Linear Algebra Solutions 51

Linear Algebra Solutions 51 - i(2-i(1 2 i =-2 i 6 = 0 Hence...

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Section 5.8 1. (i) ( - 3 + i )(14 - 2 i ) = ( - 3)(14 - 2 i ) + i (14 - 2 i ) = { ( - 3)14 - ( - 3)(2 i ) } + i (14) - i (2 i ) = ( - 42 + 6 i ) + (14 i + 2) = - 40 + 20 i. (ii) 2 + 3 i 1 - 4 i = (2 + 3 i )(1 + 4 i ) (1 - 4 i )(1 + 4 i ) = ((2 + 3 i ) + (2 + 3 i )(4 i ) 1 2 + 4 2 = - 10 + 11 i 17 = - 10 17 + 11 17 i. (iii) (1 + 2 i ) 2 1 - i = 1 + 4 i + (2 i ) 2 1 - i = 1 + 4 i - 4 1 - i = - 3 + 4 i 1 - i = ( - 3 + 4 i )(1 + i ) 2 = - 7 + i 2 = - 7 2 + 1 2 i. 2. (i) iz + (2 - 10 i ) z = 3 z + 2 i z ( i + 2 - 10 i - 3) = 2 i = z ( - 1 - 9 i ) = 2 i z = - 2 i 1 + 9 i = - 2 i (1 - 9 i ) 1 + 81 = - 18 - 2 i 82 = - 9 - i 41 . (ii) The coefcient determinant is f f f f 1 + i 2 - i 1 + 2 i 3 + i f f f f = (1 + i )(3 +
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Unformatted text preview: i )-(2-i )(1 + 2 i ) =-2 + i 6 = 0 . Hence Cramer’s rule applies: there is a unique solution given by z = f f f f-3 i 2-i 2 + 2 i 3 + i f f f f-2 + i =-3-11 i-2 + i =-1 + 5 i w = f f f f 1 + i-3 i 1 + 2 i 2 + 2 i f f f f-2 + i =-6 + 7 i-2 + i = 19-8 i 5 . 54...
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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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