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4.4-4.5 Solving Systems of Equations with Matrices

4.4-4.5 Solving Systems of Equations with Matrices. Writing a System of Equations in Matrix Equation Form. Words  System.

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4.4-4.5 Solving Systems of Equations with Matrices

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  1. 4.4-4.5 Solving Systems of Equations with Matrices

  2. 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

  3. 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

  4. 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.

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

  6. 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

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

  8. 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.

  9. 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.

  10. Elementary Row Operations

  11. Examples • System: 2x + 4y =6 x – 3y = -2 2 46 1 -3-2

  12. Examples This is what we want it to look like!! 1 0 ___ 0 1 ___ ½ R1 R1 2 46 1 -3-2 ½ R11 23  R1 1 23 1 -3-2 R2 - R1 R2 R2 1 2 3 R1-1-3-2 0 5 5R2 1 23 0 5 5

  13. Examples This is what we want it to look like!! 1 0 ___ 0 1 ___ 1 23 0 55 (1/5) R2 R2 (1/5) R20 1 1  R2 R1 - 2R2 R1 1 2 3 0 1 1 R1 1 2 3 2R2- 02 2 1 0 1R2 101 0 11 This means the solutions are: x = 1 y = 1

  14. Examples • System: -5x + 5y = 10 -2x + 2y = -4 -5 510 -2 2 -4

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

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

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

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

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

  20. Classwork/Homework

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