EGM4313HW6 - p.287, problem 6 01 10 04 Rank 0 4 0 10 01 04...

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Unformatted text preview: p.287, problem 6 01 10 04 Rank 0 4 0 10 01 04 4 0 0 10 01 00 4 24 0 12 8 60 0 0 0 6 6 1 16 7 4 0 0 2 p.287, problem 8 2 4 8 16 16 8 4 2 4 8 16 2 2 16 8 4 2 0 0 0 4 24 0 0 8 60 30 0 2 0 0 0 16 126 30 12 2 0 0 0 4 24 0 0 8 60 0 30 16 126 30 75 16 126 75 30 Rank = 4. p.287, problem 25 6 2 4 0 2 4 1 5 4 3 0 4 6 0 0 3 3 3 6 0 0 0 6 0 13 16 3 96 Therefore linearly independent. Page 300, problem 14 4 2 0 0 7 8 0 0 0 0 1 2 216 0 0 5 2 8 40 0 00 15 22 7 20 0 00 15 22 4 8 2 10 2 7 2 10 Page 308, 5: 1 0 01 0 0 2 1 00 1 0 5 4 10 0 1 100 Hence 2 1 0 541 1 0 01 0 0 010210 041501 1 1 2 3 1 0 01 0102 0 0 13 00 10 41 00 10 41 Page 308, problem 10 2 / 3 1/ 3 2/3 2/3 1/ 3 2 / 3 2 / 3 1/ 3 0 1 0 0 2/3 1 0 0 1/ 3 0 1 0 2/3 0 0 1 2/3 1 11 3/ 2 1 1 1/ 2 0 5 / 6 0 1 0 1/ 3 0 0 1 2/3 2 / 3 1/ 3 2/3 2/3 1/ 3 2 / 3 1/ 3 2/3 1/ 3 2/3 1/ 3 2/3 1 2 / 3 1/ 3 2 / 3 1 00 0 1 1 1 10 0 1/ 2 1 1/ 2 0 1 0 0 1 0 1/ 2 1 2/3 2/3 2/3 2/3 1/ 3 2/3 1 1/ 2 1 3 / 2 0 0 1 11 1 0 0 1 2 / 3 1/ 3 1 0 0 2/3 0 1 0 1/ 3 0 0 1 2/3 2/3 2/3 1/ 3 1/ 3 2/3 2/3 2/3 2/3 1/ 3 0 0 2/3 1/ 3 2/3 2/3 ...
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This note was uploaded on 12/15/2011 for the course EGM 4313 taught by Professor Mei during the Fall '08 term at University of Florida.

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