Math 141 Lecture Notes for Section 2.2
Section 2.2 -
1
Systems of Linear Equations: Unique Solutions
In the previous section, we used the method of substitution to find solutions for systems of equations. In
this section, we are going to use technique called the elimination method to solve systems of equations.
Suppose we are given a system of equations in 3 variables. Let them be
3x +2y
2x
+y
x
+y
+5z
+4z
+5z
= −3
=3
=2
We use 3 operations in the method of elimination, they are:
(1) Interchange any two equations.
(2) Replace an equation by a non-zero constant multiple of itself.
(3) Replace an equation by the sum of that equation and a constant multiple of any of the other
equations.
The first rule should be obvious, as it’s just writing the equations in a different order. The second rule
is the method of division we used in the material from section 2.1. The third method requires some
explanation.
If we have two equations:
R1 :
R2 :
x −2y
4x +3y
=3
=2
Then the difference
R2 − 4R1
following rule 3 is the value:
4(x − 2y = 3)
4x − 8y = 12
R2 − 4R1 : 4x + 3y = 2
−4x + 8y = −12
11y = −10.
4R1 :
This might seem a little strange at first, but after time this will become a powerful tool for you to use
to solve problems.
Example 2.2.1:
Use the method of elimination to solve this system of equations:
R1 :
R2 :
R3 :
3x
2x
x
+2y
+y
+y
+5z
+4z
+5z
= −3
=3
=2
Math 141 Lecture Notes for Section 2.2
2
Example 2.2.2:
Write down the augmented matrix for this system of equations
R1 : 2x +4y
R2 : 3x
−y
R3 : 6x
+2z
+4z
+2z
=1
=6
=4
Example 2.2.3:
Convert the following matrix into a system of linear equations:


1 1 3 5
 2 4 2 0 
7 0 6 3
Math 141 Lecture Notes for Section 2.2
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We can use an augmented matrix to solve a system of linear equations in the same way that we use
the elimination method. The rules for matrices are effectively the same as they are for equations, but
will be rewritten here for clarity’s sake.
(1) Interchange any two rows.
(2) Replace a row by a non-zero constant multiple of itself.
(3) Replace a row by the sum of that row and a constant multiple of any of the other rows.
This method is called Gauss-Jordan Elimination. Using Gauss-Jordan elimination, it is our goal to
apply the three rules until we have a matrix in row-reduced form, that is:
(1) Each row consisting of entirely zeroes lies below all rows having nonzero entries.
(2) The first nonzero entry in each (nonzero) row is 1.
(3) In any two successive (nonzero) rows, the leading 1 in the lower row lies to the right of the leading
1 in the upper row.
(4) If the column in the matrix contains a leading 1 then all entries to the left of the | are 0.
Math 141 Lecture Notes for Section 2.2
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Example 2.2.4:
Determine if the following matrices are in row-reduced form and convert them to the solution or
equations they represent:


1 0 0 3
 0 1 0 2 
0 0 1 1

1 0 0
 0 0 0
0 1 0

2
0 
3

1 1 0
 0 1 0
0 0 1

4
2 
5


6
0 
0
1 0 0
 0 0 0
0 0 0
Math 141 Lecture Notes for Section 2.2
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Example 2.2.5:
Use an augmented matrix to solve this system of equations:
2x + 8y + 4z = 4
x + 2y − 2z = 4
x + 4y + z = 6.
Math 141 Lecture Notes for Section 2.2
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Suggested Homework Problems: 3, 7, 13, 15, 23, 29, 37, 49, 51, 59, 63, 67, 71, 73, 75.