1. Section 2.2 Echelon Forms Goal: Develop systematic methods for the method of elimination that uses matrices for the solution of linear systems. The methods start with the augmented matrix of the given linear system and obtain a matrix of a certain form. This new matrix represents an equivalent linear system that has exactly the same solutions as the given system but is easier to solve. The task of this section is to manipulate the augmented matrix representing a given linear system into a form from which the solution can easily be found. The manipulations we use on the augmented matrix are called row operations. We proceed to choose row operations to find an equivalent linear system with upper triangular coefficient matrix then apply back substitution. reduce [A | b] to upper triangular form [U | c] The upper triangular form can vary depending on the way the row operations are used.

2. There are a number of strategies involving the use of row operations to get the augmented matrix to upper triangular form. However, there are two so called standard forms, that are frequently used. Definition A matrix is in reduced row echelon form (RREF) if all of the following are true. 1. All zero rows, if there are any, appear at the bottom of the matrix. 2. The first entry of a nonzero row is a 1. This is called a leading 1. 3. For each nonzero row, the leading 1 appears to the right and below any leading 1's in preceding rows. (Staircase downward to the right.) 4. Any column in which a leading 1 appears has zeros in every other entry. A matrix in RREF appears as a staircase ("echelon") pattern of leading 1's descending from the upper left corner of the matrix. The columns containing the leading 1's are columns of an identity matrix. This is reminiscent of the upper triangular form for linear systems.

3. Each of the following are in RREF where x, y, z, or w can be nonzero. Any zeros rows last. Leading ones in “staircase” downward. Columns with leading ones are columns of an identity matrix Definition A matrix is in row echelon form (REF) if properties 1, 2, and 3 in the definition of RREF are satisfied. A matrix is in row echelon form, REF, provided zero rows come last, the nonzero rows have leading 1's, and the leading 1's form a staircase descending to the right. We do not require that the columns containing the leading 1's have zeros above the leading 1's. Each of these matrices are in REF where x, y, z, or w can be nonzero.

4. When these equivalent matrix strategies to solve a linear system Ax = b we are manipulating the augmented matrix [A | b]. In this case we apply row operations to get the coefficient matrix into RREF and REF and then solve the system by back substitution. So the result is [U | c], U in upper triangular form. From the definitions we see that any matrix in RREF is also in REF. The RREF is unique, but the REF of a matrix and a general upper triangular form are not unique. Determine which of the following matrices are in general upper triangular form, REF, RREF or none of these forms.

5. Any of the three forms, general upper triangular, REF, or RREF provide the same information about the solution set of a linear system. In particular, we can determine whether the linear system is consistent or inconsistent and for a consistent system whether there is a unique solution or infinitely many solutions. Determine which of the following linear systems is consistent and for the consistent systems determine whether or not there is a unique solution.

6. Computational procedure for RREF As we proceed we will want to determine certain entries of the matrix called pivots. The pivots are used to construct row operations in order to generate zeros in the pivotal column below the pivot. (Pivots must be nonzero in order to perform such steps.) Strategy for RREF: To get to RREF we use a two step procedure. Forward Sweep: determine the pivots; choose a row operation to make a pivot a 1; choose row operations to eliminate below a pivot. Backward Sweep: Starting with the last pivot, choose row operations to eliminate above the pivots. Example: Use the two step approach to determine the RREF of the augmented matrix of a linear system Ax = b shown below.

7. Determine the first pivot: The (1,1)-entry of the matrix is zero so we will do a row interchange to obtain a nonzero value in the (1,1)-entry. The interchange could involve any of the remaining three rows. Here we choose Row(1) ↔ Row(2). We obtain the matrix Now we use the (1,1)-entry as a pivot, but we first apply row operation (-1/2)R 1 to make the pivot a 1. We obtain First pivot

8. We next choose row operations to eliminate below the pivot. We begin by eliminating the (3,1) entry. We are now ready to select the second pivot. Since the current (2,2) entry is zero, we need to use another row interchange.

9. Continuing with the forward sweep we eliminate below the second pivot.

10. It follows that the third pivot is the (3,3) entry and it is already a 1. From the current row equivalent matrix we see that all the pivots are 1 and only zeros appear below them. Hence we are done with the forward sweep.

11. We continue to eliminate above the last pivot in the following matrix.

12. Observe that this last matrix has zeros below and above each pivot. The pivots are all 1 and are staircase downward to the right. All zeros rows come last. Summary: the reduced row echelon form of matrix is the matrix Thus the solution of the linear system Ax = b is

14. Using row operations and echelon forms to solve linear systems. Theorem: Every m × n matrix is row equivalent to a matrix in reduced row echelon form. Strategy: 1. For linear system Ax = b, form augmented matrix [A | b]. 2. Compute the RREF of [A | b], call it [U | c]. 3. Solve system Ux = c; its solution will be the same as the solution of Ax = b. (We use back substitution to solve Ux = c.) SOLVE:

17. Application: Street Networks The central business area of many large cities is a network of one way streets. Any repairs to these thoroughfares, closings for emergency or civic functions, or even traffic accidents disrupts the normal flow of traffic. In addition many cities do not permit parking on such busy streets during working hours. The daily flow of traffic through these downtown street networks can be studied by local governments to assist with road repairs, city planning, and the planning of emergency evacuation routes. For one way street networks there is a simple rule: vehicles entering an intersection from a street must also exit the intersection by another street. So for each intersection we have an equilibrium equation or, put simply, an input-equals-output equation. Thus after some data collection involving entry and exit volumes at intersections, a city traffic commission can construct network models for traffic flow patterns.

18. Example The city traffic commission has collected data for vehicles entering and exiting the network shown in the figure. The direction of the traffic flow between intersections is indicated by the arrows. The intersections are labeled A through D and the average number of vehicles per hour on portions of the streets is indicated by x 1 through x 5. The average number of vehicles that enter or exit on a street appears near the street. For instance, at intersection A, 300 vehicles exit and at intersection C a total of x 3 + x 4 + 200 vehicles enter. In the table we develop an input-equals-output equation for each intersection. The resulting linear system models the traffic flow for the street network shown in the figure.

19. From this set of equations we construct the augmented matrix and compute the reduced row echelon form. Solving for the unknowns corresponding to the leading 1's we find that x 4 and x 5 are free variables; Since each of the traffic flows x i  0, we have that x 1 = x 4 - 100  0 so x 4  100 and x 2 = x 5 - 200  0 so x 5  200. If the streets that carry traffic flows x 4 and x 5 cannot accommodate this level of traffic per hour or had to be closed then the road network might experience gridlock. This information can be useful for a traffic commission's planning efforts.