MA307 Solution to Assignment 6 - Mathematics 307 Solution to Assignment 6 1(a Use Gauss-Jordan elimination to nd the inverse of 2 3 2 8 1 3 2 6 A= 5 1 3

MA307 Solution to Assignment 6 - Mathematics 307 Solution...

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Mathematics 307 — Solution to Assignment 6 1.(a) Use Gauss-Jordan elimination to find the inverse ofA=-232813-265-139238-1.(b) Compute the condition number ofArelative tok · k1.(c) Compute the condition number ofArelative tok · k.(d) Compute the condition number ofArelative tok · kf. Solution: (a) Eliminating and normalizing on the larger augmented matrix gives - 2 3 2 8 | 1 0 0 0 1 3 - 2 6 | 0 1 0 0 5 - 1 3 9 | 0 0 1 0 2 3 8 - 1 | 0 0 0 1 - 2 3 2 8 | 1 0 0 0 0 9 2 - 1 10 | 1 2 1 0 0 0 13 2 8 29 | 5 2 0 1 0 0 6 10 7 | 1 0 0 1 - 2 0 8 3 4 3 | 2 3 - 2 3 0 0 0 9 2 - 1 10 | 1 2 1 0 0 0 0 85 9 131 9 | 16 9 - 13 9 1 0 0 0 34 3 - 19 3 | 1 3 - 4 3 0 1 - 2 0 0 - 236 85 | 14 85 - 22 85 - 24 85 0 0 9 2 0 981 85 | 117 170 72 85 9 85 0 0 0 85 9 131 9 | 16 9 - 13 9 1 0 0 0 0 - 119 5 | - 9 5 2 5 - 6 5 1 - 2 0 0 0 | 758 2023 - 618 2023 - 288 2023 - 236 2023 0 9 2 0 0 | - 747 4046 2106 2023 - 963 2023 981 2023 0 0 85 9 0 | 725 1071 - 1285 1071 95 357 655 1071 0 0 0 - 119 5 | - 9 5 2 5 - 6 5 1 1 0 0 0 | - 379 2023 309 2023 144 2023 118 2023 0 1 0 0 | - 83 2023 468 2023 - 214 2023 218 2023 0 0 1 0 | 145 2023 - 257 2023 57 2023 131 2023 0 0 0 1 | 9 119 - 2 119 6 119 - 5 119 . 1
Therefore, A - 1 = - 379 2023 309 2023 144 2023 118 2023 - 83 2023 468 2023 - 214 2023 218 2023 145 2023 - 257 2023 57 2023 131 2023 9 119 - 2 119 6 119 - 5 119 .

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