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9.3 - Math 2940 Solutions Fall 2011 Section 9.3 3 I A = 1 1...

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Math 2940 Solutions, Fall 2011 Section 9 . 3 3 ) I - A = - 1 1 1 - 1 has - 2 as an eigenvector. Since | - 2 | > 1 the iteration diverges. 7 ) In Gauss-Seidel S - 1 T = 3 0 - 1 3 - 1 0 1 0 0 = 1 / 3 0 1 / 9 1 / 3 0 1 0 0 = 0 1 / 3 0 1 / 9 . Here | λ max | = 1 / 9. For Jacobi, S - 1 T = 3 0 0 3 - 1 0 1 1 0 = 1 / 3 0 1 / 3 0 so | λ max | = 1 / 3. 16 ) u 1 = 2 - 1 - 1 2 1 0 = 2 - 1 . Then u 2 = 2 - 1 - 1 2 2 - 1 = 5 - 4 and u 3 = 2 - 1 - 1 2 5 - 4 = 14 - 13 . The eigenvalues of our matrix are 1 and 3. An eigenvector for the larger eigenvalue is
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