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Solving Systems of Linear Equations - Gaussian Elimination

Solving Systems of Linear Equations - Gaussian Elimination. The method of solving a linear system of equations by Gaussian Elimination is outlined below:. (1) Given a system of linear equations, write the corresponding augmented matrix . .

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Solving Systems of Linear Equations - Gaussian Elimination

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  1. Solving Systems of Linear Equations - Gaussian Elimination • The method of solving a linear system of equations by Gaussian Elimination is outlined below: (1) Given a system of linear equations, write the corresponding augmented matrix. (2) Use elementary row operations to write the matrix in triangular form. (3) Use back substitution to solve for the variables. • All of the above concepts should be familiar. This presentation shows how to use the elementary row operations to write the matrix in triangular form.

  2. Solving Systems of Linear Equations - Gaussian Elimination • The following notation will be used with the elementary row operations: • Switch rows 1 and 3. • Multiply row 2 by - 3 and • put the result in place of • row 2. • Multiply row 1 by 5, add • the result to row 2, and • put this result in place of • row 2. Slide 2

  3. Solving Systems of Linear Equations - Gaussian Elimination • Example: • Solve the following system • of equations by the method • of Gaussian Elimination. • Write the augmented matrix. Slide 3

  4. Solving Systems of Linear Equations - Gaussian Elimination • First switch rows 1 and 2 to get the row 1 column 1 entry to be 1. • This is denoted as R1  R2. • Now proceed with elementary row operations to get triangular form. - 2R1 + R2 R2 Slide 4

  5. Solving Systems of Linear Equations - Gaussian Elimination 3R1 + R3 R3 R2 + R3 R3 Slide 5

  6. Solving Systems of Linear Equations - Gaussian Elimination • Note the triangle of 0’s in the • lower left hand corner. Since the • diagonal (1, 5, 4) is not all 1’s, the • matrix is not in true triangular form. • As mentioned in an earlier presentation • on triangular form, this is actually easier • to complete the solution. • Using back substitution the solution to the system is found to be (3, 1, 2). Slide 6

  7. Solving Systems of Linear Equations - Gaussian Elimination END OF PRESENTATION Click to rerun the slideshow.

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