Download 7.3 multivariable linear systems

7.3
MULTIVARIABLE LINEAR SYSTEMS
Copyright © Cengage Learning. All rights reserved.
What You Should Learn
• Use back-substitution to solve linear systems in
row-echelon form.
• Use Gaussian elimination to solve systems of
linear equations.
• Solve nonsquare systems of linear equations.
• Use systems of linear equations in three or more
variables to model and solve real-life problems.
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Row-Echelon Form and
Back-Substitution
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Row-Echelon Form and Back-Substitution
System of Three Linear Equations in Three Variables:
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
Equivalent System in Row-Echelon Form:
x – 2y + 3z = 9
y + 3z = 5
z=2
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Example 1 – Using Back-Substitution in Row-Echelon Form
Solve the system of linear equations.
x – 2y + 3z = 9
Equation 1
y + 3z = 5
Equation 2
z=2
Equation 3
Solution:
y + 3(2) = 5
y = –1.
Substitute 2 for z.
Solve for y.
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Example 1 – Solution
x – 2(–1) + 3(2) = 9
x = 1.
cont’d
Substitute –1 for y and 2 for z.
Solve for x.
The solution is x = 1, y = –1, and z = 2, which can be
written as the ordered triple (1, –1, 2).
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Gaussian Elimination
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Gaussian Elimination
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Example 3 – Using Gaussian Elimination to Solve a System
Solve the system of linear equations.
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
Equation 1
Equation 2
Equation 3
Solution:
Add Equation 1 to Equation 2.
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Example 3 – Solution
cont’d
x – 2y + 3z = 9
y + 3z = 5
2x – 5y + 5z = 17
Multiply Equation 1 by –2.
Write Equation 3.
Add revised Equation 1 to Equation 3.
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Example 3 – Solution
cont’d
x – 2y + 3z = 9
y + 3z = 5
–y – z = –1
x – 2y + 3z = 9
y + 3z = 5
2z = 4
x – 2y + 3z = 9
y + 3z = 5
z=2
x = 1,
y = –1,
and
z = 2.
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Gaussian Elimination
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Gaussian Elimination
Solution: one point
Solution: one line
Solution: one plane
Solution: none
Solution: none
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Nonsquare Systems
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Nonsquare Systems
Square System: # equations = # unknowns
Nonsquare System: the number of equations differs from
the number of variables.
A system of linear equations cannot have a unique solution
unless there are at least as many equations as there are
variables in the system.
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Example 6 – A System with Fewer Equations than Variables
Solve the system of linear equations.
x – 2y + z = 2
2x – y – z = 1
Equation 1
Equation 2
Solution:
x – 2y + z = 2
3y – 3z = –3
x – 2y + z = 2
y – z = –1
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Example 6 – Solution
cont’d
y = z – 1.
x – 2y + z = 2
x – 2(z – 1) + z = 2
x – 2z + 2 + z = 2
x=z
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Example 6 – Solution
cont’d
Let z = a, where a is a real number, you have the solution
x = a,
y = a – 1,
and
z = a.
So, every ordered triple of the form
(a, a – 1, a)
a is a real number.
is a solution of the system.
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Example
 2x  4 y  z  7

2 x  4 y  2 z  6
 x  4y  z  0

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Applications
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Example 7 – Vertical Motion
The height at time t of an object that is moving in a
(vertical) line with constant acceleration a is given by the
position equation
The height s is measured in feet, the acceleration a is
measured in feet per second squared, t is measured in
seconds, v0 is the initial velocity (at t = 0), and s0 is the
initial height.
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Example 7 – Vertical Motion
cont’d
Find the values of a, v0, and s0 if s = 52 at t = 1, s = 52 at
t = 2, and s = 20 at t = 3, and interpret the result.
(See Figure 7.17.)
Figure 7.17
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Example 7 – Solution
By substituting the three values of t and s into the position
equation, you can obtain three linear equations in a, v0, and
s0.
When t = 1:
a(1)2 + v0(1) + s0 = 52
When t = 2:
a(2)2 + v0(2) + s0 = 52
2a + 2v0 + s0 = 52
When t = 3:
a(3)2 + v0(3) + s0 = 20
9a + 6v0 + 2s0 = 40
a + 2v0 + 2s0 = 104
This produces the following system of linear equations.
a + 2v0 + 2s0 = 104
2a + 2v0 + s0 = 52
9a + 6v0 + 2s0 = 40
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Example 7 – Solution
cont’d
Now solve the system using Gaussian elimination.
a + 2v0 + 2s0 = 104
– 2v0 – 3s0 = –156
9a + 6v0 + 2s0 = 40
a + 2v0 + 2s0 = 104
– 2v0 – 3s0 = –156
– 12v0 – 16s0 = –896
a + 2v0 + 2s0 = 104
– 2v0 – 3s0 = –156
2s0 = 40
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Example 7 – Solution
cont’d
a + 2v0 + 2s0 = 104
v0 + s0 = 78
s0 = 20
So, the solution of this system is a = –32, v0 = 48, and
s0 = 20, which can be written as (–32, 48, 20).
This solution results in a position equation of
s = –16t2 + 48t + 20 and implies that the object was thrown
upward at a velocity of 48 feet per second from a height of
20 feet.
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Group Assignment
x 2  y 2  Dx  Ey  F  0
Find the equation of the circle
that passes through the points (-3,-1), (2,4), (-6, 8)
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