EE103 Section 4 cont

EE103 Section 4 cont - EE103 Lecture Notes, Winter 2009,...

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EE103 Lecture Notes, Winter 2009, Prof S. Jacobsen Section 4 74 Example: 1 97641 35894 61998 58741 95292 AP I ⎛⎞ ⎜⎟ =⇒ = ⎝⎠ The IGE pivot in the 1 st column, and the resulting matrices are 11 1 1 1 0000 9 7 6 4 1 1/ 3 1 0 0 0 0 8 / 3 6 23 / 3 11/ 3 , 2/3 0 1 0 0 0 11/3 5 19/3 22/3 5 / 9 0 0 1 0 0 37 / 9 11/ 3 16 / 9 4 / 9 1 0001 0 2 4 5 1 EA E P A == = −− We interchange the 2 nd and 4 th rows using the appropriate permutation matrix 2 P 22 1 97 6 4 1 0000 1 03 7 / 91 1 / 31 6 / 94 / 9 000 0 1 , 01 1 / 351 9 / 3 2 2 / 3 00 1 08 / 3 6 2 3 / 3 1 1 / 3 0 0 1 02 45 1 1 PP A The resulting pivot matrix 2 E and 2 1 1 AE P E P A = are 2 2 1 1 1 0 9 7 6 4 1 0 1 7 / 9 1 1 / 3 1 6 / 9 4 / 9 , 0 33 / 37 1 0 0 0 0 306 / 37 293 / 37 286 / 37 0 24 / 37 0 1 0 0 0 134 / 37 241/ 37 125 / 37 0 18 / 37 0 0 1 0 0 82 / 37 217 / 37 45 / 37 E P E P A = We see that there is no need to interchange rows at this point; i.e., 3 PI = . The resulting IGE 3 E pivot matrix and the matrix 3 A are
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EE103 Lecture Notes, Winter 2009, Prof S. Jacobsen Section 4 75 33 3 3 2 2 1 1 10 0 00 9 7 6 4 1 0 1 0 0 0 0 37 / 9 11/ 3 16 / 9 4 / 9 , 0 0 1 0 0 0 0 306 / 37 293 / 37 286 / 37 0 0 67 /153 1 0 0 0 0 466 /153 1/153 0 0 41/153 0 1 0 0 0 1222 /153 503 /153 EA E P E P E P A ⎛⎞ ⎜⎟ == = −− ⎝⎠
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This note was uploaded on 03/30/2009 for the course EE 103 taught by Professor Vandenberghe,lieven during the Winter '08 term at UCLA.

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EE103 Section 4 cont - EE103 Lecture Notes, Winter 2009,...

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