2018assg2_sol

2018assg2_sol - ERG 2018 Assignment 2 Solution 1(a) Using...

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ERG 2018 Assignment 2 Solution 1(a) Using Gauss-Jordan procedure on 1 | AI -   , 2 1 31 3 1 2 111 10 0 111 1 0 0 111 1 00 112 01 0 00 1 11 0 1 110 1 1 1 00 1 1 1 1 0 0 1 0 2 0 101 R R RR R -- -    - -  → 1 2 12 1 10 0 0 1 20 1 1 1 0 010 1 2 0 1 2 1 2 0 1 2 0 1 2 1 1 1 0 1 1 10 R R ↔- - - →- - 13 100 3 2 1 1 1 1 - - - 3 1 2 0 1 A - =- - (b) Let i E be the elementary matrix corresponding the th i row operations in (a), then E1 0 01 , E2 0 10 = - , E3 001 = , E4 0 1 - , E5 1 - = , E6 1 01 - = By the row operations in (a), 6 5 1 6 5 1 6 51 [ |] [ E E EA I E E E A E EE = L LL [|] IB = 1 6 5 1 6 E E E AI B E EEA - =⇔ == Thus A can be calculated using Gauss-Jordan procedure. (c) ( 29 1 1 1 1 6 5 1 6 5 1 1 26 E E E AIA E E E E - - - - L ( 29( 29 1 111 3 456 E EEE - --- = Notice that 123 10010010 0 11001000 1 00110101 0 and 10 011010 1 0 2001001 0 0 20 00 100100 1 = - , which is upper triangular. Let P EEE = and U EEE = , then 1 A PU - = = 100111 1010 20 110001 - , P is not a lower triangular matrix.
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2(a) 12 (1)(0 ) ( 2)(1 ) (1) ( 1 ) (3)(2 )3 uu = + - + - += vv 3 3 10 cos 10 141911 4 1 56 q = = == + + + ++ q = 1 10 cos 10 - = 71.56505 (in degree) (b) Define the angle q between u v and v v in n R be such that cos uv q = , 2 ( ) () u v uvu v u u v v u v vu - = - - =+-- uuuuv vvv v v v v v vv g gggg 2 2 22 2 2 cos u v u v q =+- =+- v v v v g Hence the Cosine law is valid in n R . Let U be the angle between v v and vu - , V be the angle between u v and - , 2 2 si n si n ( 1 co s ) ( 1 co s) v U u V v U uV - = - -- u uuv u  - •-  =- -+ --  uuuu v v ( 29( 29 2 2 v u u  - - - - - •-   = - v v v v Notice that 2 u u u - v v v v 2 ( ) ( ) u v vvuuu vvvuuuv = - - - •-+• - •--•- v
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This note was uploaded on 11/13/2010 for the course SEEM SEEM2018 taught by Professor Chan during the Spring '10 term at CUHK.

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2018assg2_sol - ERG 2018 Assignment 2 Solution 1(a) Using...

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