MA232 Mid_Term Review III

MA232 Mid_Term Review III - = & 3(13) + 2(13) = &...

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1. Find the roots to: z 3 = 8 + 8 i Solution: j 8 + 8 i j = p 8 2 + 8 2 = 8 p 2 = r Both real and imaginary parts are positive so: = Re ( z 0 ) where z 0 = 8 + 8 i 8 p 2 = 8 cos = 1 = p 2 = ±= 4 So: z 0 = 8 e 4 i = 8 + 8 i ; z 3 = 8 e 4 i Roots: z k = 8 1 3 e i ( 4 + k 2 ± ) 1 3 for k = 0 ; 1 ; 2 w 0 = 2 e i 3 ( 4 +(0)2 ± ) = 2 e i 12 w 1 = 2 e i 3 ( 4 +(1)2 ± ) = 2 e 3 4 i w 2 = 2 e i 3 ( 4 +(2)2 ± ) = 2 e 17 12 i 2. Put in complex form: 3 e 6 i +4 Solution: 3 e 6 i +4 = 3 e 4 e 6 i = 3 e 4 ( cos ± 6 + isin ± 6 ) = 3 e 4 ( p 3 = 2 + i= 2) = 3 e 4 ( p 3+ i 2 ) 3. Find: det 0 B B @ 1 2 3 4 3 3 2 1 2 3 1 2 3 4 2 1 1 C C A Solution 1
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det 0 B B B @ 1 2 3 4 3 3 2 1 2 3 1 2 3 4 2 1 1 C C C A = det 0 B B B @ 1 2 3 4 0 3 7 11 0 1 5 6 0 2 7 11 1 C C C A = det 0 @ 3 7 11 1 5 6 2 7 11 1 A = 3(55 42) + 1(77 77) 2(42 55)
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Unformatted text preview: = & 3(13) + 2(13) = & 39 + 26 = & 13 4. Find: @ 2 1 3 4 2 1 3 1 2 1 A @ 1 2 3 2 3 1 3 2 1 1 A Answer: @ 13 13 10 11 16 15 11 13 12 1 A 5. Solve for x,y,z: 1 x + & 3 y + 2 z = 6 1 x + 4 y + & 1 z = 4 5 x + 6 y + 1 z = 20 Answer: This is inconsistent and therefore has no solu-tion. 2...
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This note was uploaded on 04/23/2008 for the course MA 232 taught by Professor Toland during the Fall '08 term at Clarkson University .

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MA232 Mid_Term Review III - = & 3(13) + 2(13) = &...

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