Transcript A, B - HCC Learning Web

Chapter 5
Systems and Matricies
Copyright © 2004 Pearson Education, Inc.
5.1
Systems of Linear Equations
Copyright © 2004 Pearson Education, Inc.
Linear Systems


Any equation of the form
a1x1 + a2x2 +    + anxn = b,
for all real numbers a1,a2,…,an (not all of which
are 0) and b, is a linear equation or a firstdegree equation in n unknowns.
A set of equations is called a systems of
equations. The solutions of a system of
equations must satisfy every equation in the
system. If all the equations in a system are
linear, the system is a system of linear
equations, or a linear system.
Copyright © 2004 Pearson Education, Inc.
Slide 5-4
Linear Systems
Consistent
 The graphs of two
equations intersect a
single point.
 The coordinates of this
point give the only
solution of the system.
Copyright © 2004 Pearson Education, Inc.
Slide 5-5
Linear Systems
Inconsistent
 The graphs are distinct
parallel lines.
 The equations are
independent. That is,
there is no solution
common to both
equations.
Copyright © 2004 Pearson Education, Inc.
Slide 5-6
Linear Systems
Dependent
 The graphs are the same
line.
 Any solution of one
equation is also the
solution of the other. Thus
there are infinite number
of solutions.
Copyright © 2004 Pearson Education, Inc.
Slide 5-7
Substitution Method
In a system of two equations with two variables,
the substitution method involves using one
equation to find an expression for one variable in
terms of the other, then substituting into the
other equation of the system.
Example: Solve the system.
4x + 2y = 8
(1)
3x  7y = 11
(2)
Copyright © 2004 Pearson Education, Inc.
Slide 5-8
Solution


4x + 2y = 8
3x  7y = 11
Begin by solving one of the
equations for one of the
variables.
4x + 2y = 8
2y = 4x + 8
y = 2x + 4
(3)
Now replace y with 2x + 4 in
the second equation and solve
for x.
3x  7(2x + 4) = 11
3x + 14x  28 = 11
17x = 17
x=1
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(1)
(2)


Replace x with 1 in equation
(3) to obtain y = 2(1) + 4 = 2.
The solution of the ordered
pair is (1,2).
Check the solution in both
equations (1) and (2).
4x + 2y = 8
4(1) + 2(2) = 8
8=8
3x  7y = 11
3(1) 7(2) = 11
11 =  11
Slide 5-9
Elimination Method

The elimination method uses multiplication
and addition to eliminate a variable from one
equation. To eliminate a variable, the
coefficients of that variable in the two
equations must be additive inverses. To
achieve this, we use properties of algebra to
change the system to an equivalent system,
one with the same solution set.
Copyright © 2004 Pearson Education, Inc.
Slide 5-10
Equivalent Systems
Transformations of a Linear System
 Interchange any two equations of the system.
 Multiply or divide any equation of the system by
a nonzero real number.
 Replace any equation of the system by the sum
of that equation and a multiple of another
equation in the system.
Copyright © 2004 Pearson Education, Inc.
Slide 5-11
Example
Solve the system using the elimination method.
6x + 2y = 4
10x + 7y = 8
If we multiply the first equation by 5 and the second equation
by 3, we will be able to eliminate the x variable.
30x + 10y = 20
30x  21y = 24
 11y = 44
y=4
Copyright © 2004 Pearson Education, Inc.
Substituting:
6x + 2y = 4
6x + 2(4) = 4
6x  8 = 4
6x = 12
The solution is (2,  4)
x=2
Slide 5-12
Solving an Applied Problem by Writing
a System of Equations






Step 1 Read the problem carefully until you understand what
is given and what is to be found.
Step 2 Assign variables to represent the unknown values,
using diagrams or tables as needed. Write down what
each variable represents.
Step 3 Write a system of equations that relates the
unknowns.
Step 4 Solve the system of equations.
Step 5 State the answer to the problem. Does it seem
reasonable?
Step 6 Check the answer in the words of the original
problem.
Copyright © 2004 Pearson Education, Inc.
Slide 5-13
Solving Systems with Three Unknowns
(Variables)

The graph of a linear equation in three unknowns requires a threedimensional coordinate system. Some of the possible intersections
of planes representing three equations are shown below.
Copyright © 2004 Pearson Education, Inc.
Slide 5-14
Systems of Three Equations with Three
Variables


To solve a system with three unknowns, first eliminate a
variable from any two of the equations. Then eliminate
the same variable from a different pair of equations.
Eliminate a second variable using the resulting two
equations in two variables to get an equation with just
one variable whose value you can now determine. Find
the values of the remaining variables by substitution.
Solutions of the systems are written as ordered triples.
Example: Solve the system.
3x + 9y + 6z = 3 (1)
2x + y  z = 2 (2)
x+y+z=2
Copyright © 2004 Pearson Education, Inc.
(3)
Slide 5-15
Solution


3x + 9y + 6z = 3
2x + y  z = 2
x+y+z=2
Eliminate z by adding
equations (2) and (3) to
get 3x + 2y = 4 (4)
To eliminate z from
another pair of equations,
multiply both sides of
equations (2) by 6 and
add the result to equation
(1).
3x + 9y + 6z = 3
12x + 6y  6z = 12 (1)
15x + 15y
Copyright © 2004 Pearson Education, Inc.
= 15 (5)

(1)
(2)
(3)
To eliminate x from the
equations (4) and (5),
multiply both sides of
equation (4) by 5 and add
the result to equation (5).
Solve the new equation for
y.
15x  10y = 20
15x + 15y = 15
5y = 5
y = 1
Slide 5-16
Solution continued


Using y = 1, find x from
equation (4) by substitution.
3x + 2(1) = 4 (4)
x=2
Substitute 2 for x and 1 for
y in equation (3) to find z.
2 + (1 ) + z = 2
(3)
z=1
Copyright © 2004 Pearson Education, Inc.


Verify that the ordered triple
(2, 1, 1) satisfies all three
equations in the original
system.
The solution set is
{(2, 1, 1)}.
Slide 5-17
Using Curve Fitting to Find an Equation
Through Three Points
Example: Find the equation of the parabola
y = ax2 + bx + c that passes through (2,4), (1,
1), and (2,5).
Solution: Since the three points lie on the graph of the
equation y = ax2 + bx + c, they must satisfy the
equation. Substituting each ordered pair into the
equation gives three equations with three variables.
4 = a(2)2 + b (2) + c
or 4 = 4a + 2b + c
(1)
1 = a(1)2 + b(1) + c or 1 = a  b + c
(2)
5 = a(2)2 + b(2) + c or 5 = 4a  2b + c
(3)
Copyright © 2004 Pearson Education, Inc.
Slide 5-18
Using Curve Fitting to Find an Equation
Through Three Points continued

This system can be solved by
the elimination method. First
eliminate c using equations (1)
and (2).
4 = 4a + 2b + c (1)
1 = a + b  c
3 = 3a + 3b

Solving systems of equations
(4) and (5) in two variables by
eliminating a.
3 = 3a + 3b (4)
4 = 3a + b
1 =
4b
1
 b
4
(4)
Now, use equations (2) and (3) to
eliminate the same variable (c).
1 = a  b + c (2)
5 = 4a + 2b  c
4 = 3a + b
(5)
Copyright © 2004 Pearson Education, Inc.


(5)
1
Find a by substituting 4 for b in
equation (4), which is equivalent
to 1 = a + b.
1=a+b
1=a 1
4
5
a
4

Slide 5-19
Using Curve Fitting to Find an Equation
Through Three Points continued

Finally, find c by substituting a =
(2).
5
and
4
b=

1
in
4
equation
1 a bc
5  1
 c
4  4
6
1 c
4
1
 c
2
1

An equation of the parabola is y 
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5 2 1
1
x  x .
4
4
2
Slide 5-20
5.2
Matrix Solution of Linear
Systems
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Definitions

A matrix is a rectangular array of numbers
enclosed in brackets.

Each number is called an element of the matrix.

The size of a matrix is determined by the
number of row (horizontal) and columns
(vertical).
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Slide 5-22
Definitions continued

Linear System
x  4 y  3z  2
2x  5 y  4z  1
x yz 3

Augmented matrix
1 4 3 2 
 2 5 4 1 


1 1 1 3 
rows
columns
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Slide 5-23
Matrix Row Transformations




For any augmented matrix of a system of linear
equations, the following row transformations will
result in the matrix of an equivalent system.
1. Interchange any two rows.
2. Multiply or divide the elements of any row by
a nonzero real number.
3. Replace any row of the matrix by the sum of
the elements of that row and a multiple of the
elements of another row.
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Slide 5-24
Using the Gauss-Jordan Method to Put a
Matrix into Diagonal Form

Step 1

Step 2

Step 3

Step 4

Step 5
Obtain 1 as the first element of the first
column.
Use the first row to transform the remaining
entries in the first column to 0.
Obtain 1 as the second entry in the second
column.
Use the second row to transform the
remaining entries in the second column to 0.
Continue in this manner as far as possible.
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Slide 5-25
Example

Solve the system
5x  3y = 14
4x + y = 18
Write the augmented matrix.
5 3 14
 4 1 18


Work with the columns and rows so that it is
transformed into the form
1 0 k 
0 1 j 


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Slide 5-26
Example continued

Multiply row 1 by 1/5 to get a 1 in the first position.
1

4



1 18
3
5
1
5
14
5
R1
Introduce 0 in the second row by multiplying each element of R1 by 4
and adding to R2.
1

0

3
5
17
5
14
5
34
5


 4 R1  R 2
Obtain 1 in the second row, second column by multiplying each element
of the second row by 5/17.
1

0
3
5
1


2
14
5
Copyright © 2004 Pearson Education, Inc.
5
17
R2
Slide 5-27
Example continued

Finally, to get 0 in the first row, second column multiply each
element of the second row by 3/5 and add to the first row.
1 0 4 
0 1 2 



3
5
R 2  R1
This last matrix corresponds to the system
x=4
y=2
that has the solution set (4, 2).
Check the solution in both equations of the original system.
Copyright © 2004 Pearson Education, Inc.
Slide 5-28
Example

Solve the system.
x  y  2z  0
2x  2 y  z  6
x  3 y  3z  4

1 1 2 0 
2 2 1 6 
 

1 3 3 4 
1 in the first row, first column. Introduce 0 in the second row of the first
column by multiplying each element of row 1 by 2 and adding to row 2.
1 1 2 0 
0 4 3 6  2 R1  R 2


1 3 3 4 
Copyright © 2004 Pearson Education, Inc.
Slide 5-29
Example continued

To change the third element in the first column to 0, multiply
the first row by 1 and add.
1 1 2 0 
0 4 3 6 


0 2 1 4  1R1  R3

Use the same procedure to transform the second and third
columns. For both of these columns, perform the additional
step of getting 1 in the appropriate position of each column.
Do this by multiplying the elements of the row by the
reciprocal of the number in that position.
Copyright © 2004 Pearson Education, Inc.
Slide 5-30
Example continued
1 1
0 1

0 2
1 0

0 1
0 2
1 0

0 1
0 0
2
3
4
1
5
4
3
4
1
5
4
3
4
1
2
0
3 1
R2
2 4
4 


 R 2  R1
4 
3
2
3
2



1 2 R 2  R3
3
2
3
2
continued
1 0

0 1
0 0
5
4
3
4
1



2 2 R3
3
2
3
2
1 0 54 23 

3
0
1
0
3

 4 R3  R 2
0 0 1 2 
1 0 0 1 45 R3  R1
0 1 0 3 


0 0 1 2 
The solution set is (1, 3, 2).
Copyright © 2004 Pearson Education, Inc.
Slide 5-31
Special Systems

When a row consists entirely of 0’s, the
equations are dependent and the system is
equivalent.
1 0 4 6 
0 1 4 8 


0 0 0 0 
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Slide 5-32
Special Systems continued

When we obtain a row whose only nonzero entry
occurs in the last column, we have an
inconsistent system of equations. For example,
1 0 4 6 
0 1 4 8 


0 0 0 9 
the last row corresponds to the false equation
0 = 9, so we know the original system has no
solution.
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Slide 5-33
5.3
Determinant Solution of Linear
Systems
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Determinants

Every n  n matrix A is associated with a real
number called the determinant, of A, written |A|.
The determinant of a 2  2 matrix is defined as
follows.
a11 a12
 a11 a12 
If A = 
, then A 
 a11a22  a21a12 .

a21 a22
 a21 a22 

Note: Matrices are enclosed with square
brackets, while determinants are denoted with
vertical bars.
Copyright © 2004 Pearson Education, Inc.
Slide 5-35
Example


Let
 4 2
A
.

7 5 
Find |A|.
A  4  5  (2)  7
 20  (14)
 34
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Slide 5-36
Determinant of a 3  3 Matrix
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Slide 5-37
Cofactor

Let Mij be the minor for the element aij in an
n  n matrix. The cofactor if aij written Aij, is
Aij  (1)i  j  M ij .

Find the cofactor of the element (2).
M 32 

6 4
8 3
 14
6 2 4
8 9 3 


1 2 0 
23
(

1)
(14)  (1)(14)  14.
The cofactor is
Copyright © 2004 Pearson Education, Inc.
Slide 5-38
Finding the Determinant of a Matrix

Multiply each element in any row or column of
the matrix by its cofactor. The sum of these
products give the value of the determinant.
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Slide 5-39
Example

Evaluate
3 4 3
2 5 0 ,
2 0 4
expanding by the third row.’
4 3
M 31 
 0  15  15
5 0
M 32 
3
3
2
0
 0  (6)  6
3 4
M 33 
 15  8  23
2 5
Copyright © 2004 Pearson Education, Inc.
Slide 5-40
Example continued

Now find the cofactor of each element of these minors.
A31  (1)31  M 31  (1) 4 (15)  15
A32  (1)3 2  M 32  (1)5 (6)  6
A33  (1)3 3  M 33  (1) 6 (23)  23

Find the determinant by multiplying each cofactor by its
corresponding element in the matrix and finding the sum of
these products.
3
4 3
2 5
0  2(15)  0(6)  4(23)
2
4
0
 30  0  92  62
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Slide 5-41
Cramer’s Rule for Two Equations in
Two Variables
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Slide 5-42
Example

Use Cramer’s Rule to
solve the system.
7x + 3y = 15
2x + 9y = 12
Find D first, since if D = 0,
Cramer’s Rule does not
apply.
The solution set is {(3, 2)}.
Copyright © 2004 Pearson Education, Inc.

7 3
D
 7(9)  3(2)  57
2 9
Dx 
15
3
12 9
 15(9)  3(12)  171
7 15
Dy 
 7(12)  2(15)  114
2 12
Dx 171

3
D
57
Dy 114
y

 2
D
57
x
Slide 5-43
General Form of Cramer’s Rule
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Slide 5-44
Example

Use Cramer’s Rule to
solve the system.
x  y  2z  0
3x  y  2 z  8
x  3y  4z  7
1 1 2
D  3 1 2  8
1 3 4

0
1
2
Dx  8 1
7 3
2  16
4
1 0 2
Dy  3 8 2  8
1 7
4
1 1 0
Dz  3 1 8  4
1 3 7
Copyright © 2004 Pearson Education, Inc.
Slide 5-45
Example continued

Thus:
Dx 16
x

2
D
8
Dy 8
y

 1
D
8
Dz 4 1
z
 
D 8 2

1

The solution set is 2, 1, .
2

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Slide 5-46
5.4
Partial Fractions
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Decomposition of Rational Expressions


The sums of rational expressions are found by
combining two or more rational expressions into
one rational expression.
A special type of sum of rational expressions is
called the partial fraction decomposition, each
term in the sum is a partial fraction.
Add rational expressions
2
3 5x  3
 
x  1 x x( x  1)
Partial fraction decomposition
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Slide 5-48
Partial Fraction Decomposition
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Slide 5-49
Partial Fraction Decomposition continued
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Slide 5-50
Distinct Linear Factors
Example: Find the partial fraction decomposition of
2 x4  8x2  5x  2
.
3
x  4x
Solution: The given fraction is not a proper fraction; the
numerator has greater degree than the denominator.
Perform the division.
2x
x3  4 x 2 x 4  8 x 2  5 x  2
2 x4  8x2
5x  2
Copyright © 2004 Pearson Education, Inc.
Slide 5-51
Distinct Linear Factors continued
The quotient is
2 x4  8x2  5x  2
5x  2

2
x

.
3
3
x  4x
x  4x
Now, work with the remainder
fraction. Factor the denominator as x3  4x = x(x + 2)(x  2).
Since the factors are distinct
linear factors, use Step 3(a) to
write the decomposition as
5x  2 A
B
C
 

,
x3  4 x x x  2 x  2
Copyright © 2004 Pearson Education, Inc.
Where A, B, and C are
constants that need to be
found. Multiply both sides of
the equation (1) by
x(x + 2)(x  2) to obtain
5x  2 = A(x + 2)(x  2) + Bx(x
 2) + Cx(x + 2). Equation (1)
is an identity as is equation (2).
Equation (1) holds for all
values of x except 0, 2, and 2.
However, equation (2) holds
for all values of x.
Slide 5-52
Distinct Linear Factors continued
1
2.
Substituting 0 for x in equation (2), A = When
3

choosing x = 2, B = 2 . When choosing x = 2, C = 1.
The remainder rational expression can be written as the
following sum of partial factions:
5x  2
1
3
1



,
3
x  4x 2x 2  x  2 x  2
And the given rational expression can be written as
2 x4  8x2  5x  2
1
3
1
 2x 


.
3
x  4x
2x 2  x  2 x  2
This can be checked by combining the terms on the right.
Copyright © 2004 Pearson Education, Inc.
Slide 5-53
Distinct Linear and Quadratic Factors
Find the partial fraction decomposition of
x 2  3x  1
.
2
 x  1  x  2 
Solution: The denominator has distinct linear and
quadratic factors, where neither is repeated. Since x2 + 2
cannot be factored, it is irreducible. The partial fraction
decomposition is
x 2  3x  1
A
Bx  C

 2
.
2
 x  1  x  2  x  1 x  2
Multiply both sides by (x + 1)(x2 + 2) to get
x2 + 3x 1 = A(x2 + 2) + (Bx + C)(x + 1).
Copyright © 2004 Pearson Education, Inc.
(1)
Slide 5-54
Distinct Linear and Quadratic Factors
continued
First, substitute 1 for x to get
(1)2 + 3(1)  1 = A[(1)2 + 2] + 0
3 = 3A
A = 1.
Replace A, with 1 in equation (1) and substitute any value
for x. For instance, if x = 0, then
02 + 3(0)  1 = 1(02 + 2) + (B  0 + C)(0 + 1)
1 = 2 + C
C = 1.
Now, letting A =  1 and C = 1, substitution again in
equation (1), using another number for x.
Copyright © 2004 Pearson Education, Inc.
Slide 5-55
Distinct Linear and Quadratic Factors
continued
For x = 1,
3 = 3 + (B + 1)(2)
6 = 2B + 2
B = 2.
Using A = 1, B = 2, and C = 1, the partial fraction
decomposition is
x 2  3x  1
1 2 x  1

 2
.
2
 x  1  x  2  x  1 x  2
Again this work can be checked by combining terms on the
right side.
Copyright © 2004 Pearson Education, Inc.
Slide 5-56
Repeated Quadratic Factors
Find the partial fraction decomposition of
2x
x
2
 1
2
 x  1
.
Solution: This expression has both a linear factor and a
repeated quadratic factor. By the previously stated steps,
3(a) and 4(b),
Ax  B Cx  D
E
 2


.
2
2
2
2
 x  1  x  1 x  1  x  1 x  1
2x
Multiplying both sides by (x2 + 1)2(x  1) leads to
2x = (Ax + B)(x2 + 1)(x  1)+ (Cx + D)(x  1)+ E(x2 + 1)2. (1)
Copyright © 2004 Pearson Education, Inc.
Slide 5-57
Repeated Quadratic Factors continued
1
If x = 1, then equation (1) reduces to 2 = 4E, or E =2 .
Substituting 1 for E in equation (1) and combining terms
2
on the right gives
1
1


2 x   A   x 4    A  B  x 3   A  B  C  1 x 2    A  B  D  C  x    B  D   .
2
2



To get additional equations involving the unknowns,
equate the coefficients of like powers of x on the sides of
equation (2). Setting corresponding coefficients of
1
1
x4 equal, 0  A  or A   . From the corresponding
2
2
1
1
3
coefficients of x , 0 = A + B. Since A =  , B =  .
2
2
2
Using the coefficients of x , 0 = A  B + C + 1.
Copyright © 2004 Pearson Education, Inc.
Slide 5-58
Repeated Quadratic Factors continued
1

2

1

2
Since A =
and B =
, C = 1.
Finally, from the coefficients of x, 2 = A + B + D
 C. Substituting for A, B, and C gives D = 1.
1
1
1
With A   , B   , C  1, D  1, and E  ,
2
2
2
the given fraction has the partial decomposition
1
1
1
 x
2x
x 1
2
2
2



2
2
2
2
x
1
x

1
x

1
  
 x2  1 x  1
2x
  x  1
or
x 1
1



.
2
2
2
2
2
 x  1  x  1 2  x  1  x  1 2  x  1
Copyright © 2004 Pearson Education, Inc.
Slide 5-59
Decomposition into Partial Fractions
Method 1 For Linear Factors

Multiply both sides of the rational expression
by the common denominator.

Substitute the zero of each factor in the
resulting equation. For repeated linear factors,
substitute as many other numbers as
necessary to find all the constants in the
numerators. The number of substitutions
required will equal the number of constants A,
B,….
Copyright © 2004 Pearson Education, Inc.
Slide 5-60
Decomposition into Partial Fractions
continued
Method 2 For Quadratic Factors

Multiply both sides of a rational expression by
the common denominator.

Collect like terms on the right side of the
resulting equation.

Equate the coefficients of the like terms to get
a system of equations.

Solve the system to find the constants in the
numerators.
Copyright © 2004 Pearson Education, Inc.
Slide 5-61
5.5
Nonlinear Systems of
Equations
Copyright © 2004 Pearson Education, Inc.
Nonlinear System


A system of equations in which at least one
equation is not linear is called a nonlinear
system.
2
Solve the system x  2 x  6  y
x  y  2

(1)
(2)
When one of the equations in a nonlinear
system is linear, it is usually best to begin by
solving the linear equation for one of the
variables.
Copyright © 2004 Pearson Education, Inc.
Slide 5-63
Nonlinear System continued


Solve (2) for y. y = 2  x
Substitute this result for y in equation (1).
x 2  2 x  6  2  x
x 2  3x  4  0
( x  4)( x  1)  0
x40
or
x 1  0
x  4
x 1
Copyright © 2004 Pearson Education, Inc.
or
Slide 5-64
Nonlinear System continued

Substitute for x in equation (2) gives y = 2 and
y = 3. The solution set of the given system is
{(4, 2), (1, 3)}.
Copyright © 2004 Pearson Education, Inc.
Slide 5-65
Example

Solve the system.

x2  y 2  9
3x 2  y 2  7

22  y 2  9
Add the two equations to
eliminate y. 2
2
x  y 9
3x 2  y 2  7
4x2
Find y by substituting
back into equation (1).
4  y2  9
y2  5
y 5
 16
x2  4
x  2
Copyright © 2004 Pearson Education, Inc.
Slide 5-66
Example continued

The solutions of the given
system are:
 2, 5  ,  2,  5  ,
 2, 5  and  2,  5 .
Copyright © 2004 Pearson Education, Inc.
Slide 5-67
Example

2
2
x

3
xy

y
 22 (1)
Solve the system.
x 2  xy  y 2  6

(2)
Solution
x 2  3 xy  y 2  22 (1)
 x 2  xy  y 2  6 Multiply (2) by  1.
4 xy  16
4
y
x
Copyright © 2004 Pearson Education, Inc.
Slide 5-68
Example continued

Substitute y into either equation (1) or (2).
2
4 4
2
x  x      6
 x  x
16
2
x 4 2 6
x
x 4  4 x 2  16  6 x 2
x 4  10 x 2  16  0
( x 2  2)  0
or
( x 2  8)  0
x2  2
or
x2  8
x 2
or
x  2 2
Copyright © 2004 Pearson Education, Inc.
Slide 5-69
Example continued

Substitute the x-values into y = 4/x to find y values.
x 2
4
y
2 2
2
x 2
4
y
 2 2
 2
Copyright © 2004 Pearson Education, Inc.
x2 2
4
y
 2
2 2
x  2 2
4
y
 2
2 2
The solution set of
the system is






 2,2 2 ,  2, 2 2 , 




 2 2, 2 , 2 2,  2 


Slide 5-70
Example



Solve the
2
2
x

y
9
system.
x  y3
(1)
(2)
Solve equation (2) for |x| gives |x| = 3  y.
Since |x| > 0 for all x, 3  y  0 and thus y  3. In
equation (1), the first term is x2, which is the
same as |x|2.
Copyright © 2004 Pearson Education, Inc.
Slide 5-71
Example continued

(3  y ) 2  y 2  9

x 2  02  9
9  6y  y  y  9
2
2
2y  6y  0
2 y ( y  3)  0
y0
or y  3
x2  9
x  3
2

Solve for corresponding x
values, use either
equation (1) or (2).
Copyright © 2004 Pearson Education, Inc.
If y = 0, then

If y = 3, then
x 2  32  9
x2  0
x0
Slide 5-72
Example continued


The solution set, {(3, 0),
(3, 0), (0, 3)}, includes the
points of intersection
shown in the figure at the
right.
Be sure to check the
solutions in the original
system.
Copyright © 2004 Pearson Education, Inc.
Slide 5-73
Application

Find two numbers whose sum is 17 and whose
product is 42.

Step 1: Read the problem.
Step 2: Assign variables. Let x represent one
number and y represent the other
number.
Step 3: Write a system of equations.




The sum of the two numbers is 17: x + y = 17
The product of the two numbers is 42: xy = 42
Copyright © 2004 Pearson Education, Inc.
Slide 5-74
Application continued

Step 4 Solve the system. Solve equation (1) for y.
y = 17  x, and substitute into equation (2).
x(17  x)  42
17 x  x 2  42
x 2  17 x  42  0
( x  3)( x  14)  0
x  3 or
x  14


Step 5 State the answer. The two numbers whose sum
is 17 and whose product is 42 are 3 and 14.
Step 6 Check the answers.
Copyright © 2004 Pearson Education, Inc.
Slide 5-75
5.6
Systems of Inequalities and
Linear Programming
Copyright © 2004 Pearson Education, Inc.
Definitions

A line divides a plane into
three sets of points: the
points of the line itself
and the points belonging
to the two regions
determined by the line.
Each of these two regions
is called a half-plane.
The line is called the
boundary.
Copyright © 2004 Pearson Education, Inc.
Slide 5-77
Graphing Inequalities


For a function f, the graph of y < f(x) consists of
all the points that are below the graph of y = f(x);
the graph of y > f(x) consists of all the points that
are above the graph of y = f(x).
If the inequality is not or cannot be solved for y,
choose a test point not on the boundary. If the
test point satisfies the inequality, the graph
includes all points on the same side of the
boundary as the test point. Otherwise, the graph
includes all points on the other side of the
boundary.
Copyright © 2004 Pearson Education, Inc.
Slide 5-78
Example



Graph x + 2y < 6.
Solution The boundary is a straight line
x + 2y = 6. Since the points on the line do not
satisfy the inequality the line is dashed.
To decide which half-plane represents the
solution set, solve for y.
x  2y  6
2 y  x  6
1
y  x3
2
Copyright © 2004 Pearson Education, Inc.
Slide 5-79
Example continued


Since y is less than, the
graph of the solution set
is the half-plane below
the boundary.
As a check, choose a
point not on the boundary
line and substitute into
the inequality.

Since the point (0, 0) is
below the boundary line, the
points that satisfy the
inequality must be
contained in the region.
x  2y  6
0  2(0)  6
06
Copyright © 2004 Pearson Education, Inc.
True
Slide 5-80
Systems of Inequalities

The solution set of a system of inequalities is
the intersection of the solution sets of its
members.

Graph all solution sets on the same coordinate
axes and identify, by shading, the region
common to all graphs.
Copyright © 2004 Pearson Education, Inc.
Slide 5-81
Example

Graph the solution set of the system.
x  2y  6
x2  2  y

Solve each equation for y.
1
y  x3
2
y  x2  2

Graph the solution set for each equation.
Copyright © 2004 Pearson Education, Inc.
Slide 5-82
Example continued

Be sure to use the correct
lines (dashed or solid) for
each graph.
Copyright © 2004 Pearson Education, Inc.
Slide 5-83
Example

Graph the solution set of
the system. x  3
y x 2
y 1

If you graph each
equation and shade
appropriately you can see
that the three inequalities
have NO points in
common.
Copyright © 2004 Pearson Education, Inc.
Slide 5-84
Fundamental Theorem of Linear
Programming

If an optimal value for a linear programming
problem exists, it occurs at the vertex of the
region of feasible solutions.
Copyright © 2004 Pearson Education, Inc.
Slide 5-85
Solving a Linear Programming Problem

Step 1


Step 2
Step 3
Step 4

Step 5

Write the objective function and all
necessary constraints.
Graph the region of feasible solutions.
Identify all vertices or corner points.
Find the value of the objective function
at each vertex.
The solution is given by the vertex
producing the optimal value of the
objective function.
Copyright © 2004 Pearson Education, Inc.
Slide 5-86
Example

A tray of banana muffins requires 4 cups of milk
and 3 cups of wheat flour. A tray of pumpkin
muffins requires 2 cups of milk and 3 cups of
wheat flour. There are 16 cups of milk and 15
cups of wheat flour available, and the baker
makes $3 per tray profit on banana muffins and
$2 per tray profit on pumpkin muffins. How many
trays of each should the baker make in order to
maximize profits?
Copyright © 2004 Pearson Education, Inc.
Slide 5-87
Example continued

Step 1 We let x = the number of banana muffins and
y = the number of pumpkin muffins. Then the
profit P is given by the function P = 3x + 2y

We know that x muffins require 4 cups of milk and y
muffins require 2 cups of milk. Since there are no more
than 16 cups of milk, we have one constraint.
4x + 2y  16

Similarly, the muffins require 3 and 3 cups of wheat flour.
There are no more than 15 cups of flour available, so we
have a second constraint. 3x + 3y  15
Copyright © 2004 Pearson Education, Inc.
Slide 5-88
Example continued


We also know x  0 and y  0 because the baker cannot
make a negative number of either muffin.
Thus we want to maximize the objective function
P = 3x + 2y subject to the constraints
4x + 2y  16,
3x + 3y  15,
x  0,
y0
We graph the system of inequalities and determine the
vertices. Next, we evaluate the objective function P at
each vertex.
Copyright © 2004 Pearson Education, Inc.
Slide 5-89
Example continued
Vertices
Profit
P = 3x+ 2y
(0, 0)
P = 3(0) + 2(0) = 0
(4, 0)
P = 3(4) + 2(0) = 12
(0, 5)
P = 3(0) + 2(5) = 10
(3, 2)
P = 3(3) + 2(2) = 13
The baker will make a maximum profit when 3
tray’s of banana muffins and 2 trays pumpkin
muffins are produced.
Copyright © 2004 Pearson Education, Inc.
Maximum
Slide 5-90
5.7
Properties of Matrices
Copyright © 2004 Pearson Education, Inc.
Basic Definitions





It is customary to use capital letters to name matrices.
Also, subscript notation is often used to name elements
of a matrix, as shown.
a1n 
 a11 a12 a13
a
a22
 21
 a31 a32


 am1 am 2
a23
a33
am3
a2 n 

a3n 


amn 
A n  n matrix is a square matrix.
A matrix with just one row is a row matrix.
A matrix with just one column is a column matrix.
Two matrices are equal if they are the same size and if
corresponding elements, position by position, are equal.
Copyright © 2004 Pearson Education, Inc.
Slide 5-92
Example

Find the values of the variables which makes the
statement true.
3 4  a b
 x y    2 7

 


From the definition of equality, the only way that
the statement can be true is if a = 3, b = 4, x = 2
and y = 7.
Copyright © 2004 Pearson Education, Inc.
Slide 5-93
Addition and Subtraction of Matrices

To add two matrices of the same size, add
corresponding elements. Only matrices of the
same size can be added.

If A and B are two matrices of the same size,
then A  B = A + (B).
Copyright © 2004 Pearson Education, Inc.
Slide 5-94
Examples


Add and subtract the following.
Add
Subtract
7 3  6 5 
 4 2    3 2 

 

3  5 
7  (6)


4

3

2

(

2)


7 3  6 5 
 4 2    3 2 

 

3  5 
7  (6)


4

3

2

(

2)


1 2 


7

4


13 8


1
0


Copyright © 2004 Pearson Education, Inc.
Slide 5-95
Examples continued


Add or subtract, if possible.
a)  2   3   5 
 4    2    6 
     
 3   1  2 

b)  4 9  2
7 3  5 

  
The matrices have different sizes so they
cannot be added or subtracted.
Copyright © 2004 Pearson Education, Inc.
Slide 5-96
Properties of Scalar Multiplication

If A and B are matrices of the same size and c
and d are scalars, then
(c + d)A = cA + dA
c(A)d = cd(A)
c(A + B) = cA + cB
(cd)A = c(dA)
Example: Find the product.
 7 2  3(7) 3(2)   21 6 
3






1
6
3
(

1)
3
(6)

3
18

 
 

Copyright © 2004 Pearson Education, Inc.
Slide 5-97
Matrix Multiplication

If the number of columns of an m  n matrix A is
the same as the number of rows of an n  p
matrix B (i.e., both n). The element cij of the
product matrix C = AB is found as follows:
cij  ai1b1 j  ai 2b2 j  ...  ainbnj .

Matrix AB will be an m  p matrix.
Copyright © 2004 Pearson Education, Inc.
Slide 5-98
Example





Suppose A is a 3  4 matrix, while B is a 4  2
matrix.
a) Can the product AB be calculated?
b) If AB can be calculated, what size is it?
c) Can BA be calculated?
d) If BA can be calculated, what size is it?
Copyright © 2004 Pearson Education, Inc.
Slide 5-99
Solutions

a) AB can be calculated, because the number of
columns of A is equal to the number of rows of
B.
34
42
must match
Size of AB



b) The product is a 3  2 matrix.
c) BA cannot be calculated, the number of
columns and rows do not match.
d) Since BA cannot be calculated we cannot
determine the size.
Copyright © 2004 Pearson Education, Inc.
Slide 5-100
Multiply the Matrices

For
 0 4 
 2 4
 2 2 2


A
, B  2 7 , and C  





1
0
1
0
4




 1 3 
find each of the following.
a) AB
b) BA
c) AC
Copyright © 2004 Pearson Education, Inc.
Slide 5-101
Solution AB

A is a 2  3 matrix and B is a 3  2 matrix, so
AB will be a 2  2 matrix.
 0 4 
 2 2 2  

AB  
2

7

1
0
4

  1 3 


 2(0)  2(2)  2(1) 2(4)  2( 7)  ( 2)(3)   2 28




1
(0)

0
(2)

4
(

1)
1
(

4)

0
(

7
)

4
(3)

4
8

 

Copyright © 2004 Pearson Education, Inc.
Slide 5-102
Solution BA

B is a 3  2 matrix and A is a 2  3 matrix, so
BA will be a 3  3 matrix.
 0 4 
 2 2 2 


BA  2 7 

 1 0 4 
 1 3 
 0(2)  (4)(1) 0(2)  (4)(0) 0(2)  (4)(4)   4 0 16 
  2(2)  (7)(1) 2(2)  (7)(0) 2(2)  (7)(4)    3 4 32 

 

 1(2)  3(1)
1(2)  3(0)
1(2)  3(4)   1 2 14 
Copyright © 2004 Pearson Education, Inc.
Slide 5-103
Solution AC


The product AC is not defined because the
number of columns of A, 3, is not equal to the
number of rows of C, 2.
Note that AB  BA. Multiplication of matrices is
generally not commutative.
Copyright © 2004 Pearson Education, Inc.
Slide 5-104
Properties of Matrix Multiplication

If A, B, and C are matrices such that all of the
following products and sums exist, then

(AB)C = A(BC)

A(B + C) = AB + AC

(B + C)A = BA + CA.
Copyright © 2004 Pearson Education, Inc.
Slide 5-105
5.8
Matrix Inverses
Copyright © 2004 Pearson Education, Inc.
Identity Matrices

By the identity property for real numbers,
a  1 = a and 1  a = a
for any real number a. If there is to be a multiplicative
identity matrix I, such that
AI = A
and
IA = A
for any matrix A, then A and I must be square matrices of
the same size.
Copyright © 2004 Pearson Education, Inc.
Slide 5-107
Identity Matrices
2  2 Identity Matrix
If I2 represents the 2  2 identity matrix, then
1 0 
I2  
.

0 1 
Copyright © 2004 Pearson Education, Inc.
Slide 5-108
Identity Matrices
Copyright © 2004 Pearson Education, Inc.
Slide 5-109
Stating and Verifying the 3  3 Identity
Matrix
Example:
 2 2 1
Let A   3 5 4  . Give the 3  3 identity


 5 6 4  matrix I and show that AI = A.
Solution: By the definition of matrix multiplication,
 2 2 1 1 0 0  2 2 1
 3 5 4  0 1 0   3 5 4 


 

 5 6 4  0 0 1   5 6 4 
Copyright © 2004 Pearson Education, Inc.
Slide 5-110
Multiplicative Inverses


If A is an n  n matrix, then its multiplicative
inverse, written A1, must satisfy both
AA1 = In and A1A = In
This means that only a square matrix can have a
multiplicative inverse.
1
Caution: Although a  for any nonzero real
a
number a, if A is matrix, A1  1 .
1
A
Copyright © 2004 Pearson Education, Inc.
Slide 5-111
Finding an Inverse Matrix
To obtain A1 for any n  n matrix A for which
A1 exists, follow these steps.
Step 1 Form the augmented matrix  A I n  , where
In is the n  n identity matrix..
Step 2 Perform row transformations on  A I n  to
obtain a matrix of theI B .


n
Step 3 Matrix B is A1.

Copyright © 2004 Pearson Education, Inc.
Slide 5-112
Example
Find A1 if A =
 1 0 1
 2 1 3 .


 1 1 1
Solution: Use row transformations as follows.
Step 1 Write the augmented matrix  A I3 .
 1 0 1 1 0 0
 2 1 3 0 1 0


 1 1 1 0 0 1 
Step 2 Since 1 is already in the upper left hand corner, we
begin by using row transformation that will result in
0 for the first element in the second row.
Copyright © 2004 Pearson Education, Inc.
Slide 5-113
Example continued
 1 0 1 1 0 0
 0 1 1 2 1 0

 R2  2R1
 1 1 1 0 0 1 
The following is done to obtain 0 as
the first element in the third row.
1 0 1 1 0 0 
0 1 1 2 1 0 


0 1 2 1 0 1 
R3 + R1
1 0 1 1 0 0 
0 1 1 2 1 0


0 0 1 3 1 1 
R2  R3
The following is done to obtain 0 for
the third element in the first row.
1 0 0 2 1 1
0 1 1 2 1 0 


0 0 1 3 1 1 
R1  R3
The following is done to obtain 1 as
the third element in the third row.
Copyright © 2004 Pearson Education, Inc.
Slide 5-114
Example continued
The following is done to
obtain 0 for the third
element in the second row.
1 0 0 2 1 1
0 1 0 5 2 1 R2  R3


0 0 1 3 1 1 
Copyright © 2004 Pearson Education, Inc.
Step 3
The last transformation
shows that the inverse is
 2 1 1
A1   5 2 1 .
 3 1 1 
Slide 5-115
Solution of the Matrix Equation AX = B

If A is an n  n matrix with inverse A1, X is an
n  1 matrix of variables, and B is an n  1
matrix, then the matrix equation
AX = B
has the solution
GX = A1 B.
Copyright © 2004 Pearson Education, Inc.
Slide 5-116
Solving Systems of Equations Using
Matrix Inverses
Example:
3x + 4y = 5
5x + 7y = 9
Solution: To represent the system as a matrix equation,
use one matrix for the coefficients, one for the variables,
and one for the constants, as follows
3 4 
A
,

5 7 
 x
X   ,
 y
5 
and B   
9 
The system can then be written in matrix form as the
equation AX = B, since
3 4   x  3 x  4 y  5 
AX  

 B




 5 7   y   5 x  7 y  9 
Copyright © 2004 Pearson Education, Inc.
Slide 5-117
Solving Systems of Equations Using
Matrix Inverses continued
To solve the system, first find A1. Then find A1 B.
 7 4
A 


5
3


1
Since X = A1 B,

 7 4 5  1
A B
 



 5 3  9  2 
1
 x   1
X   
 y  2 
The final matrix shows the solution set of the
system is {(1, 2)}
Copyright © 2004 Pearson Education, Inc.
Slide 5-118