4-6 Row Operations and Augmented Matrices Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

Warm Up Solve. 1. 2. 3. What are the three types of linear systems? (4, 3) (8, 5) consistent independent, consistent dependent, inconsistent

Objective Use elementary row operations to solve systems of equations.

Vocabulary augmented matrix row operation row reduction reduced row-echelon form

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 Write the augmented matrix for the system of equations. Step 2 Write the augmented matrix, with coefficients and constants. Step 1 Write each equation in the Ax + By + Cz =D 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. Step 2 Write the augmented matrix, with coefficients and constants. –x – y = 0 –x – y = –2

Check It Out! Example 1b Write the augmented matrix. Step 2 Write the augmented matrix, with coefficients and constants. 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 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 1

Example 2A Continued Step 3 Subtract row 1 from row 2. Write the result in row 2. – 1 2 Although row 2 is now –7y = –21, an equation easily solved for y, row operations can be used to solve for both variables

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. 5f + 6c + 2t = 1040 c = flea collars f – 4c = 0 f = bags of cat food c + t = 100 t = catnip toys Enter the 3  4 augmented matrix as A.

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.

Example 1: Solving a Linear System in Three Variables Use elimination to solve the system of equations. 5x – 2y – 3z = –7 1 2x – 3y + z = –16 2 3x + 4y – 2z = 7 3 Step 1 Eliminate one variable. 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.

Use equations and to create a second equation in x and y. Example 1 Continued 5x – 2y – 3z = –7 5x – 2y – 3z = –7 1 Multiply equation - by 3, and add to equation . 1 2 3(2x –3y + z = –16) 2 6x – 9y + 3z = –48 11x – 11y = –55 4 Use equations and to create a second equation in x and y. 3 2 1 3x + 4y – 2z = 7 3x + 4y – 2z = 7 3 Multiply equation - by 2, and add to equation . 3 2 2(2x –3y + z = –16) 4x – 6y + 2z = –32 2 7x – 2y = –25 5

You now have a 2-by-2 system. 7x – 2y = –25 Example 1 Continued 11x – 11y = –55 4 You now have a 2-by-2 system. 7x – 2y = –25 5

You can eliminate y by using methods from Lesson 3-2. Example 1 Continued Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from Lesson 3-2. Multiply equation - by –2, and equation - by 11 and add. 4 5 –2(11x – 11y = –55) –22x + 22y = 110 4 1 11(7x – 2y = –25) 77x – 22y = –275 5 55x = –165 1 x = –3 Solve for x.

Step 3 Use one of the equations in your 2-by-2 system to solve for y. Example 1 Continued Step 3 Use one of the equations in your 2-by-2 system to solve for y. 11x – 11y = –55 4 1 11(–3) – 11y = –55 Substitute –3 for x. 1 y = 2 Solve for y.

Substitute –3 for x and 2 for y. 2(–3) – 3(2) + z = –16 Example 1 Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 2x – 3y + z = –16 2 Substitute –3 for x and 2 for y. 2(–3) – 3(2) + z = –16 1 z = –4 Solve for y. 1 The solution is (–3, 2, –4).

Use elimination to solve the system of equations. Check It Out! Example 1 Use elimination to solve the system of equations. –x + y + 2z = 7 1 2x + 3y + z = 1 2 –3x – 4y + z = 4 3 Step 1 Eliminate one variable. In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1.

Example 2: Business Application 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. Orchestra Mezzanine Balcony Total Sales Fri 200 30 40 $1470 Sat 250 60 50 $1950 Sun 150 $1050

Write a system of equations to represent the data in the table. Example 2 Continued Step 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 Friday’s sales. 1 250x + 60y + 50z = 1950 Saturday’s sales. 2 150x + 30y = 1050 Sunday’s sales. 3 A variable is “missing” in the last equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.

The systems in Examples 1 and 2 have unique solutions 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. Consistent means that the system of equations has at least one solution. Remember!

Example 3: Classifying Systems with Infinite Many Solutions or No Solutions Classify the system as consistent or inconsistent, and determine the number of solutions. 2x – 6y + 4z = 2 1 –3x + 9y – 6z = –3 2 5x – 15y + 10z = 5 3

Example 3 Continued The elimination method is convenient because the numbers you need to multiply the equations are small. First, eliminate x. Multiply equation by 3 and equation by 2 and add. 2 1 3(2x – 6y + 4z = 2) 6x – 18y + 12z = 6 1 2(–3x + 9y – 6z = –3) –6x + 18y – 12z = –6 2 0 = 0 

Example 3 Continued Multiply equation by 5 and equation by –2 and add. 3 1 5(2x – 6y + 4z = 2) 10x – 30y + 20z = 10 1 –2(5x – 15y + 10z = 5) –10x + 30y – 20z = –10 3  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.

Classify the system, and determine the number of solutions. Check It Out! Example 3a Classify the system, and determine the number of solutions. 3x – y + 2z = 4 1 2x – y + 3z = 7 2 –9x + 3y – 6z = –12 3

Check It Out! Example 3a Continued The elimination method is convenient because the numbers you need to multiply the equations by are small. First, eliminate y. Multiply equation by –1 and add to equation . 1 2 3x – y + 2z = 4 3x – y + 2z = 4 1 –1(2x – y + 3z = 7) –2x + y – 3z = –7 3 x – z = –3 4

Check It Out! Example 3a Continued Multiply equation by 3 and add to equation . 3 2 3(2x – y + 3z = 7) 6x – 3y + 9z = 21 2 –9x + 3y – 6z = –12 –9x + 3y – 6z = –12 3 –3x + 3z = 9 5 Now you have a 2-by-2 system. x – z = –3 4 –3x + 3z = 9 5

Check It Out! Example 3a Continued Eliminate x. 3(x – z = –3) 3x – 3z = –9 4 –3x + 3z = 9 –3x + 3z = 9 5 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.

Classify the system, and determine the number of solutions. Check It Out! Example 3b Classify the system, and determine the number of solutions. 2x – y + 3z = 6 1 2x – 4y + 6z = 10 2 y – z = –2 3

Check It Out! Example 3b Continued Use the substitution method. Solve for y in equation 3. y – z = –2 Solve for y. 3 y = z – 2 4 Substitute equation in for y in equation . 4 1 2x – y + 3z = 6 2x – (z – 2) + 3z = 6 2x – z + 2 + 3z = 6 2x + 2z = 4 5

Check It Out! Example 3b Continued Substitute equation in for y in equation . 4 2 2x – 4y + 6z = 10 2x – 4(z – 2) + 6z = 10 2x – 4z + 8 + 6z = 10 2x + 2z = 2 6 Now you have a 2-by-2 system. 2x + 2z = 4 5 2x + 2z = 2 6

Check It Out! Example 3b Continued Eliminate z. 2x + 2z = 4 5 –1(2x + 2z = 2) 6 0  2  Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.