Unformatted text preview: 2. If the top entry of the column found in step 1 is zero, swap the top row with another row so that the top entry in this column is nonzero. 3. Multiply the top row by a scalar so that the entry in the column from step 1 is 1. 4. Eliminate all other entries in this column by adding a suitable multiple of the top row to the other rows. 5. Move one row down and one column to the right from the row and column that we focused on in the previous steps. If the entry in this new row and new column is zero and all entries below it are also zero then move one column to the right (but keep the same row). Repeat this step as necessary. 6. Return to step 1. The algorithm stops when we run out of rows or columns. Next two examples show how to apply GaussJordan elimination to ±nd the reducedrowechelon form of a matrix. 3...
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 Spring '10
 STAFF
 Linear Algebra, Algebra, Multiplication, GaussJordan Elimination, Row

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