ch7studentnotes2.doc

7/27/2019 Ch7StudentNotes2.doc

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Day Assignment-worth 10 points, due March 14

On Day, bring enough food or drink (appropriate for school) for 4 people. Before Day on a sheet of paper, write

down why your food or drink involve by doing at least one calculation of your food/drink with in your final answer.

The following is my example. NO ONE can use it in their assignment!

I brought 4 cans of lemonade. I measured one can and found its height is about 12.4 cm and its diameter is 6.6 cm. The

surface area of a can is approximately the same as the surface area of a cylinder, which is 222 rrH + where r=radius

and H=height.

22

_ 62.103)3.3(2)4.12)(3.3(2 cmcmcmcmSA canone +

So the total surface area of all 4 cans of lemonade I brought is about22 48.414)62.103(4 cmcm .

Make sure you leave your final answer in terms of . Write your name, period, and column number on your paper to turn

in at the beginning of the period on Day.

Quiz 6.3 topics: Only scientific calculator is allowed!Section 6.1-6.3Ch 5 Test mistakesword problems involving bearing, Law of Sines, Law of Cosines, areaProofs of Law of Sines and Law of CosinesFind parallel or orthogonal vectorsVectors: magnitude, direction, angle in between, unit vector

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7.1 Solving Systems of Equations

7.1 day 1 p481#1-31(o),35 (graph paper). 7.1 day 2, p482#51-75(o)Many problems in science, business, and engineering involve two or more equations in two or more variables. To solvesuch problems, you will need to find the solutions of systems of equations.

Here are several examples of system of equations:

a) 2 13 4

x yx y

=

+ =b)

2 0

2 2 3 7

5 7 11

x y z

x y z

x y z

+ =

+ + = + + =

c) 96

x y zx z

+ + =

+ =

A solution of these systems is an ordered pair that satisfies each equation in the system.

Here are four of the many methods to solve these systems of equations.1. substitution2. graphical3. elimination4. Gaussian Elimination

The Method of Substitution

Example #1: Solve the system2 1

3 4

x y

x y

=

+ =using the method of substitution.

Example #2: Solve the system2 1

3 4

x y

x y

=

+ =using the method of graphing.

The equations in a system can be either linear or non-linear.

Example #3: Solve the system 2 2

3 2 0

4

x y

x y

=

+ =.

Exploration: For each of the following systems of equations. Solve by the methods of substitution and graphing. Whatdo you notice about the number of solutions and the number of intersections between the graphs of the equations?

2

x

y


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a)

2 4

1

y x

y x

=

= +b)

2

3

y x

y x

=

= c) 2

3

3

y

y x

=

= +

Example #4: Solve the system of equations graphically or algebraically.

a)

3( 1)

1

y x

y x

= +

= b)

2 4

0xx y

e y

+ =

=

3

x

y y

x

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7.2 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES

7.2 day 1, p491#1-39(o) 7.2 day 2, p492#55-69(o),73,75,77In Section 7.1, we learned how to solve systems of equations using the methods of substitution and graphing. In thissection, we will learn how to solve by the method of elimination.

The Method of Elimination

To use the method of elimination to solve a system of two linear equations in x and y, perform the following steps.

1. Obtain coefficients for x (or y) that differ only in sign by multiplying all terms of one or both equations by suitablychosen constants.

2. Add the equations to eliminate one variable, solve the resulting equation.3. Back-substitute the value obtained in Step 2 into either of the original equations and solve the other variable.4. Check your solution in both of the original equations.

Example #1: Solve the system of equations by the method of elimination

a)7 12

3 5 10

x y

x y

+ =

=b)

0.2 0.5 27.8

0.3 0.4 68.7

x y

x y

=

+ =

Exploration: Use a graphing utility to graph each system of equations. Determine the number of solutions each systemhas.

a)5 1

5

y x

y x

= +

= b)

3 4 1

8 2 6

y x

x y

=

+ = c) 1

2

2 3

4

y x

y x

= +

= +

Graphical Interpretation of Solutions

For a system of two linear equations in two variables, the number of solutions is one of the following.

Number of Solutions Graphical Intepretations1. Exactly one solution The two lines intersect at one point2. Infinitely many solutions The two lines are identical3. No solution The two lines are parallel

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Ex 2 Solve the system of equations.

a)8 14 5

2 3.5 1.25

x y

x y

=

=b)

2 3 3

2 4 10

x y

x y

=

+ =

Application:

1. Five hundred gallon of 89-octane gasoline is obtained by mixing 87-octane gasoline with 92-octane gasoline.

a. Write a system of equations. One equation represents the amount of final mixture required and the otherrepresents the amount of 87-and 92-octane gasoline in the final mixture. Let x and y represent thenumber 87-octane gallons and 92-octane gallons, respectively.

b. How much of each type of gasoline is required to obtain the 500 gallons of 89-octane gasoline?

2. A total of $32,000 is invested in two municipal bonds that pay 5.75% and 6.25% simple interest. The investorwants an annual interest income of $1900 from the investments. What is the most that can be invested in the 5.75%bond?

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7.3 MULTIVARIABLE LINEAR SYSTEMS

day 1, p505#1-25(o) day 2, p506#27-51(o) day 3, p506#57-77(o) day 4, p507#83-95(o)

The method of elimination can be applied to a system of linear equations in more than two variables. The goal in using themethod of elimination is to rewrite the system of equations in a form to which back-substitution can be applied easily.

The following system of equation is in Row-Echelon Form. Use back-substitution to solve the system.

3 2 43 2

1

x y zy z

z

+ = =

=

To solve a system that is not in row-echelon form, one can convert it to an equivalent system that is in row-echelon form byusing elementary row operations.

Gaussian Elimination

Elementary Row Operations1. Interchange two equations2. Multiply one of the equations by a nonzero constant.3. Add a multiple of one equation to another equation.

Example #1: Use Gaussian Elimination to solve the system.2 4 4

2 4 6 13

4 2 6

x y z

x y z

x y z

+ + =

+ =

+ =

Example #2: Use Gaussian Elimination to solve the system.

a)

5 3 _ 2 3

2 4 7

11 4 3

x y z

x y z

x y z

=

+ =

+ =

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b)

4 13

4 2 7

2 2 7 19

x z

x y z

x y z

+ =

+ =

=

The Number of Solutions of a Linear System

For a system of linear equations, exactly one of the following is true.1. There is exactly one solution.2. There are infinitely many solutions.3. There is no solution.

A system of linear equations is called consistentif it has at least one solution. A consistent system with exactly onesolution is independent. A consistent system with infinitely many solutions is dependent. A system of linear equations iscalled inconsistent if it has no solution.

A system of equations is called a square system when the number of equations equals the number of variables. In anonsquare system of equations, the number of equations differs from the number of variables. Such systems do not haveunique solutions.

Example #3: Solve the system of linear equations.

3 2 18

5 13 12 80

x y z

x y z

+ =

+ =

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Partial Fraction Decomposition

A rational expression can be expressed as the sum of two or more simpler rational expressions.

Decomposition ofN(x)/D(x) into Partial Fractions

1. Divide if improper: IfN(x)/D(x) is an improper fraction (degree ofN(x) > degree ofD(x), divide the denominatorinto the numerator to obtain

1( )( ) ( )( ) ( )

N xN xpolynomial

D x D x= +

and apply Steps 2, 3, and 4 (below) to the proper rational expression1( )

( )

N x

D x.

2. Factor denominator: Completely factor the denominator into factors of the form ( )mpx q+ and 2( )nax bx c+ +

where 2( )ax bx c+ + is irreducible.

3. Linear factors: For each factor of the form ( )mpx q+ , the partial fraction decomposition must include the following

sum of m fractions.

1 2 2 ..........( ) ( ) ( )

m

m

AA A

px q px q px q+ + ++ + +

4. Quadratic factors: For each factor of the form 2( )n

ax bx c+ + , the partial fraction decomposition must include the

following sum of n fractions.

1 1 2 22 2 2 2

..........( ) ( )

n n

n

B x CB x C B x C

ax bx c ax bx c ax bx c

++ ++ + +

+ + + + + +

Example #4: Write the partial fraction decomposition for2

5

6x x+ .

Example #5: Write the partial fraction decomposition for

2

2 3

6 1

( 1)

x

x x

+

.

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Example #6: Write the partial fraction decomposition for3 2

2

2 1

3 4

x x x

x x

+ +

+ .

Example #7: Find a quadratic equation, 2y ax bx c= + + , whose graph passes through the points (1, 3), (-1, 1), and

(-2, 6)

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7.4: Matrices and systems of equations hwk: p521#6,12,18-26E,30,34,44,48-60E,68,72,76,82,84and rev p570#5,18,28,38,45-59odds (Delete rev if not having quiz 7.4)

matrix A=

654

321has 6 elements, 2 rows, and 3 columns.

Order of a matrix=its dimensions: row by column: 2 by 3Position given by subscript: a12=element in row 1, column2=2

1

0

0

0

1

0

0

0

1

has ones in its principle diagonal.

Operations that produce equivalent systems:1. Two rows are interchanged.2. A row is multiplied by a nonzero constant.3. A constant multiple of one row is added to another row.

Reduced matrix (aka redueced row-echelon form):1. Each row consisting entirely of zeros is below any row having at least one nonzero element.2. The leftmost nonzero element is 1.3. All other elements in the column containing the leftmost 1 of a given row are zeros.4. The leftmost 1 in any row is to the right of the leftmost 1 in the row above.

Ex1 Which of the following matrix is not reduced? Why?

,4

2

1

0

0

0

0

1,

4

2

01

10

=

= BA

=

=

=

0

0

0

0

1

0

0

0

0

0

0

1

,

100

010

001

,

000

100

001

EDC

Gauss-Jordan Elimination Procedure

Step1. Choose the leftmost nonzero column and use appropriate row operations to get a 1 at the top.Step 2. Use multiples of the row containing the 1 from step 1 to get zeros in all the remaining places in the columncontaining this 1.Step 3. Repeat step 1 with next column.Step 4. Repeat step 2 using new column.Step 5. Repeat until entire matrix is in reduced form.

Ex2 Solve:3x1+x2-2x3=2

x1-2x2+x3=32x1-x2-3x3=3

=

3

3

2

312

121

213

3

2

1

x

x

x

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312

121

213

is called the coefficient matrix,

33

2

is called the constant matrix, and

3

3

2

312

121

213

is called the augmented matrix.

One method to solve is to get the augmented matrix to look like:

l

k

j

100

010

001

We can checkour solution on the graphing calculator by typing in the original augmented matrix as a 3 by 4 matrix intomatrix A. Back to the main screen, which is MATRIX, MATH B: rref(A).

Be sure you know how to reduce a matrix by hand, because you will be expected to do so on the non-graphing calculatorsection of the quiz.Possible final forms:

n

m

10

01one solution, consistent, independent. Graph is 2 lines that intersect once. This for for 2-D. In 3-D, the

graph looks like 3 planes that intersect at one point.

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000

1 nminfinitely many solutions, consistent, dependent. Graph is the same line.

This is for 2-D. In 3-D, there are 2 cases: either all 3 planes are really the same plane or the 3 planes intersect in a singleline.

p

nm

00

1no solution, inconsistent. Graph is 2 parallel lines that never intersect.

This is for 2-D. In 3-D, there are 2 cases: either 3 planes are parallel or the 3 planes intersect NOT in a single line.

Ex3 Solve:3x1+6x2-9x3=152x1+4x2-6x3=10-2x1-3x2+4x3=-6

6

10

15

432

642

963

Reduced matrix is

0

4

3

000

210

101

. How do we read this? Throw away the all zero row. We are left with 2

equations:

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7.5 Operations with matrices hwk: p536#4,8,14,24,26,30,32,34,38,52,56,60,68-72E,80,82,84,87-94,102,104

Add and subtract matrices with matching elements: only works when dimensions are the same.

Ex 1 +

278

610

294

996

125

Ex 2

31

43

22

30

05

23

Matrices are commutative and associative under addition, but NOT under subtraction nor multiplication nor division!Commutative: A+B=B+AAssociative: (A+B)+C=A+(B+C)

Zero matrix has all zero elements in it.

When we multiply a scalar k by matrix M, we distribute k to every element of M.

Ex 3

=

1812

2410

96

1252

When we multiply matrices, we must first check that the column number of the first matrix matches with the row number ofthe second matrix. When match, continue. When mismatch, not defined!

Ex 4

2

1

76

54

Ex 5

=

20

14A ,

=

06

70

35

B . Find AB and BA. What do you notice?

Notice that matrix multiplication is NOT commutative: in general: ABBA, and order matters!

If IM=MI=M and I and M are matrices, then I is the identity matrix. The identity matrix is always a square matrix withonly 1s in its principle diagonal and zeros everywhere else.

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...

1000

0100

0010

0001

1

0

0

0

1

0

0

0

1

10

01=

=

=

=I

Ex 6 Multiply

if

c

he

b

gd

a

1

0

0

0

1

0

0

0

1

If time: do ex 13.

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7.6 Inverse of a square matrix p547#4,6,14-18E,30,32,38,42,46,52,56,68,74,82-90E

Inverse of a square matrix: Let M be a square matrix of order n and I be the identity matrix of order n. If there exists amatrix M-1 st M-1 M=I=MM-1, then M-1 is called the multiplicative inverse of M.

Ex 1 Show that B is the inverse of A, where

A =1 1

1 2

,B =2 1

1 1

Ex 2 Find the inverse of M=

0

1

1

3

2

1

2

0

1

.

Ex 3 Checkthe results of ex 2 on graphing calc.

Be sure you know how to find the inverse of a matrix by hand, because you will be expected to do so on the non-graphingcalculator section of the test.

Here is yet another method for finding the inverse of a 2x2 matrix. This method only works for the 2x2. All the previousmethods will work forany square matrices.

A=

dc

ba. Determinant of A=detA=|A|=

dc

ba=ad-bc.

A-1 =

bcad

a

bcad

cbcad

b

bcad

d

.

If det=0, then inverse dne! Why?

Ex 4 Find the inverse of A=

63

42.

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Ex 5 Find the inverse of A=

11

34.

Solve using an inverse matrix:AX=B

A-1AX=A-1B Be careful! Order matters since mult is NOT comm: A-1BBA-1.IX=A-1BX=A-1B

Ex 6 Solve using an inverse matrix:

x1-x2+x3=12x2-x3=1

2x1+3x2=1

(x1,x2,x3)=(5,-3,-7). This is one solution, consistent, independent.

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7.7 Determinant of a square matrixp556#6,12,26,30,38,40,44,52,58,64,66-68,70-76E

For a 1x1 matrix: detA=element of A.If A=[-5.12], then detA=-5.12.

Recall from section 7.6: For a 2x2 matrix:

If A=

dc

ba, then the determinant of A=detA=|A|=

dc

ba=ad-bc.

Notice the notation: a matrix is denoted with brackets[ ], whereas a determinant is denoted with | |.

For a 3x3 matrix: detA=|A|=a11M11-a12M12+a13M13

For a 4x4 matrix: detA=|A|=a11M11-a12M12+a13M13-a14M14

For a 5x5 matrix: detA=|A|=a11M11-a12M12+a13M13-a14M14+a15M15

For a nxn even matrix: detA=|A|=a11M11-a12M12+a13M13--a1nM1nFor a nxn odd matrix: detA=|A|=a11M11-a12M12+a13M13-+a1n M1n

Ex Find the det of A=

1

0

2

4

5

4

1

3

3

.

A triangular matrix is a square matrix with all zero entries either below or above it principle diagonal. An uppertriangular matrix has all zeros below its principle diagonal. A lower triangular matrix has all zeros above its principlediagonal. The determinant of a triangular matrix of any order is the product of the entries of its principle diagonal.

Ex 4 Find the determinant of the triangular matrix.4 29 75 38

0 10

6 37

0 0 5 16

0 0 0 1

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7.8 Applications of matrices and determinants hwk: p567#2,6-14E,18,22,28,30-34E and print out ch 8 notes

Area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3) is

1

1

1

2

1

33

22

11

yx

yx

yx

A = .

Choose to get positive area.

Ex1 Find the area of a triangle with verticesa) (0,0), (4,1), (2,5)b) (3,-1), (0,-3), (9,3)

Test for collinear points:

3 points (x1,y1), (x2,y2), (x3,y3) are collinear iff .0

1

1

1

33

22

11

=

yx

yx

yx

Cramers Rule is yet another way of solving matrices:

D

Dx

D

Dx

D

Dx

xxx 321

321,, ===

Cramers Rule does not work when determinant is zero. Why?

Ex 2: Use Cramers Rule, if possible to solve the system of linear equations.

a) 3x+2y=-2 b) -7x+11y=-1 c) 3x+3y+5z=16x+4y=4 3x-9y=9 3x+5y+9z=2

5x+9y+17z=4

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Cryptography: My method is NOT the book method; its more efficient.

To encode a message, pick a square key and multiply it to the message.

Sample I pick A=

11

34as the key. Lets say my message is do your hwk.

Let A=1, B=2, C=3,,X=24, Y=25, Z=26, space=0.So do your hwk is 4-15-0-25-15-21-18-0-8-23-11-0.Write this string of number as a matrix with the same # of row as your key.

Message=

0230

11818

21

15

25

0

15

4. Use 0 as filler. (If you having a longer message to encode, then choose a

larger key.)

To encode this message so no one can read it unless they have the key:

multiply key to message:

11

34

0230

11818

21

15

25

0

15

4

=

113118

4410172

36

123

25

75

19

61

How do we decode a message?Since A[message]=[code],then A-1 A[message]=A-1 [code],so I[message]=A-1 [code],therefore [message]=A-1 [code].

Ex3 Decode: 115-30-4-1-107-28-53-16-62-19-16-4 given that key=

11

34.

__________ Rev, p576#1-18__________ Ch 7 group quiz__________ Correct group quiz__________ Chapter 7 test

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