equations
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Unformatted text preview: f (1) = a + b + c = 2 f (2) = a + 2 b + 4 c =-1 f (3) = a + 3 b + 9 c =-8 . Solving the system, we get a + b + c = 2 a + 2 b + 4 c =-1 a + 3 b + 9 c =-8 = ⇒ a + b + c = 2 b + 3 c =-3 2 b + 8 c =-10 = ⇒ a + b + c = 2 b + 3 c =-3 2 c =-4 a + b = 4 b = 3 c =-2 = ⇒ a = 1 b = 3 c =-2 . Answer: f ( t ) = 1 + 3 t-2 t 2 . 1...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.

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