Homework9 - Homework 9 Solutions 1. ( + 1 i + 2 j + 3 k ) 2...

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Unformatted text preview: Homework 9 Solutions 1. ( + 1 i + 2 j + 3 k ) 2 = ( 2- 2 1- 2 2- 2 3 ) + 2 1 i + 2 2 j + 2 3 k Hence, ( + 1 i + 2 j + 3 k ) 2 =- 1 2- 2 1- 2 2- 2 3 =- 1 and 1 = 2 = 3 = 0 2 1 + 2 2 + 2 3 = 1 (and = 0). Thus there are infinitely many solutions (1 corresponding to each point on the unit sphere). 2. Call the proposed field R , to simplify matters. additive identity = 0 0 0 0 R commutative under addition since R is. closed under addition since a + c b + d- b- d a + c R addition is associative - clear multiplicative identity = 1 0 0 1 R for A R , det ( A ) = a 2 + b 2 = 0 a = b = 0 thus every element has an inverse except the additive identity (as desired). a b- b a c d- d c = ac- bd ad + bc- ad- bc ac- bd = c d- d c a b- b a So commuta- tive under multiplication....
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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.

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Homework9 - Homework 9 Solutions 1. ( + 1 i + 2 j + 3 k ) 2...

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