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Title: Chapter 9: Systems of Equations and Inequalities; Matrices

1
Chapter 9 Systems of Equations and Inequalities
Matrices
2
9.1 Systems of Equations
• A set of equations is called a system of
equations.
• The solutions must satisfy each equation in the
system.
• A linear equation in n unknowns has the form
• where
the variables are of first-degree.
• If all equations in a system are linear, the
system is a system of linear equations, or a
linear system.

3
9.1 Linear System in Two Variables
• Three possible solutions to a linear system in
two variables
• One solution coordinates of a point,
• No solutions inconsistent case,
• Infinitely many solutions dependent case.

4
9.1
• Characteristics of a system of two linear
equations in two variables.

Graphs of solutions Classification
Nonparallel lines one consistent
Identical lines infinite dependent consistent
Parallel lines No solutions inconsistent
5
9.1 Substitution Method
• Example Solve the system.
• Solution

(1) (2)
Solve (2) for y.
Substitute y x 3 in (1).
Solve for x.
Substitute x 1 in y x 3.
Solution set (1, 4)
6
9.1 Solving a Linear System in Two Variables
Graphically
• Example Solve the system graphically.
• Solution Solve (1) and (2) for y.

(1) (2)
7
• Solve using the method of graphing.
• x2 y2 25
• x2 y 19

8
Solving a Nonlinear System of Equations
• Example Solve the system.
• Solution Choose the simpler equation, (2), and
• solve for y since x is squared in (1).
• Substitute for y into (1) .

(1) (2)
9
Solving a Nonlinear System of Equations
• Substitute these values for x into (3).
• The solution set is

10
9.2 Elimination Method
• Example Solve the system.
• Solution To eliminate x, multiply (1) by 2 and
(2)
• by 3 and add the resulting equations.

(1) (2)
(3) (4)
11
9.2 Elimination Method
• Substitute 2 for y in (1) or (2).
• The solution set is (3, 2).
• Consistent System

12
Solving an Inconsistent System
• Example Solve the system.
• Solution Eliminate x by multiplying (1) by 2 and
• adding the result to (2).
• Solution set is ?.

(1) (2)
Inconsistent System
13
Solving a System with Dependent Equations
• Example Solve the system.
• Solution Eliminate x by multiplying (1) by 2 and
• result to (2).
• Consistent
Dependent System
• Solution set is all Rs.

(1) (2)
14
9.3 Graphing Systems of Inequalities
15
Look at the two graphs. Determine the
following A. The equation of each line. B. How
the graphs are similar. C. How the graphs are
different.
1. The equation of each line is y x 3.
2. The lines in each graph are the same and
represent all of the solutions to the equation y
x 3.
3. The graph on the right is shaded above the line
and this means that all of these points are
solutions as well.

16
Inequalities with Greater Than
Point (-4, 5)
Pick a point from the shaded region and test that
point in the equation y x 3.
This is incorrect. Five is greater than or equal
to negative 1.
If a solid line is used, then the equation would
be 5 -1. If a dashed line is used, then the
equation would be 5 gt -1. The area above the line
17
Inequalities with Less Than
Point (1, -3)
Pick a point from the shaded region and test that
point in the equation y -x 4.
This is incorrect. Negative three is less than
or equal to 3.
If a solid line is used, then the equation would
be -3 3. If a dashed line is used, then the
equation would be -3 lt 3. The area below the line
18
Graphing Linear Inequalities
1. Write the inequality in slope-intercept form.
2. Use the slope and y-intercept to plot two points.
3. Draw in the line. Use a solid line for less than
or equal to (?) or greater than or equal to ().
Use a dashed line for less than (lt) or greater
than (gt).
4. Pick a point above the line or below the line.
Test that point in the inequality. If it makes
the inequality true, then shade the region that
contains that point. If the point does not make
the inequality true, shade the region on the
other side of the line.
5. Systems of inequalities Follow steps 1-4 for
each inequality. Find the region where the
solutions to the two inequalities would overlap
and this is the region that should be shaded.

19
Example
Graph the following linear system of inequalities.
Use the slope and y-intercept to plot two points
for the first inequality.
Draw in the line. For ? use a solid line.
Pick a point and test it in the inequality.
20
Example
Graph the following linear system of inequalities.
The region above the line should be shaded.
Now do the same for the second inequality.
21
Example
Graph the following linear system of inequalities.
Use the slope and y-intercept to plot two points
for the second inequality.
Draw in the line. For lt use a dashed line.
Pick a point and test it in the inequality.
22
Example
Graph the following linear system of inequalities.
The region below the line should be shaded.
23
Example
Graph the following linear system of inequalities.
The solution to this system of inequalities is
the region where the solutions to each inequality
overlap. This is the region above or to the left
of the green line and below or to the left of the
blue line. Shade in that region.
24
You Try It
Graph the following linear systems of
inequalities.
25
Problem 1
Use the slope and y-intercept to plot two points
for the first inequality.
Draw in the line.
26
Problem 1
Use the slope and y-intercept to plot two points
for the second inequality.
Draw in the line.
27
Problem 1
The final solution is the region where the two
28
Sect. 9.4 Linear Programming
Goal 1 Find Maximum and Minimum values of a
function Goal 2 Solve Real World Problems
with Linear Programming
29
When graphing a System of Linear Inequalities
Each linear inequality is called a Constraint
The intersection of the graphs is called the
Feasible Region
When the graphs of the constraints is a
polygonal region, we say the region is Bounded.
30
Sometimes it is necessary to find the Maximum or
Minimum values that a linear function has for the
points in a feasible region.
The Maximum or Minimum value of a related
function Always occurs at one of the Vertices of
the Feasible Region.
31
Graph the system of inequalities. Name the
coordinates of the vertices of the feasible
region. Find the Maximum and Minimum values of
the function f(x, y) 3x 2y for this
polygonal region.
32
The polygon formed is a quadrilateral with
vertices at (0, 4), (2, 4), (5, 1), and (- 1, -
2).
33
Use a table to find the maximum and minimum
values of the function.
(x, y) 3x 2y f(x, y)
(0, 4) 3(0) 2(4) 8
(2, 4) 3(2) 2(4) 14
(5, 1) 3(5) 2(1) 17
(- 1, - 2) 3(- 1) 2(- 2) - 7
The maximum value is 17 at (5, 1). The Minimum
value is 7 at (- 1, - 2).
34
Bounded Region
Graph the system of inequalities. Name the
coordinates of the vertices of the feasible
region. Find the Maximum and Minimum values of
the function f(x, y) 2x - 5y for this
polygonal region.
35
The polygon formed is a triangle with vertices at
(- 2, 12), (- 2, - 3), and (4, - 3)
(x, y) 2x 5y F(x, y)
(- 2, 12) 2(-2) 5(12) - 64
(- 2, - 3) 2(- 2) 5(- 3) 11
(4, - 3) 2(4) 5(- 3) 23
The Maximum Value is 23 at (4, - 3). The Minimum
Value is 64 at (- 2, 12).
36
Sometimes a system of Inequalities forms a region
that is not a closed polygon.
In this case, the region is said to be Unbounded.
37
Unbounded Region
Graph the system of inequalities. Name the
coordinates of the vertices of the feasible
region. Find the Maximum and Minimum values of
the function f(x, y) 4y 3x for this region.
y 3x ? 4 y ? - 3x 4 y ? 8 x
38
There are only two points of intersection (- 1,
7) and (- 3, 5). This is an Unbounded Region.
(x, y) 4y 3x F(x, y)
(- 1, 7) 4(7) 3(- 1) 31
(- 3, 5) 4(5) 3(- 3) 29
The Maximum Value is 31 at (- 1, 7). There is no
minimum value since there are other points in the
solution that produce lesser values. Since the
region is Unbounded, f(x, y) has no minimum value.
39
Linear Programming Procedures.
Step 1 Define the Variables
Step 2 Write a system of Inequalities
Step 3 Graph the System of Inequalities
Step 4 Find the coordinates of the vertices of
the feasible region.
Step 5 Write a function to be maximized or
minimized.
Step 6 Substitute the coordinates of the
vertices into the function.
Step 7 Select the greatest or least result.
40
Ingrid is planning to start a home-based
business. She will be baking decorated cakes and
specialty pies. She estimates that a decorated
cake will take 75 minutes to prepare and a
specialty pie will take 30 minutes to prepare.
She plans to work no more than 40 hours per week
and does not want to make more than 60 pies in
any one week. If she plans to charge 34 for a
cake and 16 for a pie, find a combination of
cakes and pies that will maximize her income for
a week.
41
Step 1 Define the Variables
C number of cakes P number of pies
Step 2 Write a system of Inequalities
Since number of baked items cant be negative, c
and p must be nonnegative c ? 0 p ? 0
A cake takes 75 minutes and a pie 30 minutes.
There are 40 hours per week available. 75c 30p
? 2400 40 hours 2400 min.
She does not want to make more than 60 pies each
week p ? 60
42
Step 3 Graph the system of Inequalities
cakes
Pies
43
Step 4 Find the Coordinates of the vertices of
the feasible region.
The vertices of the feasible region are (0, 0),
(0, 32), (60, 8), and (60, 0).
Step 5 Write a function to be maximized or
minimized.
The function that describes the income is f(p,
c) 16p 34c
44
Step 6 Substitute the coordinates of the
vertices into function.
(p, c) 16p 34c f(p, c)
(0, 0) 16(0) 34(0) 0
(0, 32) 16(0) 34(32) 1088
(60, 8) 16(60) 34(8) 1232
(60, 0) 16(60) 34(0) 960
Step 7 Select the Greatest or Least result.
The maximum value of the function is 1232 at (60,
8). This means that the maximum income is 1232
when Ingrid makes 60 pies and 8 cakes per week.
45
Section 9.5Solving Systems using Matrices
• What is a Matrix?
• Augmented Matrices
• Solving Systems of 3 Equations
• Inconsistent Dependent Systems

46
ConceptA Matrix
• Any rectangular array of numbers arranged in rows
and columns, written within brackets
• Examples

47
• Matrix Row Transformations
• Streamlined use of echelon methods

48
9.5 Solution of Linear Systems by Row
Transformations
• This is called an augmented matrix where each
member of the array is called an element or
entry.
• The rows of an augmented matrix can be treated
just like the equations of a linear system.

49
ConceptAugmented Matrices
• Are used to solve systems of linear equations
• Arrange equations in simplified standard form
• Put all coefficients into a 2x3 or 3x4 augmented
matrix

50
Reduced Row Echelon Method with the Graphing
Calculator
• Example Solve the system.
• Solution The augmented matrix of the system is
• shown below.

51
9.5 Reduced Row Echelon Method with the
Graphing Calculator
• Using the rref command we obtain the row reduced
• echelon form.

52
9.5 Solving a System with No Solutions
• Example Show that the following system is
• inconsistent.
• Solution The augmented matrix of the system is

53
9.5 Solving a System with No Solutions
• The final row indicates that the system is
• inconsistent and has solution set ?.

54
9.5 Solving a System with Dependent Equations
• Example Show that the system has dependent
• equations. Express the general solution using an
• arbitrary variable.
• Solution The augmented matrix is

55
9.5 Solving a System with Dependent Equations
• The final row of 0s indicates that the system has
• dependent equations. The first two rows represent
• the system

56
9.5 Solving a System with Dependent Equations
• Solving for y we get
• Substitute this result into the expression to
find x.
• Solution set written with z arbitrary

57
Matrix Algebra Basics
• 9.6

58
Algebra
59
Matrix
A matrix is any doubly subscripted array of
elements arranged in rows and columns.
60
Row Matrix
1 x n matrix

61
Column Matrix
m x 1 matrix
62
Square Matrix
Same number of rows and columns
63
The Identity
64
Identity Matrix
Square matrix with ones on the diagonal and zeros
elsewhere.
65
A new matrix C may be defined as the additive
combination of matrices A and B where C
A B  is defined by
Note all three matrices are of the same dimension
66
If
and
then
67
68
Matrix Subtraction
C A - B Is defined by
69
Matrix Multiplication
Matrices A and B have these dimensions
r x c and s x d
70
Matrix Multiplication
Matrices A and B can be multiplied if
r x c and s x d
c s
71
Matrix Multiplication
The resulting matrix will have the dimensions
r x c and s x d
r x d
72
Computation A x B C
2 x 2
2 x 3
2 x 3
73
Computation A x B C

3 x 2
2 x 3
A and B can be multiplied
3 x 3
74
Computation A x B C

3 x 2
2 x 3
Result is 3 x 3
3 x 3