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