CSCI 2033 P3.pdf - LINEAR EQATIONS[1.1(CONTINUED 3-1...

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LINEAR EQATIONS [1.1] + (CONTINUED) 3-1
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Gaussian Elimination Back to arbitrary linear systems. Principle of the method: Since triangular systems are easy to solve, we will transform a linear system into one that is triangular. Main operation: combine rows so that zeros appear in the required locations to make the system triangular. Recall Notation: Augmented form of a system 2 x 1 + 4 x 2 + 4 x 3 = 2 x 1 + 3 x 2 + 1 x 3 = 1 x 1 + 5 x 2 + 6 x 3 = - 6 Notation: 2 4 4 2 1 3 1 1 1 5 6 - 6 Main operation used: scaling and adding rows. 3-2 Text: 1.1 – Gauss 3-2
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Examples of such operations. Example: Replace row 2 by: row 2 + row 1: 2 4 4 2 1 3 1 1 1 5 6 - 6 2 4 4 2 3 7 5 3 1 5 6 - 6 Example: Replace row 3 by: 2 times row 3 - row 1: 2 4 4 2 3 7 5 3 1 5 6 - 6 2 4 4 2 3 7 5 3 0 6 8 - 14 Example: Replace row 1 by: (0.5 * row 1) 2 4 4 2 3 7 5 3 0 6 8 - 14 1 2 2 1 3 7 5 3 0 6 8 - 14 3-3 Text: 1.1 – Gauss 3-3
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Gaussian Elimination (cont.) Go back to original system. Step 1 must eliminate x 1 from equations 2 and 3, i.e., It must transform: 2 4 4 2 1 3 1 1 1 5 6 - 6 into: * * * * 0 * * * 0 * * * row 2 := row 2 - 1 2 × row 1 : row 3 := row 3 - 1 2 × row 1 : 2 4 4 2 0 1 - 1 0 1 5 6 - 6 2 4 4 2 0 1 - 1 0 0 3 4 - 7 3-4 Text: 1.1 – Gauss 3-4
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Step 2 must now transform: 2 4 4 2 0 1 - 1 0 0 3 4 - 7 into: * * * * 0 * * * 0 0 * * row 3 := row 3 - 3 × row 2 : 2 4 4 2 0 1 - 1 0 0 0 7 - 7 System is now triangular 2 x 1 + 4 x 2 + 4 x 3 = 2 x 2 - x 3 = 0 7 x 3 = - 7 Solve -
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