Gauss-Jordan Elimination and Reduced-Row Echelon Form

Also a method of solving linear equations. Done by transforming the system's augmented matrix into Reduced-Row Echelon form by the use of Elementary Row Operations. 

*Named for Carl Friedrich Gauss and Wilhelm Jordan. Started as Gaussian Elimination and improved by Wilhelm Jordan. 

Reduced-Row Echelon Form: 
The following are the conditions that your matrix is in RREF: 
1. When it satisfies all the conditions in Row-Echelon Form. 

Gaussian Elimination Method with Back Substitution

The method of solution where in it is based on addition/elimination, there is a systematized method for solving the three-or-more variable system. This method is called "Gaussian Elimination" (with the equations ending up in what is called "row-echelon form").


Elementary Row Operations:

1. Interchange two equations.

2. Multiply an equation by a nonzero constant.

3. Add a multiple of an equation to another equation.

Row-Echelon Form of a Matrix

*The term "echelon" refers to the stair-step pattern formed by the nonzero elements of the matrix. 

Definition of Row-Echelon Form of a Matrix: 

A Matrix that is in Row-Echelon Form has the following properties: 

1. All rows consisting entirely of zeros occur at the bottom of the matrix. 

2. For each row that does not consist entirely of zeros, the first nonzero entry is 1 (called a leading 1). 

3. For two successive (nonzero) rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row. 

Gaussian Elimination w/ Back Substitution: 

1. Write the augmented matrix of the system of linear equations. 

2. Use the elementary row operations to rewrite the augmented matrix in row-echelon form. 

3. Write the system of linear equations corresponding to the matrix in row-echelon form, and use back substitution to find the solution. 

Example #1. 

Example #2. 

Matrix

*This is a rectangular array of real or complex numbers, symbols, expressions arranged in rows and columns. 

Entry/Entries
*aij
*The individual items in a matrix, wherein (i) is the row and (j) is the column. 

Rows
*Horizontal lines. 

Columns
*Vertical lines.

 Real Matrix
*A entry consisting of real numbers. 

Square Matrix
*A matrix with the same number of rows and columns. 

Sizes of Matrices:
*A matrix with m rows and n columns (an m x n matrix) is said to be of size m x n.  

*One common use of matrix is to represent systems of linear equations. 

Augmented Matrix
*The matrix derived from the coefficients and constant terms of a system of linear equations. 

Coefficient Matrix
*The matrix containing only the coefficients of the system. 

Equivalent Systems of Equations

*Two systems of linear equations are called equivalent if they have precisely the same solution set.

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.

Parametric Representation of A Solution Set

In describing the Solution Set of a linear equation, we often use parametric representation. 

Solve the Linear Equation: 
1) x+2y=4 

 *to find the solution set of a equation involving two variables, solve for one of the variables in terms of the other variable. 

                                           x=4-2y

 *In this form, the variable y is free which means it can take any real value.The variable x is not free because it's value depends on the value assigned to y. 

 *To represent the infinite number of solutions of this equation, we use a third variable j called a parameter. By letting y=j, you can now represent the solution set. 

            *Let y=j, j is a real number. 
              x=4-2j
              Let j=1
              2y=4-x 
                y=2-1/2x
                x=2
                y=1 
The Solution Set is: (2,1) or (4,2).  

2) 3x+2y-z=3
    3x=3-2y-z
           3
      x=1-2/3y-1/3z

*Let y=s, s is a real number, z=t, t is a real number. 
  x=1-2/3s+1/3t
  x=1-2/3(1)+1/3(1)=2/3
  
The Solution Set is: (2/3,1,1) 

3) 1/2x-1/3y=1 
     (1/2x-1/3y=1)6
       3x-2y=6
       3x=6+2y
              3
         x=2+2y
                   3
*Let y=t, t is any real number. 
  x=2+2/3t
  t=1 
   x=2+2/3t
   x=2+2/3 
   x=8/3

The Solution Set is: (1, 8/3) 

Linear Systems in 3 Variables

(Consistent Independent System) 
x-2y+3z=9
-x+3y    =-4
2x-5y+5z=17

First Step:
*Add equation 1 and 2. That yields to, y+3z=5. 

Second Step:
*Mutiply (-2) to equation 1. -2(x-2y+3z), that yields to, 
(-2x+4y-6z=-18).

Third Step:
*Add the result in step 2 to equation number 3 to cancel the leading coefficient.
    -2x+4y-6z=-18
 + 2x-5y+5z=17 , this results to (-y-z=-1).

Fourth Step:
*Add the result in step 2 to the result in step 1.
    -y-z=-1
+ y+3z=5 , this results to (2z=4) , (z=2).

Fifth Step: 
*Substitute the values into the equation order to get the Solution Set of the system. 
Solution: 
x-2(-1)+3(2)=9
          x+2+6=9
                   x=1 
Solution Set:
  (1,-1,2)  *There is only one solution. 


Row Echelon Form: 
x-2y+3z=9
     y+3z=5
           z=2

(Inconsistent System) 
x1-3x2+x3=1
2x1-x2-2x3=2
x1+2x2-3x3=-1

First Step: 
*Multiply (-2) to equation 1 then add to equation 2. 
  -2x1+6x2-2x3=-2
+ 2x1-x2-2x3=2
*this results to, 5x2-4x3=0. 

Second Step:
*Multiply (-2) to equation 3 then add to equation 2. 
     -2x1+4x2-6x3=2
  + 2x1-x2-2x3=2
*this results to, -5x2+4x3=4. 

Third Step: 
*Add the results in the first and second step. 
    5x2-4x3=0
+-5x2+4x3=4 
*this results to, 0=4. (There is no solution in this system) 

Row Echelon Form: 
x1-3x2+x3=1
     5x2-4x3=0
              0=4 

(Consistent Dependent) 
      x2-x3=0
x1       -x3=-1
-x1+3x2   =1

First Step: 
*Interchange equations 1 and 2. 
x1       -x3=-1
       x2-x3=0
-x1+3x2   =1

Second Step: 
*Multiply (-3) to equation 2. 
   3x2+3x3=0

Third Step:
*Add equations 1 and 2. 
   3x2-3x3=0
+ 3x2+3x3=0
              0=0
Fourth Step: 
*Rewrite in a system of linear equation. 
x1-3x3=-1
    3x2-3x3=0
              0=0


Parametric Representation: 
x1-3x3=-1
x1=-1+3x3 free variable   
Let x3=f, f is a real number. 
x1=3f-1
x2=f
x3=f, f=2

The solution set is: (5,2,2) or (2,1,1) 
  Let 

Linear Systems in 2 Variables

a) Consistent Independent System 

3x-4y=-6                     
2x+4y=16                                                                             
  

*This is a system with only one solution. 
*The graphical solution is the intersection of two straight lines. 







b) Consistent Dependent System 

x+y=1                 -2x+2y=-2     *This is a system with infinite 
2x+2y=2               2x+2y=2        solutions. 
              
                      

*The graphical solution is any point on the two identical straight lines or coincident lines.                       







This is again a Consistent Dependent System. 
2x+2y=6 
m= -1 
b= 3 
x+y=3          

c) Inconsistent System 
     x+y=3    *This is a system with no solution. 
     x+y=1

*The graphical solution has parallel lines. 





*If the equations have different slopes then the system is independent and the lines cross at a point. 
*If the equations have the same slope but different intercepts, then the system is inconsistent and the lines are parallel and never cross. 
*If the equations have the same slope and the same intercept, then the system is dependent and the lines are actually both in the same line.