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Quiz4.EGM4313.Fall2011.solution

# Quiz4.EGM4313.Fall2011.solution - EGM 4313 Mikolaitis Fall...

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EGM 4313 - Mikolaitis NAME:_____________________________ Fall 2011 Sorting Number:______________________ Quiz 4 Open book/Closed notes/No calculators or computers 30 minute time limit Find three linearly independent solutions to the homogeneous system 1 1 1 0 1 1 . 0 0 1 yy Solution: By inspection (triangular matrix) the eigenvalues are 1, 1, and 1. Now find the eigenvector(s) by Gaussian elimination: 0 1 1 0 0 1 000 is already in row echelon form. 1 2 3 0 1 1 0 0 0 1 0 0 0 0 0 has the solution 1 arbitrary and 23 0. Therefore 1 0 0 is an eigenvector and there are no other linearly independent ones. One solution is therefore (1) 1 0 0 t ye and there is a second solution of the form (2) 1 0 0 tt y t e e where 11 2 3 0 1 1 1 0 0 1 0 1 0 0 0 0 0 C with 1 C arbitrary. We will take it as zero hence 10 01 00 y t e e .

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There is a third solution of the form (3) 2 10 1 01 2 00 t t t y t e t e e where 12 2 3 0 1 1 0 0 0 1 1 1 0 0 0 0 1 C where 2 C is arbitrary. We will take it as zero hence (3) 2 1 0 0 1 0 1 1 2 0 0 1 t t t y t e t e e . Of course there are other possible sets of three solutions.
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Quiz4.EGM4313.Fall2011.solution - EGM 4313 Mikolaitis Fall...

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