Transcript Document

Chapter 9: Systems of Equations and
Inequalities; Matrices
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
a1 x1  a2 x2    an xn  b 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.
9.1 Linear System in Two Variables
•
Three possible solutions to a linear system
in two variables:
1. One solution: coordinates of a point,
2. No solutions: inconsistent case,
3. Infinitely many solutions: dependent
case.
9.1
• Characteristics of a system of two linear
equations in two variables.
Graphs
Nonparallel
lines
Identical lines
Parallel lines
# of solutions
Classification
one
consistent
infinite
dependent &
consistent
inconsistent
No solutions
9.1 Substitution Method
Example Solve the system.
3x  2 y  11
x  y  3
Solution y  x  3
3 x  2( x  3)  11
3 x  2 x  6  11
5x  5
x 1
y  1 3
y4
(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)}
9.1 Solving a Linear System in Two
Variables Graphically
ExampleSolve the system graphically.
3x  2 y  11
x  y  3
Solution Solve (1) and (2) for y.
(1)
(2)
• Solve using the method of graphing.
x2 + y2 = 25
x2 + y = 19
Solving a Nonlinear System of
Equations
Example Solve the system.
3x 2  2 y  5
x  3y   4
(1)
(2)
Solution Choose the simpler equation, (2), and
solve for y since x is squared in (1).
x  3 y  4
4 x
y
(3)
3
Substitute
4 x
3
for y into (1) .
Solving a Nonlinear System of
Equations
 4  x

3 x  2
5
 3 
9 x 2  2(4  x)  15
2
9x2  2x  7  0
7
(9 x  7)( x  1)  0  x  1 or x 
9
Substitute these values for x into (3).
 4   79 
43
y

or
3
27
The solution set is
 4   1
y
 1
3
79 ,  2743 ,  1,  1.
9.2 Elimination Method
Example Solve the system.
3x  4 y  1
2 x  3 y  12
(1)
(2)
Solution To eliminate x, multiply (1) by –2 and
(2)
by 3 and add the resulting equations.
(3)
 6 x  8 y  2
6 x  9 y  36
(4)
17 y  34
y2
9.2 Elimination Method
Substitute 2 for y in (1) or (2).
3 x  4(2)  1
3x  9
x3
The solution set is {(3, 2)}.
Consistent System
Solving an Inconsistent System
Example Solve the system.
3x  2 y  4
 6x  4 y  7
(1)
(2)
Solution Eliminate x by multiplying (1) by 2 and
adding the result to (2).
6x  4 y  8
 6x  4 y  7
0  15
Solution set is .
Inconsistent System
Solving a System with Dependent
Equations
Example Solve the system.
 4x  y  2
8x  2 y   4
(1)
(2)
Solution Eliminate x by multiplying (1) by 2
and adding the
result to (2).
 8x  2 y  4
8 x  2 y  4
0 0
Solution set is all R’s.
Consistent &
Dependent System
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.
A. The equation of each line is
y = x + 3.
B. The lines in each graph are the same and represent all of
the solutions to the equation y = x + 3.
C. The graph on the right is shaded above the line and this
means that all of these points are solutions as well.
Point: (-4, 5)
yx3
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.
5  4  3
5  1
5  1
5  1
or
5  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 > -1.
The area above the line is shaded.
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.
y  x  4
3  1  4
3  3
3  3
3  3
or
 3 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 < 3.
The area below the line is shaded.
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
(<) or greater than (>).
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.
Graph the following linear system of inequalities.
y  2x  4
Use the slope and yintercept to plot two
points for the first
inequality.
y  3x  2
y
Draw in the line. For 
use a solid line.
x
Pick a point and test
it in the inequality.
Shade the appropriate
region.
Graph the following linear system of inequalities.
y  2x  4
y  2x  4
0  2(0) - 4
0  -4
y  3x  2
y
Point (0,0)
The region above the line
should be shaded.
x
Now do the same for the
second inequality.
Graph the following linear system of inequalities.
y  2x  4
Use the slope and yintercept to plot two
points for the second
inequality.
y  3x  2
y
Draw in the line. For <
use a dashed line.
x
Pick a point and test
it in the inequality.
Shade the appropriate
region.
Graph the following linear system of inequalities.
y  2x  4
y  3x  2
y
y  3x  2
-2  3(-2) + 2
-2 < 8
Point (-2,-2)
The region below the line
should be shaded.
x
Graph the following linear system of inequalities.
y  2x  4
y  3x  2
y
x
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.
Graph the following linear systems of inequalities.
1.
y  x  4
yx2
y  x  4
yx2
y
Use the slope and yintercept to plot two
points for the first
inequality.
x
Draw in the line.
Shade in the
appropriate region.
y  x  4
yx2
y
Use the slope and yintercept to plot two
points for the second
inequality.
x
Draw in the line.
Shade in the
appropriate region.
y  x  4
yx2
y
The final solution is the
region where the two
shaded areas overlap
(purple region).
x
Sect. 9.4 Linear Programming
Goal 1
Goal 2
Find Maximum and Minimum
values of a function
Solve Real World Problems with
Linear Programming
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.
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.
Example 1
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.
y4
y-x+6
1
3
y  x
2
2
y  6x + 4
y4
6
y-x+6
4
1
3
y  x
2
2
y  6x + 4
2
-5
5
-2
-4
-6
The polygon formed is a
quadrilateral with vertices at (0, 4),
(2, 4), (5, 1), and (- 1, - 2).
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).
Example 2
1
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.
12
10
y-3
8
x-2
6
4
5
y   x7
2
2
-10
-5
5
-2
-4
10
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)
(4, - 3)
2(4) – 5(- 3)
11
23
The Maximum Value is 23 at (4, - 3).
The Minimum Value is – 64 at (- 2, 12).
Sometimes a system of Inequalities
forms a region that is not a closed
polygon.
In this case, the region is said
to be Unbounded.
Example 3
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.
10
8
y + 3x  4
6
y  - 3x – 4
4
2
y8+x
-10
-5
5
-2
-4
-6
10
There are only two points of
intersection (- 1, 7) and (- 3, 5).
This is an Unbounded Region.
(x, y)
4y – 3x
F(x, y)
4(7) – 3(31
1)
(- 3, 5)
4(5) – 3(29
The Maximum Value is 313)
at (- 1, 7). There is no
(- 1, 7)
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.
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. Answer the
problem.
Example 4
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.
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 can’t 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
Step 3: Graph the system of Inequalities
16
14
12
10
cakes
8
6
4
2
5
10
Pies
15
20
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
Step 6: Substitute the coordinates of the vertices into function.
(p, c)
(0, 0)
(0, 32)
(60, 8)
(60, 0)
16p + 34c
16(0) + 34(0)
16(0) + 34(32)
16(60) + 34(8)
16(60) + 34(0)
f(p, c)
0
1088
1232
960
Step 7: Select the Greatest or Least result. Answer the
Problem.
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.
Section 9.5
Solving Systems using Matrices
•
•
•
•
What is a Matrix? [ ]
Augmented Matrices
Solving Systems of 3 Equations
Inconsistent & Dependent Systems
Concept:
A Matrix
• Any rectangular array of numbers arranged
in rows and columns, written within brackets
• Examples:
2  2
9
1
0 17 


 2  1 4 
1 4 2
0 1 7 


 9 11 
 2 1 


 0 22 


 5  3
• Matrix Row Transformations
– Streamlined use of echelon methods
a1 x  b1 y  c1 z  d1
a2 x  b2 y  c2 z  d 2
a3 x  b3 y  c3 z  d3

  a1 b1 c1 d1 
Rows  a2 b2 c2 d 2 


  a3 b3 c3 d3 
   
Columns
9.5 Solution of Linear Systems by Row
Transformations
 a1 b1 c1 d1 
a b c d 
 2 2 2 2
 a3 b3 c3 d3 
– 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.
Concept:
Augmented 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
 9a  2b  2c  1
x  4 y  2

a
 17c  4


y7

 2a  b  4c  3

1 4 2
0 1 7 


2 2
9
1
0 17

 2  1 4
1
4

 3
Reduced Row Echelon Method
with the Graphing Calculator
Example Solve the system.
x  2 y  z  30
3 x  2 y  2 z  56
2x  3y 
z  44
Solution The augmented matrix of the system
is
shown below.
9.5 Reduced Row Echelon Method
with the Graphing Calculator
Using the rref command we obtain the row
reduced
echelon form.
x 8
This represents the system y  6 , with solution set {(8, 6, 10)}.
z 10
9.5 Solving a System with No Solutions
Example Show that the following system is
inconsistent.
x  2y  z  4
x  2 y  3z  1
2x  4 y  2z  9
Solution The augmented matrix of the system
is
 1 2 1 4
 1  2 3 1.


2 4 2 9
9.5 Solving a System with No
Solutions
 1 2 1 4
 1  2 3 1


0 0 0 1
 2R 1  R 3  R 3
The final row indicates that the system is
inconsistent and has solution set .
9.5 Solving a System with Dependent
Equations
Example Show that the system has dependent
equations. Express the general solution using
an
arbitrary variable.
2 x  5 y  3z  1
x  4 y  2z  8
4 x  10 y  6 z  2
Solution The augmented matrix is
2  5 3 1
 1 4  2 8.


4  10 6 2
9.5 Solving a System with Dependent
Equations
2  5 3 1
 1 4  2 8


0 0 0 0
 2R 1  R 3  R 3
The final row of 0s indicates that the system
has
dependent equations. The first two rows
represent 2 x  5 y  3 z  1
.
the system
x  4 y  2z  8
9.5 Solving a System with Dependent
Equations
Solving for y we get
15 7
y   z.
13 13
Substitute this result into the expression to find
x.
x  4 y  2z  8
15 7 

x  4  z   2 z  8
 13 13 
60 28
x   z  2z  8
13 13
44 2
x  z
13 13
 44132 z , 15137 z , z .
Solution set written with z arbitrary:
Matrix Algebra Basics
9.6
Algebra
Matrix
A matrix is any doubly subscripted array of
elements arranged in rows and columns.
a11 ,, a1n 
a 21 ,, a 2 n 

A
 


am1 , , amn 
Row Matrix
[1 x n] matrix
A a1 a2 ,, an
Column Matrix
[m x 1] matrix
a1 
a 2 
A 
 
 
 am 
Square Matrix
Same number of rows and columns
5 4 7
B  3 6 1 


2 1 3 
The
Identity
Identity Matrix
Square matrix with ones on the
diagonal and zeros elsewhere.
1
0
I  
0
0

0 0 0 
1 0 0 

0 1 0

0 0 1
Matrix Addition and Subtraction
A new matrix C may be defined as the
additive combination of matrices A and B
where: C = A + B
is defined by:
Cmn Amn Bmn
Note: all three matrices are of the same dimension
Addition
I
f
a11 a12 
A 

a 21 a 22 

and
b11 b12 
B 

b 21 b 22

then
a11  b11 a12  b12 
C 

a 21  b 21 a 22  b22 

Matrix Addition Example
3 4  1 2  4 6 
A  B 


 C

5 6 
 
3 4 
 
8 10 

Matrix Subtraction
C = A - B
Is defined by
Cmn Amn Bmn
Matrix Multiplication
Matrices A and B have these dimensions:
[r x c] and [s x d]
Matrix Multiplication
Matrices A and B can be multiplied if:
[r x c] and [s x d]
c=s
Matrix Multiplication
The resulting matrix will have the dimensions:
[r x c] and [s x d]
rxd
Computation: A x B = C
a11 a12 
A 

a 21 a 22 

[2 x 2]
b11 b12 b13 
B 

b 21 b 22 b 23

[2 x 3]
a11b11  a12b21 a11b12  a12b22 a11b13  a12b23 
C

a 21b11  a 22b21 a 21b12  a 22b22 a 21b13  a 22b23
[2 x 3]
Computation: A x B = C
2 3
1 1 1 

A  1 1  and B  

1
0
2




1 0 
[3 x 2]
[2 x 3]
A and B can be multiplied
2 *1  3 *1  5 2 *1  3 * 0  2 2 *1  3 * 2  8 5 2 8
C  1*1  1*1  2 1*1  1* 0  1 1*1  1* 2  3   2 1 3 
1*1  0 *1  1 1*1  0 * 0  1 1*1  0 * 2  1  111 
[3 x 3]
Computation: A x B = C
2 3
1 1 1 

A  1 1  and B  

1
0
2




1 0 
[3 x 2]
[2 x 3]
Result is 3 x 3
2 *1  3 *1  5 2 *1  3 * 0  2 2 *1  3 * 2  8 5 2 8
C  1*1  1*1  2 1*1  1* 0  1 1*1  1* 2  3   2 1 3 
1*1  0 *1  1 1*1  0 * 0  1 1*1  0 * 2  1  111 
[3 x 3]