Download Lecture slides, Ch 6

6. Systems of Equations and Inequalities
6.1 Functions and Equations in Two Variables
Heat Index – Temp and Humidity
Wind Chill – Temp and Wind Speed
GPA – Grades and Credit Hours
z  f  x, y   x  y
f  x, y   xy
f 3,4  12
m
f m, g  
g
f 120,5  24mpg
f r , h   r 2 h
f 0.5,2  0.5 2  1.57
2
Systems of Equations
A linear equation in two variables can be
written in the form ax  by  k , where a, b, and k
are constants, with a and b are not zero.
Any system of two linear equations in two
variables can be written in the form
a1 x  b1 y  c1
a2 x  b2 y  c2
The average of two numbers can be found by
x y
. Suppose the average of two numbers is
2
10, and the numbers differ by 2.
Possible solutions to systems
1.One solution: The two equations form graphs
that intersect at one point. The system is
consistent, the equations are independent.
2.Infinitely many solutions: The two equations
form graphs of the same line, the system is
consistent, but dependent.
3.No solution: The two equations form graphs
of parallel lines. The equations are
inconsistent.
x y 2
4x  y  2
2x  y  1
 x  y 1
x  2 y  3
 4 x  2 y  2
Method of substitution
1. Solve one equation for one variable
2. Sub the result into the other equation, then
solve for the other variable.
3. Solve equation from step 1 using step 2.
In first quarter 2011, Apple sold a combined
total of 35.7 million iPods and iPhones. There
were 3.3 million more iPods than iPhones. Find
the individual totals.
5 x  2 y  16
x  4 y  1
Non-linear equations
6 x  2 y  10
2 x 2  3 y  11
A circle of radius 5 is centered on the origin
and intersects the graph of y  2 x .
Solving a system with no or infinitely many solutions
x2  y  1
x 2  y  2
2x  4 y  5
 x  2y   5
2
Elimination method
2 x  y  4
4 x  y  10
x y 6
3x  y  1
 4 x  y  10
x y 3
2 x  3 y  18
5x  2 y  7
5 x  10 y  10
x  2y  2
x2  y2  4
2x2  y  7
Graphical and Numerical Methods
Modeling Roof Trusses
If a 200 lb force is applied to the peak of a truss,
then the weights W1 and W2
exerted on each rafter of the truss are
determined by the following system of linear
equations.
W1  W2  0
3
W1  W2   200
2
Determining dimensions of a cylinder
V r , h   r 2 h
S r , h   2rh excluding top and bottom
Suppose the volume is 38 cubic in, and the
surface area is 63 square in, find the radius and
the height.
Solving non-linear equations graphically
2x  y  2
3
ln x  3 y  1
2
Y1  2X 3  1
ln X 2  1
Y2 
3
Joint variation
V r , h   r 2 h
V varies jointly as h and the
square of r. The constant of variation is  .
Let m and n be real numbers. Then z varies
jointly as the nth power of x and the mth power
of y if a non-zero number k exists such that
z  kx n y m
The volume of wood in a tree varies jointly as
the 1.12 power of the height, and the 1.98 power
of the diameter. A tree with a 13.8 inch
diameter and a 64 foot height has a volume of
25.14 cubic feet. Estimate the volume of a tree
with d  11 and h  47 .
6.2 Linear Systems of Equations and
Inequalities in Two Variables
skip
a1 x  b1 y  c1
a2 x  b2 y  c2
x  y  10
x y 4
x  y  10
2 x  2 y  20
x  y  10
x y 4
Elimination method
2 x  y  4
4 x  y  10
x y 6
3x  y  1
 4 x  y  10
x y 3
Graphical solution – solve each equation for y
Title IX legislation prohibits sex discrimination in
sports. In 1997, the average spent on two varsity
athletes, ♂♀, was $6050, but average ♂ > average
♀ by $3900.
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2 x  3 y  18
5x  2 y  7
5 x  10 y  10
x  2y  2
6.2 Systems of Inequalities in Two Variables
skip
Systems of Inequalities
ax  by  c
2x  3y  6
x2  y2  9
x2  y2  4
2x2  y  7
Modeling plant growth
skip
Temp
Precip
7 P  5T  70
35P  3T  140
Grasslands occur for ordered pairs T , P 
between the two lines above. Bismarck ND has
an average temp of 40° and precip of 15”.
Determine a system of inequalities where
grasslands occur. Does Bismarck fit into that
area?
y  x2
x y 4
skip
x  3y  9
2 x  y  1
Linear Programming
Objective function
feasible
Constraints
solutions
skip
Suppose a company sells radios and CD players.
Each radio generates a $15 profit, and each CD,
a $35 profit. To meet demand, the company
must produce at least 5 but no more than 25
radios, and must make as many CDs as radios,
but no more than 30.
Fundamental Theorem of Linear Programming
skip
If an optimal value for a linear programming
problem exists, then it occurs at a vertex of the
region of feasible solutions.
Minimize C  2 x  3 y , subject to x  0 , y  0 ,
x  y  4, and 2 x  y  8 .
Solving a Linear Programming Problem
1. Read problem carefully.
2. Use a table to write the objective function
and all of the constraints.
3. Sketch a graph of the region of feasible
solutions. Identify vertices.
4. Evaluate the objective function at each one
of the vertices. Choose the min/max.
(if region unbounded, there may not be a
minimum or maximum)
skip
A breeder is buying two brands of food, A and
B, for her animals. She mixes these for feed.
Each serving should contain a minimum of 40
grams of protein and a minimum of 30 grams of
fat.
A is 90¢ per unit, has 20g protein, 10g fat
B is 60¢ per unit, has 10g protein, 10g fat
skip
6.3 Systems of Linear Equations in Three Variables
2x  3y  4z  4
 y  2z  0
3,2,1
x  5 y  6z  7
x  4 y  2 z  15
4x  y  z  1
6 x  2 y  3 z  6
 1,3,2
1,10,13
Solving with elimination and substitution
1. Eliminate one variable from two of the
equations.
2. Apply the techniques from 6.1 and 6.2 to
solve the resulting two equations.
3. Substitute back in to find the third variable.
x  y  2z  6
2 x  y  2 z  3
 x  2 y  3z  7
One thousand tickets were sold for a play,
which generated $3800. The prices of the
tickets were $3 for children, $4 for students, and
$5 for adults. There were 100 fewer student
tickets sold than adult tickets. Find the number
of each ticket sold.
Three students buy lunch in the cafeteria. One
student buys 2 hamburgers, 1 order of fries, and a
soda for $9. Another student buys 1 hamburger,
2 orders of fries, and a soda for $8. A third
student buys 3 hamburgers, 3 fries, and 2 sodas
for $18. If possible, find the cost of each item.
x  y  z  2
x  2 y  2 z  3
y  z  1
6.4 Solutions of Linear Systems Using Matrices
Carl Friedrich Gauss (1777-1855)
Gaussian Elimination with Backward Substitution
Matrix – a rectangular array of numbers
The dimension of a matrix is m n if it has m
rows and n columns. A square matrix is n n.
a1 x  b1 y  c1 z  d1
a2 x  b2 y  c2 z  d 2
a3 x  b3 y  c3 z  d 3
3x  4 y  6
 5 x  y  5
2 x  5 y  6 z  3
3x  7 y  3z  8
x  7y
5
 a1
a
 2
 a3
b1
c1
b2
c2
b3
c3
d1 
d2 

d 3 
1
0
2
2 10
1 2
1 2
2
3
3
3
3
5
4
0 1 6
7
0 0
8
1
Row-Echelon Form
1 3
0
0 1 6
0 0
1
1
1
2
1 2 0
0 1 4
1 3 1 5
0 0
1
3
0 0
0
0
1 3 1 5
0 1 1 3
0 0
1
0
1 3 5
0 0 1
Main diagonal
1s, then maybe zeros
First non-zero element is a 1 (leading 1)
Leftmost leading 1 listed first
Rows with only zeros at the bottom
All elements below main diagonal are zeros
Solve the system
1 1
3
12
0 1 2 4
0 0
1
1 1 5 5
0
1
3 3
0
0
0 0
3
Gaussian Elimination
Matrix Row Transformations
1. Any two rows may be interchanged
2. The elements of any row may be
multiplied by a non-zero constant
3. Any row may be changed by adding or
subtracting a multiple of another row.
x  y  z 1
x y z 5
y  2z  5
2x  4 y  4z  4
x  3y  z  4
 x  3 y  2 z  1
x  2 y  3z  2
2x  3y  2z  7
4 x  y  8z  8
Geometric Interpretation
Reduced Row-Echelon Form
Row-Echelon Form with elements above and
below the leading 1 are zeros
1 0
1 0
0 1
0 0
1 0 0
1 0
0 1 0
0 1 2
0 0 1
1 0 0
3
1 0
0 1 0
1
0 1 1 2
0 0 1 1
0 0
4
8
0
0
1 0 0
3
Solve the system
1 0
6
0 1 5
3
0 1 0 1
0 0 1
2
1 0 0 4
1 0 2 3
0 1 0 3
0 1
2
1
0 0 0 2
0 0
0
0
Transform to Reduced Row-Echelon Form
2 x  y  2 z  10
x
 2z  5
x  2 y  2z  1
6.5 Properties and Applications of Matrices
a11
a12
a 21
a 22
a11
a12
a13
a 21
a 22
a 23
a31
a32
a33
a11
a12
a13
a14
a 21
a 22
a 23
a 24
a31
a32
a33
a34
a 41
a 42
a 43
a 44
a11
a12
a 21
a 22
a31
a32
An element of matrix A is designated
aij
a11
a12
a13
a 21
a 22
a 23
( arc )
Two matrices A and B are equal if all corresponding
elements are equal ( aij  biji, j )
3 3
7
Let A= 1
6
2
4
2
5
a12 
b32 
3 x
B= 1
7
6 2
4 5
2
a13 
What value of x would make A=B?
a31b13  a32b23  a33b33 
Sum, Difference, Scalar Multiplication
The sum of two m n matrices A and B is the
m n matrix A  B , where each element is the
sum of corresponding elements in A and B.
aij  bij i, j  1  i  m,1  j  n
2 2 2
1 1 1
0 2 0  1 1 1
0 2 0
1 1 1
1 2 1
1 1 1
2 2 2 
1 2 1
1 1
1
1 1 1
The difference of two m n matrices A and B is
the m n matrix A  B , where each element is the
difference of corresponding elements in A and B.
aij  bij i, j  1  i  m,1  j  n
3 3 3
1 1 1
1 3 1  1 1 1
1 3 1
1 1 1
A=
7
8 1
0 1
6
A+B=
B+A=
A – B=
1 1 1
1 1 1
1 1 1  1 1 1
1 1 1
1 1 1
1 1 1
2  1 1 1
1 1 1
B=
5  2 10
3
2
4
The product of a scalar (real number) k and
m n matrix A is the m n matrix kA , where
each element is k times the corresponding
element in A.
kaij i, j  1  i  m,1  j  n
2 7
If A=  1
3
11
 5,
0 9  12
find  4 A
A=
4 2
3
5
B=
A  3B 
AC 
 2C  3D 
0 1
2 3
1 1
C=
0
7
4
2
1  3
D=
9 7
1
8
Matrix Products
A
B
$
Student1 10
7
CollegeA 60
Student 2 11 4
CollegeB 80
Cost for Student 1
Cost for Student 2
10 7  60
AB= 
  

11 4  80
The product of an m n matrix A and an n  k
matrix B is the m  k matrix AB. The elements
are  aij b jk .
j 1, n
nk
m n
1
1
A= 0
3
4 2
AB=
CA=
B=
1
2
C=
1 2 3
4 5 6
1 1
D=
0
3
2
3 2
4
5
DC=
CD=
A=
1
0
3
2 1
5 2
7
5
4 6
B= 8
0
9
7
10
1 3
AB=
Graphing Calculator
2 A  3B 3
One click paths
Page 1
Page 2
Page 3
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0
0
A
1
0

1 1 0
0 0 0

0 0 0
1 1 0
Two click paths A2
Properties of Matrices
Let A, B, and C be matrices. Assume that each
matrix operation is defined.
A B  B  A
 A  B   C  A  B  C 
 AB C  ABC 
A B  C   AB  AC
6.6 Inverses of Matrices
Translation
1 0
A  0 1

0 0
h
k

1 
A1 like f 1
 x
X   y
 
 1 
AX 
Translate the point  1,2, 3 units right, down 4.
1 0  3
A1  0 1 4 


0 0 1 
AA1 
A1 A 
The n n identity matrix, denoted by I n , has
only 1s on the main diagonal and 0s elsewhere.
2 3
4 5


Let A be an n n matrix. If there exists an n n
matrix, denoted A1, that satisfies
A1 A  I n
and
AA1  I n
then A1is the inverse of A.
If A1 exists, A is invertible or non-singular
If A is not invertible, then A is singular.
3
3
 5
 2
Let A  
B


  3  2
 3  5
B  A1?
90° clockwise rotation
 0 1 0
0  1 0 
Let A   1 0 0 and A1   1 0 0




 0 0 1
0 0 1
Rotate the point  2,0  clockwise 90° about origin
What will A1 do?
Finding Inverses Symbolically
1 4 
Find A if A  

2
9


1
 1 0 1
Find A1 if A   2 1 3


 1 1 1
Representing linear systems with matrix equations
3x  2 y  4 z  5
2 x  y  3z  9
 x  5 y  2z  5
3x  4 y  7
 x  6 y  3
x  5y
2
 3 x  2 y  z  7
4 x  5 y  6 z  10
Solving linear systems with inverses
AX  B
x  4y  3
2x  9 y  5
 9  4
 2 1 


x  3y  z  6
 2 y  z  2
 x  y  3z  4
Modeling blood pressure
P A W
113 39 142
138 53 181
152 65 191
.25 
 1.25 2
 .25  1  .25


  .5  1  .5 
P A,W   a  bA  cW