7.6 Solving Systems with Gaussian Elimination

7.6 Gaussian Elimination
Chapter 7 Systems of Equations and Inequalities

Concepts and Objectives
⚫ The objectives for this section are
⚫ Write the augmented matrix of a system of equations.
⚫ Write the system of equations from an augmented
matrix.
⚫ Perform row operations on a matrix.
⚫ Solve a system of linear equations using matrices.

Writing an Augmented Matrix
⚫ A matrix can serve as a device for representing and
solving a system of equations written in standard form.
To express a system in matrix form, we extract the
coefficients of the variables and the constants, and these
become the entries of the matrix.
⚫ We use a vertical line to separate the coefficient
entries from the constants, essentially replacing the
equal signs.
⚫ When a system is written in this form, it is called an
augmented matrix.

Writing an Augmented Matrix
⚫ For example, the following system of equations,
can be written as an augmented matrix:
⚫ Notice that the matrix is written so that the variables
line up in their own columns: x-terms go in the first
column, y-terms in the second column, etc.
⚫ If there is a missing variable term, the coefficient is 0.
3 4 7
4 2 5
x y
x y
+ =
− =
3 4 7
4 2 5
 
 
−
 

Writing a Coefficient Matrix
⚫ We can also write a matrix containing just the
coefficients. This is called the coefficient matrix.
⚫ A 3 × 3 system of equations such as
has a coefficient matrix
3 0
5
2 3 2
x y z
x y
x z
− − =
+ =
− =
3 1 1
1 1 0
2 0 3
− −
 
 
 
 
−
 

Writing a System From a Matrix
⚫ In the same fashion, we can take an augmented matrix,
and generate a system of equations. For example, the
augmented matrix
can generate the following system of equations:
1 3 5 2
2 5 4 5
3 5 4 6
 
− − −
 
− −
 
 
−
 
3 5 2
2 5 4 5
3 5 4 6
x y z
x y z
x y z
− − = −
− − =
− + + =

Performing Row Operations
⚫ Now that we can write systems of equations in an
augmented matrix form, we will examine the various
row operations that can be performed on a matrix, such
as addition, multiplication by a constant, and
interchanging rows.
⚫ We can use row operations to solve a system of
equations by converting the matrix to row-echelon form,
in which there are ones down the main diagonal from
the upper left corner to the lower right corner, and zeros
in every position below the main diagonal as shown on
the next slide.

Performing Row Operations
⚫ We use row operations corresponding to
equation operations to obtain a new matrix
that is row-equivalent in a simpler form.
Here are the guidelines to obtaining row-
echelon form:
1
1
0
0 0 1
a b
d
 
 
 
 
 
1. In any nonzero row, the first nonzero number is 1. It is called a
leading 1.
2. Any all-zero rows are placed at the bottom on the matrix.
3. Any leading 1 is below and to the right of a previous leading 1.
4. Any column containing a leading 1 has zeros in all other positions
in the column.

Row Operations and Notation
⚫ To solve a system of equations we can perform the
following row operations to convert the coefficient
matrix to a row-echelon form and do back-substitution
to find the solution.
1. Interchange rows. (Notation: )
2. Multiply a row by a constant. (Notation: )
3. Add the product of a row multiplied by a constant to
another row. (Notation: )
i j
R R

i
cR
i j
R cR
+

Gaussian Elimination
⚫ The primary tool we will be using is called Gaussian
elimination, named after the prolific German
mathematician Karl Friedrich Gauss. While there is no
definitive order in which operations are to be
performed, there are specific guidelines as to what type
of moves can be made.
⚫ To obtain a matrix in row-echelon form for finding
solutions, we use row operations to obtain a 1 as the
first entry so that row 1 can be used to convert the
remaining rows.

Procedure
Given an augmented matrix, perform row operations to
achieve row-echelon form.
1. The first equation should have a leading coefficient of 1.
Interchange rows or multiply by a constant, if necessary.
2. Obtain zeros down the first column below the first entry of 1.
3. Obtain a 1 in row 2, column 2
4. Obtain zeros down column 2, below the entry of 1.
5. Continue this process for all rows until there is a 1 in every
entry down the main diagonal and there are only zeros
below.
6. If any rows contain all zeros, place them at the bottom.

Procedure (cont.)
⚫ Example: Solve the given system by Gaussian
elimination.
2 3 6
8
x y
x y
+ =
− =

Procedure (cont.)
⚫ Example: Solve the given system by Gaussian
elimination.
1. Write the system as an augmented matrix.
2 3 6
8
x y
x y
+ =
− =
2 3 6
1 1 8
 
 
−
 

Procedure (cont.)
2. Interchange row 1 and row 2 to put a 1 in row 1,
column 1.
3. Multiply row 1 by ‒2 and add the result to row 2.
1 2
1 1 8
2 3 6
R R
 
−
 →  
 
1 2 2
1 1 8
2
0 5 10
R R R
 
−
− + = →  
−
 

Procedure (cont.)
4. Multiply row 2 by ⅕.
5. The second row of the matrix represents y = ‒2.
Substitute y = ‒2 into the second equation.
6. The solution is (6, ‒2)
2 2
1 1 8
1
5 0 1 2
R R
 
−
= →  
−
 
( )
2 8
6
x
x
− − =
=

Using a Matrix Calculator
⚫ We can also use a matrix calculator to handle all of the
tedious bits for us.
⚫ Example: Solve the system of linear equations using
matrices.
8
2 3 2
3 2 9 9
x y z
x y z
x y z
− + =
+ − = −
− − =

Using a Matrix Calculator
⚫ Example: Solve the system of linear equations using
matrices.
1. Set up the augmented matrix.
8
2 3 2
3 2 9 9
x y z
x y z
x y z
− + =
+ − = −
− − =
1 1 1 8
2 3 1 2
3 2 9 9
 
−
 
− −
 
 
− −
 

Using a Matrix Calculator (cont.)
2. Go to desmos.com/matrix (not the graphing calculator)
3. Click on

3. Click on
4. Change the rows to 3 and the columns to 4.

3. Click on
4. Change the rows to 3 and the columns to 4.
5. Enter the values from your matrix.

3. Click on
4. Change the rows to 3 and the columns to 4
5. Enter the values from your matrix
6. Click on (reduced row-echelon form) and then

3. Click on
4. Change the rows to 3 and the columns to 4
5. Enter the values from your matrix
6. Click on (reduced row-echelon form) and then
7. The solution will be in the
right-hand column:
(4, ‒3, 1)

⚫ Some notes:
⚫ If we had used Gaussian elimination to solve the
previous system, we would have ended up with the
matrix
⚫ The difference is that we are doing “ref” or row-
echelon form; Desmos does “rref” or reduced row-
echelon form, which saves us from having to
substitute for the final solution.
1 1 1 8
0 1 12 15
0 0 1 1
 
−
 
− −
 
 
 

⚫ Some notes (cont.):
⚫ Desmos defaults to giving decimal answers, but it
does have the ability to convert these to fractions.
⚫ Calculators always assume you know what you’re
doing, and that you’ve typed everything correctly.
Check your answers against the original equations!

Classwork
⚫ College Algebra 2e
⚫ 7.6: 6-14 (even); 7.5: 18, 20, 30-40 (even); 7.3: 52-58
(even)
⚫ Quiz 7.5

7.6 Solving Systems with Gaussian Elimination

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