HW3 - 9.8 (a) The system is first expressed as an augmented...

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9.8 (a) The system is first expressed as an augmented matrix: 5 . 21 5 1 1 5 . 61 2 6 3 27 1 2 10 Forward elimination: a 21 is eliminated by multiplying row 1 by –3/10 and subtracting the result from row 2. a 31 is eliminated by multiplying row 1 by 1/10 and subtracting the result from row 3. 2 . 24 1 . 5 8 . 0 0 4 . 53 7 . 1 4 . 5 0 27 1 2 10 a 32 is eliminated by multiplying row 2 by 0.8/(–5.4) and subtracting the result from row 3. 1111 . 32 351852 . 5 0 0 4 . 53 7 . 1 4 . 5 0 27 1 2 10 Back substitution: 6 351852 . 5 1111 . 32 3 = = x 8 4 . 5 ) 6 ( 7 . 1 4 . 53 2 = = x 5 . 0 10 ) 8 ( 2 ) 6 )( 1 ( 27 1 = = x (b) Check: 5 . 21 ) 6 ( 5 8 5 . 0 5 . 61 ) 6 ( 2 ) 8 ( 6 ) 5 . 0 ( 3 27 ) 6 ( ) 8 ( 2 ) 5 . 0 ( 10 = + + = + = + 10.2 (a) The coefficient a 21 is eliminated by multiplying row 1 by f 21 = –3/10 = –0.3 and subtracting the result from row 2. a 31 is eliminated by multiplying row 1 by f 31 = 1/10 = 0.1 and subtracting the result from row 3. The factors f 21 and f 31 can be stored in a 21 and a 31 . 1 . 5 8 . 0 1 . 0 7 . 1 4 . 5 3 . 0 1 2 10 a 32 is eliminated by multiplying row 2 by f 32 = 0.8/(–5.4) = –0.14815 and subtracting the result from row 3. The factor f 32 can be stored in a 32 .
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351852 . 5 14815 . 0 1 . 0 7 . 1 4 . 5 3 . 0 1 2 10 Therefore, the LU decomposition is = 1 14815 . 0 1 . 0 0 1 3 . 0 0 0 1 ] [ L = 351852 . 5 0 0 7 . 1 4 . 5 0 1 2 10 ] [ U These two matrices can be multiplied to yield the original system. For example, using MATLAB to perform the multiplication gives >> L=[1 0 0;-.3 1 0;0.1 -.14815 1]; >> U=[10 2 -1;0 -5.4 1.7;0 0 5.351852]; >> L*U ans = 10.0000 2.0000 -1.0000 -3.0000 -6.0000 2.0000 1.0000 1.0000 5.0000 (b) Forward substitution: [ L ]{ D } = { B } = 5 . 21 5 . 61 27 1 14815 . 0 1 . 0 0 1
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This note was uploaded on 01/20/2011 for the course ENG EAD 115 taught by Professor Rocke during the Spring '10 term at UC Davis.

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HW3 - 9.8 (a) The system is first expressed as an augmented...

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