An augmented matrix consists of the coefficients and constant terms of a system of linear equations. A vertical line separates the coefficients from the.

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Presentation transcript:

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 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

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 2A Continued Step 4 Multiply row 1 by 7 and row 2 by –3. 7 –3 1 2 Step 5 Subtract row 2 from row 1. Write the result in row 1. – 1 2

Example 2A Continued Step 6 Divide row 1 by 42 and row 2 by 21.  42  21 1 2 1x = 4 1y = 3 The solution is x = 4, y = 3. Check the result in the original equations.

Check It Out! Example 2b Write the augmented matrix and solve. Step 1 Write the augmented matrix. Step 2 Multiply row 1 by 2 and row 2 by 3. 2 3 1

Check It Out! Example 2b Continued Step 3 Add row 1 to row 2. Write the result in row 2. + 2 1 The second row means 0 + 0 = 60, which is always false. The system is inconsistent.

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.

Example 3 Continued Press , select MATH, and move down the list to B:rref( to find the reduced row-echelon form of the augmented matrix. There are 140 bags of cat food, 35 flea collars, and 65 catnip toys.

Check It Out! Example 3a Solve by using row reduction on a calculator. The solution is (5, 6, –2).

HW pg. 291 # 14, 15, 16, 18, 22, 23, 24

Homework set #2 HW pg. 291 # 17, 19, 20, 21, 25, 31, 34