Holt McDougal Algebra 2 4-6 Row Operations and Augmented Matrices 4-6 Row Operations and Augmented...

Holt McDougal Algebra 2

4-6 Row Operations andAugmented Matrices4-6 Row Operations and

Augmented Matrices

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz



4-6 Row Operations andAugmented Matrices

Warm UpSolve.

1.

2.

3. What are the three types of linear systems?

consistent independent, consistent dependent, inconsistent

(4, 3)

(8, 5)



Use elementary row operations to solvesystems of equations.

Objective



augmented matrixrow operationrow reductionreduced row-echelon form

Vocabulary



In previous lessons, you saw how Cramer’s rule and inverses can be used to solve systems of equations. Solving large systems requires a different method using an augmented matrix.

An augmented matrix consists of the coefficients and constant terms of a system of linear equations.

A vertical line separates the coefficients from the constants.



Example 1A: Representing Systems as Matrices

Write the augmented matrix for the system of equations.

Step 1 Write each equation in the ax + by = c form.

Step 2 Write the augmented matrix, with coefficients and constants.

6x – 5y = 14

2x + 11y = 57



Example 1B: Representing Systems as Matrices

Step 1 Write each equation in the Ax + By + Cz =D


Write the augmented matrix for the system of equations.

x + 2y + 0z = 12

2x + y + z = 14

0x + y + 3z = 16



Check It Out! Example 1a

Write the augmented matrix.

Step 1 Write each equation in the ax + by = c form.


–x – y = 0

–x – y = –2



Check It Out! Example 1b

Write the augmented matrix.

Step 1 Write each equation in the Ax + By + Cz =D


–5x – 4y + 0z = 12

x + 0y + z = 3

0x + 4y + 3z = 10



You can use the augmented matrix of a system to solve the system. First you will do a row operation to change the form of the matrix. These row operations create a matrix equivalent to the original matrix. So the new matrix represents a system equivalent to the original system.

For each matrix, the following row operations produce a matrix of an equivalent system.





Row reduction is the process of performing elementary row operations on an augmented matrix to solve a system. The goal is to get the coefficients to reduce to the identity matrix on the left side.

This is called reduced row-echelon form.

1x = 5

1y = 2



Example 2A: Solving Systems with an Augmented Matrix

Write the augmented matrix and solve.

Step 1 Write the augmented matrix.

Step 2 Multiply row 1 by 3 and row 2 by 2.

3

2

12



Example 2A Continued

Step 3 Subtract row 1 from row 2. Write the result in row 2.

Although row 2 is now –7y = –21, an equation easily solved for y, row operations can be used to solve for both variables

– 12



Example 3: Charity Application

A shelter receives a shipment of items worth $1040. Bags of cat food are valued at $5 each, flea collars at $6 each, and catnip toys at $2 each. There are 4 times as many bags of food as collars. The number of collars and toys together equals 100. Write the augmented matrix and solve, using row reduction, on a calculator. How many of each item are in the shipment?



Example 3 Continued

Use the facts to write three equations.

Enter the 3 4 augmented matrix as A.

5f + 6c + 2t = 1040

f – 4c = 0

c + t = 100

f = bags of cat food

c = flea collars

t = catnip toys







Identifying the exact solution from a graph of a 3-by-3 system can be very difficult. However, you can use the methods of elimination and substitution to reduce a 3-by-3 system to a 2-by-2 system and then use the methods that you learned in Lesson 3-2.



Use elimination to solve the system of equations.

Example 1: Solving a Linear System in Three Variables

Step 1 Eliminate one variable.

5x – 2y – 3z = –7

2x – 3y + z = –16

3x + 4y – 2z = 7

In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1 and z is easy to eliminate from the other equations.

1

2

3



Example 1 Continued

5x – 2y – 3z = –7

11x – 11y = –55

3(2x –3y + z = –16)

5x – 2y – 3z = –7

6x – 9y + 3z = –48

1

2

1

4

3x + 4y – 2z = 7

7x – 2y = –25

2(2x –3y + z = –16)3x + 4y – 2z = 74x – 6y + 2z = –32

3

2

Multiply equation - by 3, and add to equation .1

2

Multiply equation - by 2, and add to equation .3

2

5

Use equations and to create a second equation in x and y.

3 2



11x – 11y = –55

7x – 2y = –25You now have a 2-by-2 system.

4

5

Example 1 Continued



–2(11x – 11y = –55)

55x = –165

11(7x – 2y = –25) –22x + 22y = 110 77x – 22y = –275

4

5

1

1Multiply equation - by –2, and equation - by 11 and add.

4

5

Step 2 Eliminate another variable. Then solve for the remaining variable.

You can eliminate y by using methods from Lesson 3-2.

x = –3 Solve for x.

Example 1 Continued



11x – 11y = –55

11(–3) – 11y = –55

4

1

1

Step 3 Use one of the equations in your 2-by-2 system to solve for y.

y = 2

Substitute –3 for x.

Solve for y.

Example 1 Continued



2x – 3y + z = –16

2(–3) – 3(2) + z = –16

2

1

1

Step 4 Substitute for x and y in one of the original equations to solve for z.

z = –4

Substitute –3 for x and 2 for y.

Solve for y.

The solution is (–3, 2, –4).

Example 1 Continued



Use elimination to solve the system of equations.

Step 1 Eliminate one variable.

–x + y + 2z = 7

2x + 3y + z = 1

–3x – 4y + z = 4

1

2

3

Check It Out! Example 1

In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1.



The table shows the number of each type of ticket sold and the total sales amount for each night of the school play. Find the price of each type of ticket.

Example 2: Business Application

Orchestra Mezzanine Balcony Total Sales

Fri 200 30 40 $1470

Sat 250 60 50 $1950

Sun 150 30 0 $1050



Example 2 ContinuedStep 1 Let x represent the price of an orchestra seat,

y represent the price of a mezzanine seat, and z represent the present of a balcony seat.

Write a system of equations to represent the data in the table.

200x + 30y + 40z = 1470

250x + 60y + 50z = 1950

150x + 30y = 1050

1

2

3

Friday’s sales.

Saturday’s sales.

Sunday’s sales.

A variable is “missing” in the last equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.



Consistent means that the system of equations has at least one solution.

Remember!

The systems in Examples 1 and 2 have unique solutions. However, 3-by-3 systems may have no solution or an infinite number of solutions.



Classify the system as consistent or inconsistent, and determine the number of solutions.

Example 3: Classifying Systems with Infinite Many Solutions or No Solutions

2x – 6y + 4z = 2

–3x + 9y – 6z = –3

5x – 15y + 10z = 5

1

2

3



Example 3 Continued

3(2x – 6y + 4z = 2)

2(–3x + 9y – 6z = –3)

First, eliminate x.

1

2

6x – 18y + 12z = 6

–6x + 18y – 12z = –6

0 = 0

Multiply equation by 3 and equation by 2 and add.2

1

The elimination method is convenient because the numbers you need to multiply the equations are small.



Example 3 Continued

5(2x – 6y + 4z = 2)

–2(5x – 15y + 10z = 5)

1

3

10x – 30y + 20z = 10

–10x + 30y – 20z = –10

0 = 0

Multiply equation by 5 and equation by –2 and add.

3

1

Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent and has an infinite number of solutions.



Check It Out! Example 3a

Classify the system, and determine the number of solutions.

3x – y + 2z = 4

2x – y + 3z = 7

–9x + 3y – 6z = –12

1

2

3



3x – y + 2z = 4–1(2x – y + 3z = 7)

First, eliminate y.

1

3

3x – y + 2z = 4

–2x + y – 3z = –7

x – z = –3

Multiply equation by –1 and add to equation . 1

2

The elimination method is convenient because the numbers you need to multiply the equations by are small.

Check It Out! Example 3a Continued

4



3(2x – y + 3z = 7)–9x + 3y – 6z = –12

2

3

6x – 3y + 9z = 21

–9x + 3y – 6z = –12

–3x + 3z = 9

Multiply equation by 3 and add to equation . 3

2

Now you have a 2-by-2 system.

x – z = –3

–3x + 3z = 9 5

4

5




3(x – z = –3)

–3x + 3z = 9 5

4 3x – 3z = –9

–3x + 3z = 9

0 = 0

Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent, and has an infinite number of solutions.

Eliminate x.




Check It Out! Example 3b

Classify the system, and determine the number of solutions.

2x – y + 3z = 6

2x – 4y + 6z = 10

y – z = –2

1

2

3



y – z = –2y = z – 2

3Solve for y.

Use the substitution method. Solve for y in equation 3.

Check It Out! Example 3b Continued

Substitute equation in for y in equation .4 1

4

2x – y + 3z = 6

2x – (z – 2) + 3z = 6

2x – z + 2 + 3z = 6

2x + 2z = 4 5



Substitute equation in for y in equation .4 2

2x – 4y + 6z = 10

2x – 4(z – 2) + 6z = 102x – 4z + 8 + 6z = 10

2x + 2z = 2 6

Now you have a 2-by-2 system.

2x + 2z = 4

2x + 2z = 2 6

5




2x + 2z = 4

–1(2x + 2z = 2)6

5

Eliminate z.

0 2


Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.

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Holt McDougal Algebra 2 4-6 Row Operations and Augmented Matrices 4-6 Row Operations and Augmented...

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