Numerical Linear Algebra solutions

Numerical Linear Algebra solutions - Thanks to the great...

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Thanks to the great people at www.mathhelpforum.com and www.physicsforums.com who have helped me solve / check many of these problems 1
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Exercise 1.1 a) 1 - 1 0 0 0 1 0 0 0 - 1 1 0 0 - 1 0 1 E 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 F 1 0 0 0 0 1 0 0 0 0 1 2 0 0 0 0 1 G B 2 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 H 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 I 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 J 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 K E : Subtract row 2 from each of the other rows. F : Add row 3 to row 1. G : Halve row 3. H : Double column 1. I : Interchange columns 1 and 4. J : Replace column 4 by column 3. K : Delete column 1. b) 1 - 1 1 2 0 0 1 0 0 0 - 1 1 2 0 0 - 1 0 1 B 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 2
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Exercise 1.2 f 1 = k 12 ( x 2 - x 1 - l 12 ) f 2 = k 23 ( x 3 - x 2 - l 23 ) - f 1 f 3 = k 34 ( x 4 - x 3 - l 34 ) - ( f 1 + f 2 ) f 4 = f 1 + f 2 + f 3 f 1 = k 12 ( x 2 - x 1 - l 12 ) = k 12 x 2 - k 12 x 1 - k 12 l 12 k 12 l 12 = const. f 2 = k 23 ( x 3 - x 2 - l 23 - k 12 ( x 2 - x 1 - l 12 )) f 2 = x 2 ( - k 12 - k 23 ) + x 1 k 12 + x 3 k 23 - k 23 k 23 + k 12 l 12 - k 23 k 23 + k 12 l 12 = const. f 3 = k 34 ( x 4 - x 3 - l 34 ) - k 23 ( x 3 - x 2 - l 23 ) f 3 = k 23 x 2 + x 3 ( - k 23 - k 34 ) + k 34 x 4 - k 34 l 34 + k 23 l 23 - k 34 l 34 + k 23 l 23 = const. f 4 = k 34 ( x 4 - x 3 - l 34 ) = k 34 x 4 - k 34 x 3 - k 34 l 34 - k 34 l 34 = const. f = Kx + c f 1 f 2 f 3 f 4 = - k 12 k 12 0 0 k 12 ( - k 12 - k 23 ) k 23 0 0 k 23 ( - k 23 - k 34 ) k 34 0 0 - k 34 k 34 x 1 x 2 x 3 x 4 + - k 12 l 12 - k 23 l 23 + k 12 l 12 - k 34 l 34 + k 23 l 23 - k 34 l 34 3
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Exercise 1.3 The columns of R up to and including n span R n . The same is true for I . The columns of I are linear combinations of R . Since the entries of the column vectors of R - 1 are the coefficients of the linear combinations, R - 1 must be upper-triangular.
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Numerical Linear Algebra solutions - Thanks to the great...

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