Solving a System of Linear Equations

The system is in Row-Echelon form if it follows a stair-step pattern and has leading coefficients of 1.
Example:
                    x - 2y + 3z = 9
                         y + 3z = 5
                               z = 2

Using Back-Substitution to Solve a System in Row-Echelon Form.
Example      Use back-substitution to solve the system:
                5x + 4y - z = 0        Equation 1
                     10y - 3z = 11     Equation 2
                              z = 3        Equation 3
Solution       From Equation 3 you know that z = 3. By substitution this value of z into Equation 2, you get the value for y. Then you simply plug z and y into the first equation and solve for x.
                10y - 3(3) = 11
                          y = 2
               5x + 4(2) - 3 = 0
                            5x = -5
                             x = -1
        The system has exactly one solution: (-1, 2, 3)

Two systems of linear equations are called equivalent if they have precisely the same solution set. To solve a system that is not in row-echelon form, first change it to an equivalent system that is in row-echelon form by using the operations listed below:

Operations That Lead to Equivalent Systems of Equations
1. Interchange two equations
2. Multiply an equation by a nonzero constant
3. Add a multiple of an equation to another equation.

Using Gaussian Elimination to Rewrite a System in Row-Echelon Form
Example        Solve the system using Gaussian elimination
          x - 2y + 3z = 9
        -x + 3y        = -4
        2x - 5y + 5z = 17
Solution         Work from the upper left corner of the system, saving the x in the upper left position and eliminating the order x's from the first column.
          x - 2y + 3z = 9                Adding the first equation to the
                  y + 3z = 5                  second equation produces a new second
             2x - 5y + 5z = 17               equation.

            x - 2y+ 3z = 9                   Adding -2 times the first equation
                  y + 3z = 5                to the third equation produces a new
                  -y - z = -1                third equation.
Now the everything but the first x has been eliminated from the first column, work on the second column.
        x - 2y + 3z = 9                  Adding the second equation to the third
              y + 3z = 5                  equation produces a new third equation.
                   2z = 4

         x - 2y + 3z = 9                 Multipling the third equation by 1/2
              y + 3z = 5                  produces a new third equation.
                   z = 2
The system has one solution: (1, -1, 2)

Example        Solve the system.
          x - 3y + z = 1
         2x - y - 2z = 2
          x + 2y - 3z = -1
Solution     
          x - 3y + z = 1                Adding -2 times the first equation to the
              5y - 4z = 0               second equation produces a new second
          x + 2y - 3z = -1            second equation.

         x - 3y + z = 1                Adding -2 times the first equation to the
               5y - 4z = 0                 third equation produces a new third
            5y - 4z = 0                equation.
Continuing the elimination process, add -1 times the second equation to the third equation to produce a new third equation.
          x - 3y + z = 1                Adding -1 times the second equation to the
             5y + 4z = 0               third equation produces a new third equation.
                     0  = -2
Because the third "equation" is a false statement, the system has no solution.

Example        Solve the system.
                    y - z = 0
             x      - 3z = -1
             -x + 3y    = 1
Solution        Begin by rewriting the system in row-echelon form as it follows.
             x      - 3z = -1              The first two equations are interchanged.
                    y - z = 0
             -x + 3y    = 1

            x      - 3z = -1               Adding the first equation to the third
                   y - z = 0               equation produces a new third equation.
                3y - 3z = 0

            x      - 3z = -1               Adding -3 times the second equation to the third
                   y - z = 0               equation produces a new third equation.
                        0 = 0
Because the third equation is unnecessary, omit it to obtain the system shown below.
            x      - 3z = -1              
                   y - z = 0            
To represent the solutions, choose z to be the free variable and represent it by parameter t. Because y = z and x = 3z - 1, you can describe the solution set as
               x = 3t - 1        y = t       z = t           t is any real number.
The system has an infinite number of solutions.