5.1soln - Solutions to problems from section 5.1 6. Is 1- 2...

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Unformatted text preview: Solutions to problems from section 5.1 6. Is 1- 2 1 an eigenvector of 3 6 7 3 3 7 5 6 5 ? If so, find the eigenvalue. Solution: 3 6 7 3 3 7 5 6 5 1- 2 1 = - 2 4- 2 =- 2 1- 2 1 so 1- 2 1 is an eigenvector of 3 6 7 3 3 7 5 6 5 with eigenvalue- 2. 8. Is λ = 3 and eigenvalue of A = 1 2 2 3- 2 1 1 1 ? If so, find one corresponding eigenvector. Solution: Let’s look at the augmented matrix corresponding to the system ( A- 3 I 3 ) x = : - 2 2 2 3- 5 1 1- 2 NR 1 =- 1 2 R 1-→ 1- 1- 1 0 3- 5 1 1- 2 0 NR 2 = R 2- 3 R 1-→ 1- 1- 1- 2 4 1- 2 NR 2 =- 1 2 R 2-→ 1- 1- 1 0 1- 2 0 1- 2 0 NR 3 = R 3- R 2 NR 1 = R 1 + R 2-→ 1 0- 3 0 0 1- 2 0 0 0 0 0 . Thus there is a nontrivial solution to ( A- 3 I 3 ) x = (i.e. an eigenvector of eigenvalue 3) so 3 is a n eigenvalue. We can also list all 3-eigenvectors using our reduced matrix above as all3 is a n eigenvalue....
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This note was uploaded on 01/28/2010 for the course MATH 307 taught by Professor Axenovich during the Fall '08 term at Iowa State.

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5.1soln - Solutions to problems from section 5.1 6. Is 1- 2...

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