chapter 9: systems of equations and inequalities; matrices

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 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.

bxaxaxa nn 2211

• 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 Linear System in Two Variables

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

9.1 Substitution Method

Example Solve the system.

Solution

4

31

1

55

11623

11)3(23

3

y

y

x

x

xx

xx

xy Solve (2) for y.

Substitute y = x + 3 in (1).

Solve for x.

Substitute x = 1 in y = x + 3.

Solution set: {(1, 4)}

3

1123

yx

yx (1)

(2)

9.1 Solving a Linear System in Two Variables Graphically

Example Solve the system graphically.

Solution Solve (1) and (2) for y.

3

1123

yx

yx (1)

(2)

• Solve using the method of graphing.

x2 + y2 = 25

x2 + y = 19

Solving a Nonlinear System of Equations


Solution Choose the simpler equation, (2), and

solve for y since x is squared in (1).

Substitute for y into (1) .

43

523 2

yx

yx (1)

(2)

(3)3

4

43

xy

yx

34 x

Solving a Nonlinear System of Equations

Substitute these values for x into (3).

The solution set is

97

or 10)1)(79(

0729

15)4(29

53

423

2

2

2

xxxx

xx

xx

xx

1

314

or2743

34 9

7

yy

.1,1,, 2743

97

9.2 Elimination Method


Solution To eliminate x, multiply (1) by –2 and (2)

by 3 and add the resulting equations.

3696

286

yx

yx (3)

(4)

2

3417

y

y

1232

143

yx

yx (1)

(2)

9.2 Elimination Method

Substitute 2 for y in (1) or (2).

The solution set is {(3, 2)}.

Consistent System

3

93

1)2(43

x

x

x

Solving an Inconsistent System


Solution Eliminate x by multiplying (1) by 2 and

adding the result to (2).

Solution set is .

746

423

yx

yx (1)

(2)

746

846

yx

yx

150 Inconsistent System

Solving a System with Dependent Equations


Solution Eliminate x by multiplying (1) by 2 and adding the

result to (2).

Consistent & Dependent

SystemSolution set is all R’s.

42824

yxyx (1)

(2)

428

428

yx

yx

00

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.

Pick a point from the shaded region and test that point in the equation y = x + 3.

Point: (-4, 5)

y x

3

5 4 3

5 1

This is incorrect. Five is greater than or equal to negative 1.

5 1 5 1 5 1 o r

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.

Pick a point from the shaded region and test that point in the equation y = -x + 4.

Point: (1, -3)

y x

4

3 1 4

3 3

This is incorrect. Negative three is less than or equal to 3.

3 3 3 3 3 3 o r

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 x

y x

2 4

3 2

x

y

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. Shade the appropriate region.


y x

2 4

3 2y x

2 4 P o in t (0 ,0 )

0 2 (0 ) - 4

0 -4

The region above the line should be shaded.

x

y

Now do the same for the second inequality.


y x

2 4

3 2

x

y

Use the slope and y-intercept to plot two points for the second inequality.

Draw in the line. For < use a dashed line.

Pick a point and test it in the inequality. Shade the appropriate region.


y x

2 4

3 2

x

y

y x

3 2

3

P o in t (-2 ,-2 )

-2 (-2 ) + 2

-2 < 8

The region below the line should be shaded.


y x

2 4

3 2

x

y

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

y x

4

2

y x

y x

4

2

x

y Use the slope and y-intercept to plot two points for the first inequality.

Draw in the line.

Shade in the appropriate region.

x

y

y x

y x

4

2

Use the slope and y-intercept to plot two points for the second inequality.

Draw in the line.

Shade in the appropriate region.

x

y

y x

y x

4

2

The final solution is the region where the two shaded areas overlap (purple region).

Sect. 9.4 Linear Programming

Goal 1 Find Maximum and Minimum values of a function

Goal 2 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

y

y 6x + 42

3

2

1x

y 4

y - x + 6

y

y 6x + 42

3

2

1x

6

4

2

-2

-4

-6

-5 5

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 1Example 2

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.

y - 3

x - 2

y 72

5 x

Bounded Region

12

10

8

6

4

2

-2

-4

-10 -5 5 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) 11

(4, - 3) 2(4) – 5(- 3) 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

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.

Unbounded Region

y + 3x 4

y - 3x – 4

y 8 + x

10

8

6

4

2

-2

-4

-6

-10 -5 5 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)

(- 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.

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 Inequalities16

14

12

10

8

6

4

2

5 10 15 20

Pies

cakes

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

(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 6: Substitute the coordinates of the vertices into function.

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 9.5Solving 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:

412

1701

229

7

2

10

41

3

22

1

11

5

0

2

9

• Matrix Row Transformations– Streamlined use of echelon methods

3333

2222

1111

dzcybxa

dzcybxa

dzcybxa

Columns

Rows

3

2

1

333

222

111

d

d

d

cba

cba

cba

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.

3

2

1

333

222

111

d

d

d

cba

cba

cba

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

3

4

1

412

1701

229

7

2

10

41

7

24

y

yx

342

417

1229

cba

ca

cba

Reduced Row Echelon Method with the Graphing Calculator


Solution The augmented matrix of the system is

shown below.

4432

56223

302

zyx

zyx

zyx

9.5 Reduced Row Echelon Method with the Graphing Calculator

Using the rref command we obtain the row reduced

echelon form.

.10)}6,{(8,set solution with ,

10

6

8

system therepresents This

z

y

x

9.5 Solving a System with No Solutions

Example Show that the following system is inconsistent.

Solution The augmented matrix of the system is

9242

132

42

zyx

zyx

zyx

.

9

1

4

242

321

121

9.5 Solving a System with No Solutions

The final row indicates that the system is inconsistent and has solution set .

1

1

4

000

321

121

331 RRR2

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

26104

824

1352

zyx

zyx

zyx

.

2

8

1

6104

241

352


The final row of 0s indicates that the system has

dependent equations. The first two rows represent

the system

0

8

1

000

241

352

331 RRR2

.824

1352

zyx

zyx


Solving for y we get

Substitute this result into the expression to find x.

Solution set written with z arbitrary:

.137

1315

zy

zx

zzx

zzx

zyx

132

1344

821328

1360

82137

1315

4

824

.,, 13715

13244 zzz

Matrix Algebra Basics

9.6

Algebra

Matrix

,,

,,

,,

1

221

111

mnm

n

n

aa

aa

aa

A

A matrix is any doubly subscripted array of elements arranged in rows and columns.

Row Matrix

[1 x n] matrix

,, 2 1 naaaA

Column Matrix

2

1

ma

a

a

A

[m x 1] matrix

Square Matrix

B

5 4 7

3 6 1

2 1 3

Same number of rows and columns

The Identity

Identity Matrix

I

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

Square matrix with ones on the diagonal and zeros elsewhere.

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:

mnmnmn BAC

Note: all three matrices are of the same dimension

Addition

A a11 a12

a21 a22

B b11 b12

b21 b22

C a11 b11 a12 b12

a21 b21 a 22 b22

If

and

then

Matrix Addition Example

A B 3 4

5 6

1 2

3 4

4 6

8 10

C

Matrix Subtraction

C = A - BIs defined by

mnmnmn BAC

Matrix Multiplication

Matrices A and B have these dimensions:

[r x c] and [s x d]


Matrices A and B can be multiplied if:

[r x c] and [s x d]

c = s


The resulting matrix will have the dimensions:

[r x c] and [s x d]

r x d

Computation: A x B = C

A a11 a12

a21 a22

B b11 b12 b13

b21 b22 b23

232213212222122121221121

2312131122121211 21121111

babababababa

babababababaC

[2 x 2]

[2 x 3]

[2 x 3]


A

2 3

1 1

1 0

and B

1 1 1

1 0 2

[3 x 2] [2 x 3]A and B can be multiplied

1 1 1

3 1 2

8 2 5

12*01*1 10*01*1 11*01*1

32*11*1 10*11*1 21*11*1

82*31*2 20*31*2 51*31*2

C

[3 x 3]


1 1 1

3 1 2

8 2 5

12*01*1 10*01*1 11*01*1

32*11*1 10*11*1 21*11*1

82*31*2 20*31*2 51*31*2

C

A

2 3

1 1

1 0

and B

1 1 1

1 0 2

[3 x 2] [2 x 3]

[3 x 3]

Result is 3 x 3

chapter 9: systems of equations and inequalities; matrices

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