4.4-4.5 Solving Systems of Equations with Matrices

Writing a System of Equations in Matrix Equation Form Words  System The school that Imani goes to is selling tickets to the annual dance competition. On the first day of ticket sales the school sold 3 senior citizen tickets and 3 child tickets for a total of $69. The school took in $91 on the second day by selling 5 senior citizen tickets and 3 child tickets. What is the price each of one senior citizen ticket and one child ticket? Let x = price of Senior Citizen Tickets Let y = price of Child Tickets

Writing a System of Equations in Matrix Equation Form Words  System  Matrix Let x = price of Senior Citizen Tickets Let y = price of Child Tickets Variables Constants Coefficients

Writing a System of Equations in Matrix Equation Form Try this one on your own: Amanda and Ndiba are selling flower bulbs for a school fundraiser. Customers can buy packages of tulip bulbs and bags of daffodil bulbs. Amanda sold 6 packages of tulip bulbs and 12 bags of daffodil bulbs for a total of $198. Ndiba sold 7 packages of tulip bulbs and 6 bags of daffodil bulbs for a total of $127. Find the cost each of one package of tulips bulbs and one bag of daffodil bulbs.

Writing a System of Equations in Matrix Equation Form Let x = Tulips Let y = Daffodils 6x + 12y = 198 7x + 6y = 127

Solving Systems Using Row Operations The row reduction method is used to solve systems of equations. The row reduction method is performed on an augmented matrix. An augmented matrix consists of the coefficients and constant terms in the system of equations. Ex: 3x + y = 5 -x+2y = 3

Solving Systems Using Row Operations 3x + y = 5 -3x + 6y = 9 7y = 14 3x = 5 y = 2 x=1 y=2 What’s special about this matrix? Identity Matrix!!

Solving Systems Using Row Operations The goal of the row-reduction method is to transform, if possible, the coefficients columns into columns that form an identity matrix. This is called reduced row-echelon form.

Elementary Row Operations The following operations produce equivalent matrices, and may be used in any order and as many times as necessary to obtain reduced row-echelon form. interchange two rows. Multiply all entries in one row by a nonzero number. Add a multiple of one row to another row.

Elementary Row Operations Notation Interchange rows 1 and 2 R1↔R2 Multiply each entry in row 3 by -2 -2R3  R3 Replace row 1 with the sum of row 1 and 4 times each entry in row 2 4R2 + R1  R1

Examples System: 2x + 4y =6 x – 3y = -2 2 4 6 1 -3 -2

This is what we want it to look like!! Examples 1 0 ___ 0 1 ___ This is what we want it to look like!! 2 4 6 1 -3 -2 ½ R1  R1 ½ R1 1 2 3  R1 1 2 3 1 -3 -2 R2 - R1  R2 R2 1 2 3 R1 - 1 -3 -2 0 5 5  R2 1 2 3 0 5 5

This means the solutions are: Examples 1 0 ___ 0 1 ___ This is what we want it to look like!! 1 2 3 0 5 5 (1/5) R2  R2 (1/5) R2 0 1 1  R2 1 2 3 0 1 1 R1 - 2R2  R1 1 0 1 0 1 1 R1 1 2 3 2R2 - 0 2 2 1 0 1  R2 This means the solutions are: x = 1 y = 1

Examples System: -5x + 5y = 10 -2x + 2y = -4 -5 5 10 -2 2 -4

Examples System : x + y + z =21 2x + y = 23 y + 3z = 25 1 1 1 21 1 1 1 21 2 1 0 23 0 1 3 25

This is what we want it to look like!! Examples 1 0 0 ___ 0 1 0 ___ 0 0 1 ___ This is what we want it to look like!! 1 1 1 21 2 1 0 23 0 1 3 25 -2R1+R2  R2 -2R1 -2 -2 -2 -42 R2 + 2 1 0 23 0 -1 -2 -19  R2 1 1 1 21 0 -1 -2 -19 0 1 3 25 -1R2  R2 -1R2 0 1 2 19  R2 1 1 1 21 0 1 2 19 0 1 3 25

This is what we want it to look like!! Examples 1 0 0 ___ 0 1 0 ___ 0 0 1 ___ This is what we want it to look like!! 1 1 1 21 0 1 2 19 0 1 3 25 R2 – R3  R3 R2 0 1 2 19 R3 - 0 1 3 25 0 0 -1 -6  R3 1 1 1 21 0 1 2 19 0 0 -1 -6 -1R3  R3 -1R3 0 0 1 6  R3 1 1 1 21 0 1 2 19 0 0 1 6

This is what we want it to look like!! Examples 1 0 0 ___ 0 1 0 ___ 0 0 1 ___ This is what we want it to look like!! 1 1 1 21 0 1 2 19 0 0 1 6 R1 – R2  R1 R1 1 1 1 21 R2 - 0 1 2 19 1 0 -1 2  R1 1 0 -1 2 0 1 2 19 0 0 1 6 R1 + R3  R1 R1 1 0 -1 2 R3 + 0 0 1 6 1 0 0 8  R1 1 0 0 8 0 1 2 19 0 0 1 6

This means the solutions are: Examples 1 0 0 ___ 0 1 0 ___ 0 0 1 ___ This is what we want it to look like!! 1 0 0 8 0 1 2 19 0 0 1 6 R2 – 2R3  R2 R2 0 1 2 19 2R3 - 0 0 2 12 0 1 0 7  R2 1 0 0 8 0 1 0 7 0 0 1 6 This means the solutions are: x = 8 y = 7 z = 6

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