1. Vocabulary multivariable linear system row-echelon form Gaussian elimination augmented matrix coefficient matrix reduced row-echelon form Gauss-Jordan elimination

2. Key Concept 1

3. Example 1 Gaussian Elimination with a System Write the system of equations in triangular form using Gaussian elimination. Then solve the system. x + 3y + 2z = 5 3x + y – 2z = 7 2x + 2y + 3z = 3 Step 1Eliminate the x-term in Equation 2. [–3(Equation 1) + Equation 2] (–3)(x + 3y + 2z= 5) →  3x  9y  6z =  15 (+) 3x + y  2z= 7  8y  8z=  8

4. Example 1 Gaussian Elimination with a System x + 3y + 2z= 5 –8y – 8z = –8 2x + 2y + 3z= 3 Step 2 Eliminate the x-term in Equation 3. [  2(Equation 1) + Equation 3] (–2)(x + 3y + 2z = 5) →–2x – 6y  4z= –10 (+) 2x + 2y + 3z = 3 –4y – z = –7 x + 3y + 2z= 5 –8y – 8z = –8 4y – z= –7

5. Example 1 Gaussian Elimination with a System Step 3 Eliminate the y-term in Equation 3. [–2(Equation 3) + Equation 2] (  2)(  4y  z =  7) → 8y + 2z = 14 (+)  8y  8z=  8  6z= 6 x + 3y + 2z = 5  8y  8z =  8  6z = 6

6. Example 1 Answer: x – y – 2z = 1 y + z = 1 z = –1; (1, 2, –1) Gaussian Elimination with a System Step 4 Simplify Equations 2 and 3. Divide Equation 2 by  8 and divide Equation 3 by  6. x + 3y + 2z = 5 y + z = 1 z=  1 You can use substitution to find that y = 2 and x = 1. So, the solution of the system is x = 1, y = 2, and z = –1.

7. Example 1 Write the system of equations in triangular form using Gaussian elimination. Then solve the system. x – 2y + z = 3 2x – 3y + z = 3 x – 2y + 2z = 5 A.x –2y + z = 3 y – 2z = –3 z = 2; (–1, –1, 2) B.x + 2y  z = 3 y – 2z = –5 z = 2; (–1, –1, 2) C.x + 2y  z = 3 y – 2z = –5 z = 2; (7, –1, 2) D.x – 2y + z = 3 y + 2z = –5 z = 2; (–17, –9, 2)

8. Example 2 Write an Augmented Matrix Write the augmented matrix for the system of linear equations. x + y – z = 5 2w + 3x – z = –2 2w – x + y = 6 While each linear equation is in standard form, each of the four variables of the system is not represented in each equation, so the like terms do not align. Rewrite the system, using the coefficient 0 for the missing terms. Then write the augmented matrix.

9. Example 2 Write an Augmented Matrix System of EquationsAugmented Matrix 0w + x + y – z = 5 2w + 3x + 0y – z = –2 2w – x + y + 0z = 6 Answer:

10. Example 2 Write the augmented matrix for the system of linear equations. w + 2x + z = 2 x + y – z = 1 2w – 3x + y + 2z = –2 w + 3y – z = 5 A. B. C. D.

11. Key Concept 3

13. Example 3 Identify an Augmented Matrix in Row-Echelon Form There is a zero below the leading one in the first row. The second row also contains a leading 1. The matrix is in row-echelon form. A. Determine whether is in row-echelon form. Answer:yes

14. Example 3 Identify an Augmented Matrix in Row-Echelon Form The second row, which contains all zeros, needs be the last row in the matrix. The matrix is not in row-echelon form. B. Determine whether is in row-echelon form. Answer: no

15. Example 3 Identify an Augmented Matrix in Row-Echelon Form There is a zero below the leading one in the first row and the leading one in the second row. The row containing all zeros is the last row. The matrix is in row-echelon form. C. Determine whether is in row-echelon form. Answer: yes

16. Example 3 A.A B.B C.C D.D Which matrix is not in row-echelon form? A.B. C.D.

17. Example 4 Gaussian Elimination with a Matrix RESTAURANTS Three families ordered meals of hamburgers (HB), French fries (FF), and drinks (DR). The items they ordered and their total bills are shown below. Write and solve a system of equations to determine the cost of each item.

18. Example 4 Gaussian Elimination with a Matrix Write the information as a system of equations. Let x, y, and z represent hamburgers, French fries, and drinks, respectively. 6x + 5y + 5z = 55 2x + y + z = 15 x + 3y + 2z = 18 Next, write the augmented matrix and apply elementary row operations to obtain a row-echelon form of the matrix.

19. Example 4 Gaussian Elimination with a Matrix Augmented Matrix Switch R 3 with R 1. 3R 2 – R 3 

20. Example 4 Gaussian Elimination with a Matrix R 3 – 6R 1  13R 2 + R 3  R 2 

21. Example 4 Gaussian Elimination with a Matrix You can use substitution to find that y = 3 and x = 5. Therefore, the solution of the system is that hamburgers cost $5, French fries cost $3, and drinks cost $2. Answer:hamburgers: $5, French fries: $3, drinks: $2 R 3 

22. Example 4 MOVIES A movie theater sells three kinds of tickets: senior citizen (SC), adult (A), and children (C). The table shows the number of each kind of ticket purchased by three different families and their total bills. Write and solve a system of equations to determine the cost of each kind of ticket. A.senior citizen: $5, adult: $8, children: $2 B.senior citizen: $5, adult: $7, children: $4 C.senior citizen: $6, adult: $8, children: $2 D.senior citizen: $4, adult: $7, children: $6

23. Example 5 Use Gauss-Jordan Elimination Solve the system of equations. x – y + z = 3 –x + 2y – z = 2 2x – 3y + 3z = 8 R 1 + R 2  Augmented Matrix

24. Example 5 Use Gauss-Jordan Elimination R 1 + R 2  R 3 – 2R 1  3R 2 + R 3 

25. Example 5 Answer:x = 1, y = 5, z = 7 Use Gauss-Jordan Elimination R 1 – R 3  The solution of the system is x = 1, y = 5, and z = 7. Check this solution in the original system of equations.

26. Example 5 Solve the system of equations. 2x – y + z = 7 x + y – z = –4 y + 2z = 4 A.x = 1, y = 2, z = 3 B.x = –1, y = –2, z = –3 C.x = –1, y = 2, z = –3 D.x = 1, y = –2, z = 3

27. Example 6 A. Solve the system of equations. x + 2y + z = 8 2x + 3y – z = 13 x + y – 2z = 5 Write the augmented matrix. Then apply elementary row operations to obtain a reduced row-echelon matrix. No Solution and Infinitely Many Solutions Augmented Matrix

28. Example 6 No Solution and Infinitely Many Solutions R 2 – 2R 3  R 3 – R 1  R 3 + R 2 

29. Example 6 Answer: (5z + 2, –3z + 3, z) No Solution and Infinitely Many Solutions Because the value of z is not determined, this system has infinitely many solutions. Solving for x and y in terms of z, you have x = 5z + 2 and y = –3z + 3. So, a solution of the system can be expressed as (5z + 2, –3z + 3, z), where z is any real number. Write the corresponding system of linear equations for the reduced row-echelon form of the augmented matrix. x + 2y + z= 8 y + 3z= 3

30. Example 6 B. Solve the system of equations. 2x + 3y – z = 1 x + y – 2z = 5 x + 2y + z = 8 Write the augmented matrix. Then apply elementary row operations to obtain a reduced row-echelon matrix. No Solution and Infinitely Many Solutions Augmented Matrix

31. Example 6 Because 0 ≠ –12, the system has no solution. Answer: no solution No Solution and Infinitely Many Solutions R 1 – R 3  R 1 – R 2 

32. Example 6 Solve the system of equations. x + 2y + z = 2 x + y – 2z = 0 2y + 6z = 8 A.(–2, 0, 4) B.no solution C.(–3z, 2z, z) D.(–5, 5, –3)

33. Example 7 Infinitely Many Solutions Solve the system of equations. 4w + x + 2y – 3z = 10 3w + 4x + 2y + 8z = 3 w + 3x + 4y + 11z = 11 Write the augmented matrix. Then apply elementary row operations to obtain leading 1s in each row and zeros below these 1s in each column. Augmented Matrix

34. Example 7 Infinitely Many Solutions Interchange R 1 with R 3. R 2 – 3R 1  R 2 

35. Example 7 Infinitely Many Solutions R 3 – 4R 1  R 3 + 11R 2  R 3 

36. Example 7 Infinitely Many Solutions R 2 – 2R 3  R 1 – 3R 2  R 1 – 4R 3 

37. Example 7 Infinitely Many Solutions Write the corresponding system of linear equations for the reduced row-echelon form of the augmented matrix. w – 2z = 1 x + 3z = 2 y + z = 4 This system of equations has infinitely many solutions because for every value of z there are three equations that can be used to find the corresponding values of w, x, and y. Solving for w, x, and y in terms of z, you have w = 2z + 1, x = –3z – 2, and y = –z + 4.

38. Example 7 Infinitely Many Solutions Answer: (2z + 1, –3z – 2, –z + 4, z) So, a solution of the system can be expressed as (2z + 1, –3z – 2, –z + 4, z), where z is any real number. Check Using different values for z, calculate a few solutions and check them in the original system of equation. For example, if z = 1, a solution of the system is (3, –5, 3, 1). This solution checks in each equation of the original system. 4(3) + (–5) + 2(3) – 3(1) = 10 3(3) + 4(–5) + 2(3) + 8(1) = 3 3 + 3(–5) + 4(3) + 11(1) = 11

39. Example 7 Solve the system of equations. –w – x + y – z = 2 2w – x + 2y – z = 7 w – 2x – y – 2z = 1 A.(1, –z – 1, 2, z) B.(–1, z + 1, –2, –z) C.(z + 2, z + 1, z, z) D.no solution