Table of Contents Solving Linear Systems - Elementary Row Operations A linear system of equations can be solved in a new way by using an augmented matrix and what is called elementary row operations. Consider the system at the right... Recall the various procedures for solving a system by the addition method. (1) Switch two rows, and the solution is unchanged...

Table of Contents Slide 2 Solving Linear Systems - Elementary Row Operations (2) Multiply any row by a nonzero real number and the solution is unchanged. In this case, multiply row #1 by 2... (3) Add the nonzero multiple of a row to any other row and the solution is unchanged. Adding the two rows yields...

Table of Contents Slide 3 Solving Linear Systems - Elementary Row Operations When solving a system by the addition method, we usually just write equation #2. Equation #1 was included to help illustrate later methods. Using equation #2 we have... Substituting in either of the original equations yields...

Table of Contents Slide 4 Solving Linear Systems - Elementary Row Operations In solving the previous system, three operations were used. These operations will now be used in conjunction with augmented matrices and will be referred to as... ELEMENTARY ROW OPERATIONS (1) Switch any two rows (2) Multiply any row by a nonzero number. (3) Add a nonzero multiple of any row to any other row. Using any combination of these elementary row operations results in a system of equations that is equivalent to the original system.

Table of Contents Slide 5 Solving Linear Systems - Elementary Row Operations Example: Consider the system of equations given by... It is easily verified that the solution to the system is (3, 1), or x = 3, y = 1 Write the augmented matrix that represents the system...

Table of Contents Slide 6 Solving Linear Systems - Elementary Row Operations The following operations are not intended to solve the system, but rather to illustrate the elementary row operations. Switch row #1 and row #2... The solution to the original system was (3, 1). The solution to the revised system is still (3, 1)...

Table of Contents Slide 7 Solving Linear Systems - Elementary Row Operations Thus, using the first elementary row operation did not change the solution to the system. This is true of all three of the elementary row operations. Multiply row #2 by (-3)... Note that (3, 1) is still a solution to row #2...

Table of Contents Slide 8 Solving Linear Systems - Elementary Row Operations Now for the third elementary row operation. Multiply row #1 by 2, add the result to row #2, and put this result in the place of row #2. Note - this operation will be done a column at a time.

Table of Contents Slide 9 Solving Linear Systems - Elementary Row Operations Now for the third elementary row operation. Multiply row #1 by 2, add the result to row #2, and put this result in the place of row #2. Note - this operation will be done a column at a time.

Table of Contents Slide 10 Solving Linear Systems - Elementary Row Operations Now for the third elementary row operation. Multiply row #1 by 2, add the result to row #2, and put this result in the place of row #2. Note - this operation will be done a column at a time.

Table of Contents Slide 11 Solving Linear Systems - Elementary Row Operations Once again note that the original solution (3, 1) is still a solution...

Table of Contents Slide 12 Solving Linear Systems - Elementary Row Operations (1) Switch any two rows (2) Multiply any row by a nonzero number. (3) Add a nonzero multiple of any row to any other row. Using any combination of the elementary row operations... will always result in an equivalent system. This fact will be used later to solve linear systems of equations.

Table of Contents Solving Linear Systems - Elementary Row Operations