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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. x + 2 y + 0 z = 12. 2 x + y + z = 14. 0 x + y + 3 z = 16. Example 1B: Representing Systems as Matrices.
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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.
x + 2y + 0z = 12 2x + y + z = 14 0x + y + 3z = 16 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
Check It Out! Example 1a Write the augmented matrix. Step 2 Write the augmented matrix, with coefficients and constants. Step 1 Write each equation in the ax + by = c form. –x – y = 0 –x – y = –2
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
3 1 2 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.
– 1 2 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
7 –3 – 1 2 1 2 Example 2A Continued Step 4 Multiply row 1 by 7 and row 2 by –3. Step 5 Subtract row 2 from row 1. Write the result in row 1.
1x = 4 42 21 1y = 3 1 2 Example 2A Continued Step 6 Divide row 1 by 42 and row 2 by 21. The solution is x = 4, y = 3. Check the result in the original equations.
2 1 2 3 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 1 Check It Out! Example 2b Continued Step 3 Add row 1 to row 2. Write the result in row 2. 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. c = flea collars 5f + 6c + 2t = 1040 f – 4c = 0 f = bags of cat food c + t = 100 t = catnip toys Enter the 3 4 augmented matrix as A.
Press , select MATH, and move down the list to B:rref( to find the reduced row-echelon form of the augmented matrix. Example 3 Continued 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