1. How To Find The Reduced Row Echelon Form

2. Reduced Row Echelon Form A matrix is said to be in reduced row echelon form provided it satisfies the following properties:

3. All zero rows, if there are any, appear as bottom rows.

4. The first nonzero entry in a nonzero row is a 1; it is called a leading 1.

5. For each nonzero row, the leading 1 appears to the right and below any leading 1s in preceding rows.

6. If a column contains a leading 1, then all other entries in that column are zero.

7. There are three elementary row operations allowed: Interchanging rows of a matrix; Interchanging rows of a matrix; Multiply one row by a number that is not equal to 0; Multiply one row by a number that is not equal to 0; Add any number not equal to 0) times one row of the matrix to another row (as long as the two rows are not equal to each other). Add any number not equal to 0) times one row of the matrix to another row (as long as the two rows are not equal to each other).

8. Allow the following to equal A:

9. Allow the following to equal b, the solutions to each row of A:

10. The augmented matrix (Ab) is below and this is the matrix we will be reducing to reduced row echelon form:

11. Below is the original matrix

12. Add -2 times the first row to the second row -2(R1)+R2=new R2

13. Add -4 times the first row to the third row -4(R1)+R4=new R4

14. Multiply the second row by -1/10 (-1/10)R2=new R2

15. Add 13 times the second row to the third row (13)R2+R3=new R3

16. Add -5 times the second row to the first row (-5)R2+R1=new R1 The bottom row is comprised of all 0’s which means that when we are solving the system of linear equations, that row means nothing.

17. To solve the system of equations set each row in A equal to the corresponding row in b and add in the variables of x (x1, x2 and x3):

18. This then is simplified to:

19. There are infinitely many solutions if one or more rows drop to zero, leaving one or more of the remaining equations with free variables.

20. Solving the remaining linear equations for x3 (the free variable) yields: x1=-2 x3-3 x2=-1 x3+2 x3=arbitrary

21. Solution: The system has infinitely many solutions because each of the dependant variables x1 and x2, can be written in terms of the free variable, x3.