MATH150sol4

# MATH150sol4 - Solution of Assignment 4 1(a Solve det For 3...

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Unformatted text preview: Solution of Assignment 4 1. (a) Solve det For 3 2 2 2 0. Then we have 1 . For 2 1 2 1, 2. 1, eigenvector is 3 4 1 2 2, eigenvector is 1 2. (b) Solve det For 1 2 , the eigenvector is 1 0. Then we have 1 1 2 . (1 mark) 1 2 . So the solution is 1 . The general solution is – sin 2 . (1 mark) sin 2 cos 2 1 . So the general solution 1 2. Solve det For 5 2, eigenvector is 3 1 0. Then we have 1 . For 3 1 3 cos 2 sin 2 cos 2 4, eigenvector is 2, 4. is After substituting the initial value, we have 3. (a) eigenvalues are (b) attractant : Node : 0; repellant: √ . (1 mark) √20. 0 1 . (1 mark) 1 1 1 . (1 mark) 3 1 Improper node : √20. Saddle point: impossible. Spiral point: √20 2 marks √20 and √20 . 0 0 0 0. Then 0 0 . . (1mark) 4. In this question, we would like to solve First, solve det For 0 0, eigenvector is The eigenvector associated with So the general solution is where . (1mark) 1 1 1 . 1 0 is . cos , ...
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