Gauss-Jordan Elimination

We will look at systems of equation with more than two variables
We will not require that a system have the same number of equations as variables

Possible Solutions to a System of Linear Equations
Any system of linear equations has exactly
     one solution
     no solution
               or
     an infinite number of solutions
regardless of the number of equations or the number of variables in the system

The terms unique, inconsistent, dependent, and independent
are used to describe the solutions of any system of linear equations

In the previous section (systems of linear equations and augmented matrices)
row operations
were used to transform the augmented matrix of a system of two equations in two variables


                    

into one of the simplified forms

                
          Form 1                          Form 2                           Form 3

where , , and are real numbers
Each of these reduced forms represents a system
that has a different type of solution set, and no two of these forms are row-equivalent
We consider each of these forms to be a different simplified form

Now we will look at larger systems with more equations and more variables


Definition of Reduced Matrix
A matrix is in reduced form if
     1]   Each row consisting entirely of 0’s is below any row having at least one nonzero element
     2]   The leftmost nonzero element in each row is 1
     3]   The column containing the leftmost 1 of a given row has 0’s above and below the 1
     4]   The leftmost 1 in any row is to the right of the leftmost 1 in the preceding row

      See Example 1, pages 677 – 678, of the textbook


Solving Systems of Linear equations by Gauss-Jordan Elimination
The method transforms an augmented matrix into a reduced form
The system corresponding to a reduced augmented coefficient matrix
is called a reduced system

     See Example 1, pages 677 - 678, of the textbook


Gauss-Jordan Elimination
     Step 1.
                    Choose the leftmost nonzero column and use appropriate row operations
                    to get a 1 at the top

     Step 2.
                    Use multiples of the row containing the 1 from Step 1 to get zeros
                    in all remaining places in the column containing this 1

     Step 3.
                    Repeat Step 1 with the submatrix formed (mentally) by deleting the row
                    used in Step 2 and all rows above this row

     Step 4.
                    Repeat Step 2 with the entire matrix, including the mentally deleted rows.
                    Continue this process until it is impossible to go further


Notes:
1]   Although each matrix has a unique reduced form, the sequence of steps given here
      for transforming a matrix into a reduced form is not unique
      Other sequences of steps (using row operations) can produce a reduced matrix



2]   If at any point in the process we get a row with all zeros to the left of the vertical line
      of the augmented matrix and a nonzero number to the right of the vertical line, we must
      stop, since we have a contradiction
                  
      Conclude that the system of linear equations has no solution

3]   If the number of leftmost 1’s in a reduced augmented matrix is less than the number
      of variables in the system and there are no contradictions, then the system is dependent
      and has infinitely many solutions
      Solve each equation in a reduced system for its leftmost variable and then introduce
      a different parameter for each remaining variable.
      This method gives concise representation of the solutions of a dependent system
      of linear equations


      See Examples 2 – 6, pages 679 – 686, of the textbook


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