2
SYSTEMS OF
LINEAR
EQUATIONS AND
MATRICES
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2.3
Systems of Linear Equations:
Underdetermined and Overdetermined
Systems
Copyright © Cengage Learning. All rights reserved.
Systems of Linear Equations: Underdetermined and Overdetermined Systems
In this section, we look at systems that have infinitely many
solutions and those that have no solution.
We also study systems of linear equations in which the
number of variables is not equal to the number of equations
in the system.
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Solution(s) of Linear Equations
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Solution(s) of Linear Equations
The next example illustrates the situation in which a system
of linear equations has infinitely many solutions.
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Example 1 – A System of Equations with an Infinite Number of Solutions
Solve the system of linear equations given by
x + 2y = 4
3x + 6y = 12
(9)
Solution:
Using the Gauss-Jordan elimination method, we obtain the
following system of equivalent matrices:
The last augmented matrix is in row-reduced form.
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Example 1 – Solution
cont’d
Interpreting it as a system of linear equations, we see that
the given System (9) is equivalent to the single equation
x + 2y = 4
or
x = 4 – 2y
If we assign a particular value to y—say, y = 0—we obtain
x = 4, giving the solution (4, 0) to System (9). By setting
y = 1, we obtain the solution (2, 1).
In general, if we set y = t, where t represents some real
number (called a parameter), we obtain the solution given
by (4 – 2t, t).
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Example 1 – Solution
cont’d
Since the parameter t may be any real number, we see that
System (9) has infinitely many solutions.
Geometrically, the solutions of System (9) lie on the line on
the plane with equation x + 2y = 4. The two equations in
the system have the same graph (straight line), which you
can verify graphically.
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Example 1 – A System of Equations with an Infinite Number of Solutions
Solve the system of linear equations given by
x + 2y – 3z = 4
3x - y - 2z = 12
2x + 3y – 5z = -3
Solution: Using the Gauss-Jordan elimination method, we
obtain the following system of equivalent matrices:
 1 2 3 2  1 0 1 0 
 3 1 2 1   0 1 1 1

 

 2 3 5 3 0 0 0 0 
The last augmented matrix is in row-reduced form.
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Example 1 – Solution
cont’d
Interpreting it as a system of linear equations, we see that
the given System is equivalent to the single equation
x – z = 0 or
y – z = -1 or
x=z
and
y=z-1
If we set z = t, where t is a parameter, then the system has
infinitely many solutions given by (t, t – 1, t).
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Practice
p. 98 Self-Check Exercises #1 & 2
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