holt mcdougal algebra 2 4-6 row operations and augmented matrices 4-6 row operations and augmented...
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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
Holt McDougal Algebra 2
Holt McDougal Algebra 2
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)
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
Use elementary row operations to solvesystems of equations.
Objective
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
augmented matrixrow operationrow reductionreduced row-echelon form
Vocabulary
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
Example 1B: Representing Systems as Matrices
Step 1 Write each equation in the Ax + By + Cz =D
Step 2 Write the augmented matrix, with coefficients and constants.
Write the augmented matrix for the system of equations.
x + 2y + 0z = 12
2x + y + z = 14
0x + y + 3z = 16
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
Check It Out! Example 1a
Write the augmented matrix.
Step 1 Write each equation in the ax + by = c form.
Step 2 Write the augmented matrix, with coefficients and constants.
–x – y = 0
–x – y = –2
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
Check It Out! Example 1b
Write the augmented matrix.
Step 1 Write each equation in the Ax + By + Cz =D
Step 2 Write the augmented matrix, with coefficients and constants.
–5x – 4y + 0z = 12
x + 0y + z = 3
0x + 4y + 3z = 10
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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?
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
11x – 11y = –55
7x – 2y = –25You now have a 2-by-2 system.
4
5
Example 1 Continued
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
–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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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.
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Check It Out! Example 3a Continued
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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 3a Continued
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
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
Check It Out! Example 3b Continued
Holt McDougal Algebra 2
4-6 Row Operations andAugmented Matrices
2x + 2z = 4
–1(2x + 2z = 2)6
5
Eliminate z.
0 2
Check It Out! Example 3b Continued
Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.