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
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Since we already have that the coefficinet of x is 1 in R3 , we can switch it with R1 and obtain
R1 :
R2 :
R3 :
x
2x
3x
+y
+y
+2y
+5z
+4z
+5z
=2
=3
= −3
Now it is our goal to eliminate the variable x from each of R2 and R3 . We do this by applying rule
(3) twice. First, we compute R2 − 2R1
2R1
= 2(x + y + 5z = 2)
= 2x + 2y + 10z = 4
R2 − 2R1 : 2x + y + 4z = 3
−2x − 2y − 10z = −4
= −y − 6z = −1
and replace R2 with the new equation −y − 6z = −1.
R1 :
R2 :
R3 :
x
3x
+y
−y
+2y
+5z
−6z
+5z
=2
= −1
= −3
Then we compute R3 − 3R1
3R1
= 3(x + y + 5z = 2)
= 3x + 3y + 15z = 6
R3 − 3R1 : 3x + 2y + 5z = −3
−3x − 3y − 15z = −6
= −y − 10z = −9
and replace R3 with the new equation −y − 10z = −9.
R1 : x +y
R2 :
−y
R3 :
−y
+5z
−6z
−10z
=2
= −1
= −9
We now seek to eliminate the y variable from the R3 equation so we perform the operation R3 − R2
which yields
R3 − R2 : −y − 6z = −1
+y + 10z = +9
4z = 8.
replacing R3 with the equaiton 4z = 8.
R1 : x
R2 :
R3 :
We can now use rule (2) to multiply R3 by
1
4
R1 : x
R2 :
R3 :
+y
−y
+5z
−6z
4z
=2
= −1
=8
to obtain z = 2.
+y
−y
+5z
−6z
z
=2
= −1
=2
Math 141 Lecture Notes for Section 2.2
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Now we want to eliminate the variable z from both R1 and R2 , which we can do by subtracting 5R3
and adding 6R3 respecitvely to obtain:
R1 : x
R2 :
R3 :
z
= −8
= 11
=2
z
=3
= 11
=2
+y
−y
Next we replace R1 with R1 + R2
R1 : x
R2 :
R3 :
−y
As a final step, we multiply R2 by −1 and we have
R1 : x
R2 :
R3 :
y
z
=3
= −11
=2
Leaving us with a final solution of (3, −11, 2).
Example 2.2.2:
Write down the augmented matrix for this system of equations
R1 : 2x +4y
R2 : 3x
−y
R3 : 6x

2
 3
6
4 2
−1 4
0 2
+2z
+4z
+2z
=1
=6
=4

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
R1 :
x
+y
R2 : 2x +4y
R3 : 7x
+3z
+2z
+6z
=5
=0
=3
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.
Math 141 Lecture Notes for Section 2.2
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(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.
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
Row-reduced. Corresponds to x = 3, y = 2, z = 1.

1 0 0
 0 0 0
0 1 0

2
0 
3
Not row-reduced, the second row is all 0s when it should be swapped with the 3rd row. Corresponds
to the solution set x = 2, y = 3, z = t for some parameter t.


1 1 0 4
 0 1 0 2 
0 0 1 5
Not row-reduced, the first row contains two nonzero entries before the |. This can be fixed by applying
the operation R1 − R2 . The solution that this matrix represents is x + y = 4, y = 2, and z = 5. This
can be amended to be the solution x = 2, y = 2, z = 5.


1 0 0 6
 0 0 0 0 
0 0 0 0
This is row-reduced. This corresponds to the set of solutions x = 6, y = s, z = t where s and t are
both parameters.
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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The augmented matrix for this system of equations is

2 8
4
 1 2 −2
1 4
1
The first thing we do is divide the first row by

1
 1
1

4
4 
6
2 and replace it. This is written as 12 R1 −→ R1 .

4
2 2
2 −2 4 
4
1 6
Now, we perform the operation R2 − R1 −→ R2

1
4
2
 0 −2 −4
1
4
1

2
2 
6
and then the operation R3 − R1 −→ R3

1
 0
0
4
2
−2 −4
0 −1

2
2 
4
We can now perform − 21 R2 −→ R2

1 4
 0 1
0 0
and −R3 −→ R3
2
2
−1

2
−1 
4
4 2
1 2
0 1

2
−1 
−4
4 2
1 0
0 1

2
7 
−4

1 4 0
 0 1 0
0 0 1

10
7 
−4


−18
7 
−4

1
 0
0
We then work backwars, performing R2 − 2R3

1
 0
0
followed by R1 − 2R3
as a final step we have R1 − 4R2
1 0 0
 0 1 0
0 0 1
The solution for our system then is x = −18, y = 7, z = −4 and we are done.
Suggested Homework Problems: 3, 7, 13, 15, 23, 29, 37, 49, 51, 59, 63, 67, 71, 73, 75.