Download Section 9.2

CHAPTER 9:
Systems of Equations
and Matrices
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Systems of Equations in Two Variables
Systems of Equations in Three Variables
Matrices and Systems of Equations
Matrix Operations
Inverses of Matrices
Determinants and Cramer’s Rule
Systems of Inequalities and Linear Programming
Partial Fractions
Copyright © 2009 Pearson Education, Inc.
9.2
Systems of Equations in
Three Variables



Solve systems of linear equations in three variables.
Use systems of three equations to solve applied
problems.
Model a situation using a quadratic function.
Copyright © 2009 Pearson Education, Inc.
Solving Systems of Equations in Three
Variables
A linear equation in three variables is an equation
equivalent to one of the form Ax + By + Cz = D. A, B, C, and
D are real numbers and A, B, and C are not 0.
A solution of a system of three equations in three
variables is an ordered triple that makes all three equations
true.
Example: The triple (4, 0, 3) is the solution of this
system of equations. We can verify this by substituting 4
for x, 0 for y, and 3 for z in each equation.
x  2y + 4z = 8
2x + 2y  z = 11
x + y  2z = 10
Copyright © 2009 Pearson Education, Inc.
Slide 9.2-4
Gaussian Elimination
An algebraic method used to solve systems in three
variables.
The original system is transformed to an equivalent
one of the form:
Ax + By + Cz = D,
Ey + Fz = G,
Hz = K.
Then the third equation is solved for z and backsubstitution is used to find y and then x.
Copyright © 2009 Pearson Education, Inc.
Slide 9.2-5
Operations
The following operations can be used to transform
the original system to an equivalent system in the
desired form.
1. Interchange any two equations.
2. Multiply both sides of one of the equations by a
nonzero constant.
3. Add a nonzero multiple of one equation to another
equation.
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Slide 9.2-6
Example
Solve the system
x + 3y + 2z = 9
x  y + 3z = 16
3x  4y + 2z = 28
Solution: Choose 1 variable to eliminate using
2 different pairs of equations. Let’s eliminate x
from equations (2) and (3).
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Slide 9.2-7
Example
x  3y  2z = 9
x  y + 3z = 16
4y + z = 7
Mult. (1) by 1
(2)
(4)
3x  9y  6z = 27 Mult. (1) by 3
3x  4y + 2z = 28
(3)
13y  4z = 1
(5)
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Slide 9.2-8
Example continued
Now we have…
x + 3y + 2z = 9
4y + z = 7
13y  4z = 1
(1)
(4)
(5)
Next, we multiply equation (4) by 4 to make the z
coefficient a multiple of the z coefficient in the
equation below it.
x + 3y + 2z = 9
(1)
16y + 4z = 28 (6)
13y  4z = 1
(5)
Copyright © 2009 Pearson Education, Inc.
Slide 9.2-9
Example continued
Now, we add equations 5 and 6.
13y  4z = 1
(5)
16y + 4z = 28
(6)
29y = 29
Now, we have the system of equations:
x + 3y + 2z = 9
(1)
13y  4z = 1
(5)
29y = 29
(7)
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Slide 9.2-10
Example continued
Next, we solve equation (7) for y:
29y = 29
y = 1
Then, we back-substitute 1 in equation (5)
and solve for z.
13(1)  4z = 1
13  4z = 1
4z = 12
z=3
Copyright © 2009 Pearson Education, Inc.
Slide 9.2-11
Example continued
Finally, we substitute 1 for y and 3 for z in equation
(1) and solve for x:
x + 3(1) + 2(3) = 9
x3+6=9
x=6
The triple (6, 1, 3) is the solution of this system.
Copyright © 2009 Pearson Education, Inc.
Slide 9.2-12
Graphs
The graph of a linear equation in three variables is a plane.
Thus the solution set of such a system is the intersection of
three planes.
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Slide 9.2-13
Application
A food service distributor conducted a study to
predict fuel usage for new delivery routes, for a
particular truck. Use the chart to find the rates of
fuel in rush hour traffic, city traffic, and on the
highway.
Week 1
Week 2
Week 3
Rush Hour
Hours
2
7
6
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City Traffic
Hours
9
8
18
Highway
Hours
3
3
6
Total Fuel
Used (gal)
15
24
34
Slide 9.2-14
Solution
1.
2.
3.
Familiarize. We let x, y, and z represent the hours in rush
hour traffic, city traffic, and highway, respectively.
Translate. We have three equations:
2x + 9y + 3z = 15
(1)
7x + 8y + 3z = 24
(2)
6x + 18y + 6z = 34
(3)
Carry Out. We will solve this equation by eliminating z
from equations (2) and (3).
2x  9y  3z = 15 Mult. (1) by 1
7x + 8y + 3z = 24
(2)
5x  y = 9
(4)
Copyright © 2009 Pearson Education, Inc.
Slide 9.2-15
Solution continued
Next, we can solve for x:
4x  18y  6z = 30 Mult. (1) by 2
6x + 18y + 6z = 34
(3)
2x = 4
x=2
Next, we can solve for y by substituting 2 for x in equation (4):
5(2)  y = 9
y=1
Finally, we can substitute 2 for x and 1 for y in equation (1) to
solve for z:
2(2) + 9(1) + 3z = 15
4 + 9 + 3z = 15
3z = 2
2
2
z = 3 Solving the system we get (2, 1, 3 ).
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Slide 9.2-16
Solution continued
2
4. Check: Substituting 2 for x, 1 for y, and for z, we see that
3
the solution makes each of the three equations true.
5. State: In rush hour traffic the distribution truck uses fuel at a
rate of 2 gallons per hour. In city traffic, the same truck uses 1
gallon of fuel per hour. In highway traffic, the same truck
2
used gallon of fuel per hour.
3
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Slide 9.2-17