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Math 155 - Cooley
Finite Math with Applications
OCC
Section 2.2 – Solutions of Linear Systems by the Gauss-Jordan Method
A system of equations consists of two or more equations (in x and y) where a common solution is sought. The
solution of this system will be an ordered pair (x, y) that satisfies all the equations in the system.
y  x  5
A system of equations is usually denoted by: 
, where the solution is written as an ordered pair. In
 y  2x  4
this example, the ordered pair, (1, 6) satisfies both equations. Each equation corresponds to a line, when
graphed, and the ordered pair, (1, 6), is the point of intersection of the two lines.
When a system of two linear equations is graphed, three physical situations (solutions) are possible:
Name of System:
Number of
Solutions:
Consistent , Independent
1
(Exactly 1 solution)
What to look for:
Different slopes.
What going on:
Non-parallel lines.
Inconsistent , Independent
0
(No solution)
1. Same slope.
2. Different y-intercepts.
Parallel lines.
Consistent , Dependent
Infinitely many
(Infinite solutions)
1. Same slope.
2. Same y-intercepts.
Same line.
Consistent – A system of equations that has at least one solution.
Inconsistent – A system of equations that has no solution.
Independent – A system of equations with no more than one solution.
Dependent – A consistent system of equations that has infinitely many solutions.
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Math 155 - Cooley
Finite Math with Applications
OCC
Section 2.2 – Solutions of Linear Systems by the Gauss-Jordan Method
Definition – Matrix (or Matrices for plural)
A rectangular array of numbers enclosed by brackets.
Definition – Element (or Entry)
Each number in a matrix is called an element or entry.
Definition – Coefficient Matrix & Augmented Matrix
Consider the following system of equations for a 2×2 system, 3×3 system, 4×4 system, etc,
a1 x  b1 y  c1
a2 x  b2 y  c2
a1 x  b1 y  c1z  d1
,
a2 x  b2 y  c2 z  d 2 ,
a3 x  b3 y  c3 z  d3
a1w  b1 x  c1 y  d1 z  e1
a2 w  b2 x  c2 y  d 2 z  e2
a3 w  b3 x  c3 y  d3 z  e3
,….
a4 w  b4 x  c4 y  d 4 z  e4
Then, the coefficient matrices respectively would be
 a1
a
 2
b1 
,
b2 
 a1
a
 2
 a3
b1
b2
b3
c1 
c2  ,
c3 
 a1
a
 2
 a3

 a4
b1
c1
b2
b3
c2
c3
b4
c4
d1 
d 2 
,….
d3 

d4 
and, the augmented matrices respectively would be
 a1 b1 c1 d1 e1 
 a1 b1 c1 d1 
a b c d e 
 a1 b1 c1 
2
2
a b c d  ,
 2 2 2
,
,….
2
a b c 
 2 2 2

a
b
c
d
e
3
3
3
3
3
 2 2 2
 a3 b3 c3 d3 


 a4 b4 c4 d 4 e4 
Definition – The Gaussian Elimination Method
A method used to solve a system of linear equations that use row operations to obtain simpler and simpler
augmented matrices. The goal is to get the augmented matrix to the simplest form of
1 0 a 
0 1 b  ,


1 0 0 a 
0 1 0 b  ,


0 0 1 c 
1
0

0

0
0 0 0 a
1 0 0 b 
,….
0 1 0 c

0 0 1 d
That is, we want the left side of any augmented matrix reduced to the form of the identity matrix. The solution
would simply be the constants obtained on the right side of the augmented matrix
Definition – Row Operations
For any augmented matrix of a system of equations, the following operations produce the augmented matrix of
an equivalent system:
1. Interchanging any two rows;
2. Multiplying the elements of a row by any nonzero real number;
3. Adding a nonzero multiple of the elements of one row to the corresponding elements of a nonzero multiple of
some other row.
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Math 155 - Cooley
Finite Math with Applications
OCC
Section 2.2 – Solutions of Linear Systems by the Gauss-Jordan Method
Procedure – Gauss-Jordan Method of Solving a Linear System
1. Write each equation so that variable terms are in the same order on the left side of the equal sign and
constants are on the right.
2. Write the augmented matrix that corresponds to the system.
3. Use row operations to transform the first column so that all elements except the element in the first row are
zero.
4. Use row operations to transform the second column so that all elements except the element in the second row
are zero.
5. Use row operations to transform the third column so that all elements except the element in the third row are
zero.
6. Continue in this way, when possible, until the last row is written in the form
0
0 0  0
j k ,
where j and k are constants with j ≠ 0. When this is not possible, continue until every row has more zeros on
the left than the previous row (except possibly for any rows of all zeros at the bottom of the matrix), and the
first nonzero entry in each row is the only nonzero entry in its column.
7. Multiply each row by the reciprocal of the nonzero element in that row.
 Exercises:
Use the Gauss-Jordan method to solve each system of equations.
1)
2)
x  2y  5
2 x  y  2
4x  2 y  3
2 x  3 y  1
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Math 155 - Cooley
Finite Math with Applications
OCC
Section 2.2 – Solutions of Linear Systems by the Gauss-Jordan Method
 Exercises:
Use the Gauss-Jordan method to solve each system of equations.
3)
4)
2x  3y  9
4x  6 y  7
x  y 1
 x  y  1
x  2 y  7 z  2
5)
2 x  5 y  2 z  1
3x  5 y  4 z  9
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Math 155 - Cooley
Finite Math with Applications
OCC
Section 2.2 – Solutions of Linear Systems by the Gauss-Jordan Method
 Exercises:
Use the Gauss-Jordan method to solve the system of equations.
6)
The U-Drive Rent-A-Truck company plans to spend $7 million on 200 new vehicles. Each commercial
van will cost $35,000, each small truck $30,000, and each large truck $50,000. Past experience shows
that they need twice as many vans as small trucks. How many of each type of vehicle can they buy?
Let x = the number of vans to be purchased,
y = the number of small trucks to be purchased, and
z = the number of large trucks to be purchased.
x  y  z  200
35,000 x  30,000 y  50,000 z  7,000,000
x  2y
1
1
200 
 1
35000 30000 50000 7000000


 1

2
0
0
 1 1 1 200 
35 30 50 7000 


 1 2 0
0 
1 0 0 120 
0 1 0 60 


0 0 1 20 
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Math 155 - Cooley
Finite Math with Applications
OCC
Section 2.2 – Solutions of Linear Systems by the Gauss-Jordan Method
 Exercises:
7)
 1 2 2 4
Suppose the augmented matrix  2 1 1 1  reduces to the form
 1 1 1 3 
1 0 2  2 
3


7
 0 1 1 3  by using the


 0 0 0 0 
Gauss-Jordan method. What is the solution set?
8)
1 0 13 1 
The following matrix was reduced by using the Gauss-Jordan method: 0 1 4 3  .
0 0 0 0 
What is the solution set?
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