# Table of Contents Solving Linear Systems of Equations - Triangular Form Consider the following system of equations... The system is easily solved by starting.

## Presentation on theme: "Table of Contents Solving Linear Systems of Equations - Triangular Form Consider the following system of equations... The system is easily solved by starting."— Presentation transcript:

Table of Contents Solving Linear Systems of Equations - Triangular Form Consider the following system of equations... The system is easily solved by starting with equation #3 and solving for z... Then use equation #2 and z = -1 to solve for y...  

Table of Contents Slide 2 Solving Linear Systems of Equations - Triangular Form Finally, use equation #1, y = 2 and z = -1 to solve for x...  Thus, the solution to the system is (1, 2, -1). The process just used is called back substitution.

Table of Contents Slide 3 Solving Linear Systems of Equations - Triangular Form Now consider the following augmented matrix representing a system of linear equations... Note that it is the same system used earlier...

Table of Contents Slide 4 Solving Linear Systems of Equations - Triangular Form If the goal was to solve the system represented by the matrix, we would proceed as before. Write equation #3 as... and solve... Write equation #2 as... and solve using z = -1... Write equation #1 as... and solve using y = 2, z = -2...

Table of Contents Slide 5 Solving Linear Systems of Equations - Triangular Form The augmented matrix at the right is considered to be in triangular form. The letters a - f represent real numbers. Along the diagonal, all entries are 1’s. The bottom left corner forms a triangle of 0’s. While the 0’s are essential, the author feels that the diagonal of 1’s is not necessary. Often to get the 1’s, fractions are introduced.

Table of Contents Slide 6 Solving Linear Systems of Equations - Triangular Form Example: Use the augmented matrix at the right to solve the system. Note that the matrix is in triangular form (not considering the diagonal). Solve the system using back substitution as before.