Solving Linear Systems of Equations Using Gauss

Solving Linear - (Note that if we could not get the 1 by switching the rows we would just multiply Row 1 times 1/3 Step 2 We want a 0 below the

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Solving Linear Systems of Equations Using Gauss-Jordan Elimination Example 1: Solve = - = + 13 4 4 3 y x y x First, we will write the augmented matrix: Our goal is to find a row-equivalent matrix of the form b a 1 0 0 1 . Our strategy will be to go column-by-column, getting the one first and then the zeros in the rest of the column. Remember that there are 3 elementary row operations that we may use: 1) We can interchange any two rows. (Abbreviation: 2 1 R R means switch rows 1 and 2) 2) We can multiply any row times a nonzero constant. (Abbreviation: 2 3 2 R means multiply Row 2 times 2/3) 3) We can add a multiple of one row to another row. (Abbreviation: 2 1 3 R R + - means multiply Row 1 times -3 and add the result to Row 2) Step 1 We want a 1 in the upper left hand corner. The easiest way to accomplish this is to switch rows 1 and 2.
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Unformatted text preview: (Note that if we could not get the 1 by switching the rows we would just multiply Row 1 times 1/3.) Step 2 We want a 0 below the first 1. We will use the 3 rd elementary row operation and multiply Row 1 times –3 and add the result to Row 2. Be careful that you only change the row you are adding to . Step 3 We want our 1 in the second column. We will use the second elementary row operation and multiply Row 2 times 1/7. Note that the resulting 1 is called a “pivot”. Step 4 The only remaining step is to get a 0 above our pivot in the second column. To accomplish this we will again use the third elementary row operation and add 1 times Row 2 to Row 1. Solution: _______________________ (Make sure your solution is written as an ordered pair since it is a point.)...
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This note was uploaded on 01/12/2012 for the course MATH 1204 taught by Professor Duck during the Spring '08 term at NorthWest Arkansas Community College.

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