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Chapter 9: Systems of Equations and Inequalities; MatricesPowerPoint Presentation

Chapter 9: Systems of Equations and Inequalities; Matrices

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### Matrix Algebra Basics Calculator

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.

9.1 Linear System in Two Variables

- Three possible solutions to a linear system in two variables:
- One solution: coordinates of a point,
- No solutions: inconsistent case,
- Infinitely many solutions: dependent case.

9.1

- Characteristics of a system of two linear equations in two variables.

9.1 Substitution Method

Example Solve the system.

Solution

(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

Example Solve the system graphically.

Solution Solve (1) and (2) fory.

(1)

(2)

- Solve using the method of graphing.
x2 + y2 = 25

x2 + y = 19

Solving a Nonlinear System of Equations

Example Solve the system.

Solution Choose the simpler equation, (2), and

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

Substitute for y into (1) .

(1)

(2)

9.2 Elimination Method

Example Solve the system.

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

by 3 and add the resulting equations.

(1)

(2)

(3)

(4)

9.2 Elimination Method

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

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

Consistent System

Solving an Inconsistent System

Example Solve the system.

Solution Eliminate x by multiplying (1) by 2 and

adding the result to (2).

Solution set is .

(1)

(2)

Inconsistent System

Solving a System with Dependent Equations

Example Solve the system.

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

result to (2).

Consistent & Dependent System

Solution set is all R’s.

(1)

(2)

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.

- The equation of each line is y = x + 3.
- The lines in each graph are the same and represent all of the solutions to the equation y = x + 3.
- The graph on the right is shaded above the line and this means that all of these points are solutions as well.

Inequalities with Greater Than

Point: (-4, 5)

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.

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.

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.

- Write the inequality in slope-intercept form.
- Use the slope and y-intercept to plot two points.
- 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 (>).
- 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.
- 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.

y

x

Graph the following linear system of inequalities.

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

Graph the following linear system of inequalities.

The region above the line should be shaded.

Now do the same for the second inequality.

y

x

Graph the following linear system of inequalities.

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

Graph the following linear system of inequalities.

The region below the line should be shaded.

y

x

Graph the following linear system of inequalities.

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.

y

x

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

Draw in the line.

Shade in the appropriate region.

y

x

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

Draw in the line.

Shade in the appropriate region.

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.

y 4

y - x + 6

y

y 6x + 4

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 + 4

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.

The maximum value is 17 at (5, 1). The Minimum value is – 7 at (- 1, - 2).

Example 1 function.

Example 2

y - 3

x - 2

y

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.

The polygon formed is a triangle with vertices at (- 2, 12), (- 2, - 3), and (4, - 3)

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 not a closed polygon.

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.

y + 3x 4

y - 3x – 4

y 8 + x

There are only two points of intersection (- 1, 7) and (- 3, 5). This is an Unbounded Region.

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. 5). This is an Unbounded Region.

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 5). This is an Unbounded Region.

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 5). This is an Unbounded Region.: 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 4 5). This is an Unbounded Region.: 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 5). This is an Unbounded Region.: 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 9.5 5). This is an Unbounded Region.Solving Systems using Matrices

- What is a Matrix? [ ]
- Augmented Matrices
- Solving Systems of 3 Equations
- Inconsistent & Dependent Systems

Concept: 5). This is an Unbounded Region.A Matrix

- Any rectangular array of numbers arranged in rows and columns, written within brackets
- Examples:

- Matrix Row Transformations 5). This is an Unbounded Region.
- Streamlined use of echelon methods

9.5 Solution of Linear Systems by Row Transformations 5). This is an Unbounded Region.

- 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: 5). This is an Unbounded Region.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

Reduced Row Echelon Method with the Graphing Calculator 5). This is an Unbounded Region.

Example Solve the system.

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.

9.5 Solving a System with No Solutions Calculator

Example Show that the following system is

inconsistent.

Solution The augmented matrix of the system is

9.5 Solving a System with No Solutions Calculator

The final row indicates that the system is

inconsistent and has solution set .

9.5 Solving a System with Dependent Equations Calculator

Example Show that the system has dependent

equations. Express the general solution using an

arbitrary variable.

Solution The augmented matrix is

9.5 Solving a System with Dependent Equations Calculator

The final row of 0s indicates that the system has

dependent equations. The first two rows represent

the system

9.5 Solving a System with Dependent Equations Calculator

Solving for y we get

Substitute this result into the expression to find x.

Solution set written with z arbitrary:

9.6

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