Table of Contents Solving Systems of Linear Equations - Gaussian Elimination The method of solving a linear system of equations by Gaussian Elimination.

Presentation on theme: "Table of Contents Solving Systems of Linear Equations - Gaussian Elimination The method of solving a linear system of equations by Gaussian Elimination."— Presentation transcript:

Table of Contents 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.

Table of Contents Slide 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.

Table of Contents Slide 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.

Table of Contents Slide 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 R 1  R 2. - 2R 1 + R 2  R 2 Now proceed with elementary row operations to get triangular form.

Table of Contents Slide 5 Solving Systems of Linear Equations - Gaussian Elimination 3R 1 + R 3  R 3 R 2 + R 3  R 3

Table of Contents Slide 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).