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Gaussian Elimination Digital Lesson
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 Gaussian elimination with back-substitution is a procedure for solving linear systems by applying a particular sequence of elementary row operations to the augmented matrix of the system. Definition: Augmented Matrix. A unique augmented matrix is associated with each system of linear equations. System Augmented matrix Elementary row operations are of three types. 1. Interchange two rows of a matrix. 2. Multiply a row of a matrix by a nonzero constant. 3. Add a multiple of one row of a matrix to another.
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 A row of a matrix is a zero row if it consists entirely of zeros. A row which is not a zero row is a nonzero row. A unit column of a matrix is a column with one entry a 1 and all other entries 0. Definition: Zero Rows unit column zero row non-zero row
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 A matrix is in row-echelon form if and only if: 1. All zero rows occur below all nonzero rows. 2. The first nonzero entry of any nonzero row is a one. This is called the leading 1 of the row. 3. For any two nonzero rows, the leading 1 of the higher row is to the left of the leading 1 of the lower row. Definition: Row-echelon form. 1. 2. 3.
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 A matrix is in reduced row-echelon form if and only if: 1. The matrix is in row echelon form. 2. Every column containing a leading one is a unit column. Definition: Reduced Row-echelon form. unit columns leading 1s
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 not a unit column Example: Tell if each matrix is in row-echelon form. A is in row-echelon form. A fails to be in reduced row-echelon form because the second column contains a leading 1, but is not a unit column. B is in both row-echelon form and reduced row-echelon form. C fails to be in row-echelon form because the leading 1 in the second row is to the right of the leading 1 in the third row. Since C is not in row-echelon form, it cannot be in reduced row-echelon form. Example 2: Row-echelon form. leading 1s leading 1
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 To solve a system of linear equations using Gaussian elimination with back substitution, 1. Write the augmented matrix of the system. 2. Use elementary row operations to transform the augmented matrix into row-echelon form. 3.Write the system of equations corresponding to the augmented matrix in row-echelon form 4. Use back substitution to find the solution. Definition:Gaussian elimination with back substitution.
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 –2R 1 + R 3 R1R1 Example : Gaussian elimination with back substitution. Example continued Example: Use Gaussian elimination with back substitution to solve the system
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 Row-echelon form R3R3 Example Continued: Row-echelon form. Example continued
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 The unique solution is (–1, –3, 2).
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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11 Example: Use Gaussian elimination with back substitution to solve the system. The system corresponding to this augmented matrix is: Since the second equation is false, the system has no solutions. Row-echelon form –2R 1 + R 2 Example 2: Gaussian elimination with back substitution.
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