Gaussian Elimination Example by Jose Mena

The following example will show how to apply Gaussian Elimination to a 3x4 matrix starting from the augmented form of a matrix.
Suppose we're given the following system of equations:

X + Y + Z = 6
2X -3Y + Z = 1
4Y - 7Z =-13

STEP (1):
First, in order apply the Gaussian Elimination method, we must rewrite the previous three equations in the form of an augmented matrix,
such as follows:

R1= 1 1 1 6
R2= 2 -3 1 1
R3= 0 4 -7 -13

NOTE: Notice how from the third equation we didn't have an entry for the X value, so in R3, we placed the number 0 to indicate a
place value of zero.

STEP (2):
Now we use row operations to reduce the augmented matrix to row echlon form:

(I)
1 1 1 6
2 -3 1 1 R2-2R1= New R2
0 4 -7 -13

What this means is that we do the following operations to the Second Row (R2):

First: 2-2*(1) = New entry for the first number in the second row
Second: (-3)-2*1 = New entry for the second number in the second row
Third: 1-2*1 = New entery for the third number in the second row
Last: 1-2*6 = New entry for the fourth number in the second row

Note: The row operations for the next several steps are done similarly.

(II)
1 1 1 6
0 -4 -1 -11 R2/(-4)= New R2
0 4 -7 -13

(III)
1 1 1 6
0 1 1/4 11/4
0 4 -7 -13 R3-4R2= New R3

(IV)
1 1 1 6
0 1 1/4 11/4
0 0 -8 -24 R3/(-8)= New R3

(V)
X Y Z
1 1 1 6
0 1 1/4 1/4
0 0 1 3

From this last matrix we can obtain three equations that will help us solve the system of equations at the beginning of the the example.
If you notice on the top there are three variables (X, Y, Z) that correspond to each column of the matrix except for the last one, later you will know why the last column is not labled.

Next we take each variable and multiply it by the corresponding column and develop the following equations:

1X + 1Y + 1Z = 6 (Equation 1)
1X + 1Y + (1/4)Z = 11/4 (Equation 2)
0X + 0Y + 1Z = 3 (Equation 3)

In a sense we're doing the reverse of what we did to get the Augmented matrix and forming our new equations.

LAST STEP: Now we solve for each variable:

From Equation (3):

Z=3

From Equation (2):

Y + (1/4)Z = 11/4 (We substitute Z with 3 from equation 3)
Y + 3/4 = 11/4 (We subtract (3/4) from bothe sides of the equation to isolate the variable Y)
Y = (11/4) - (3/4) = (8/4) = 2
Y = 2

From Equation (1):

X + Y + Z = 6 (We substitute the Z with 3 and the Y with 2)
X + 2 + 3 = 6 (We now subtract 2 and 3 from both sides of the equation to isolate the variable X)
X = 6 - 2 - 3
X = 6 - 5 = 1
X = 1

Now we are finished having solved and found the values of all the variables where:

X = 1
Y = 2
Z = 3

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