1. Multivariable Linear Systems and Row Operations
LESSON 6–1
Multivariable Linear Systems and Row Operations

2. Five-Minute Check (over Chapter 5) Then/Now New Vocabulary
Key Concept: Operations that Produce Equivalent Systems
Example 1: Gaussian Elimination with a System
Example 2: Write an Augmented Matrix
Key Concept: Elementary Row Operations
Key Concept: Row-Echelon Form
Example 3: Identify an Augmented Matrix in Row-Echelon Form
Example 4: Real-World Example: Gaussian Elimination with a Matrix
Example 5: Use Gauss-Jordan Elimination
Example 6: No Solution and Infinitely Many Solutions
Example 7: Infinitely Many Solutions
Lesson Menu

3. Find sec θ and cot θ if csc θ = 4 and tan < 0.B. C. D.
5–Minute Check 1

4. is equivalent to which of the following expressions?
A. csc2 t cot t
B. sec2 t cot t
C. csc t sec t
D. sin t cos t
5–Minute Check 2

5. Find the values of sin 2θ, cos 2θ, and tan 2θ if in the interval .B. C. D.
5–Minute Check 3

6. Solve sin 2θ – cos θ = 0 in the interval [0, 2π).B. C. D.
5–Minute Check 4

7. You solved systems of equations algebraically and represented data using matrices. (Lessons 0-5 and 0-6)
Solve systems of linear equations using matrices and Gaussian elimination.
Solve systems of linear equations using matrices and Gauss-Jordan elimination.
Then/Now

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

9. Key Concept 1

10. (–3)(x + 3y + 2z = 5) → 3x  9y  6z = 15 (+) 3x + y  2z = 7
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 1 Eliminate 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
Example 1

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

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

13. 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 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.
Answer: x – y – 2z = 1 y + z = 1 z = –1; (1, 2, –1)
Example 1

14. 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)
Example 1

15. 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.
Example 2

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

17. Write the augmented matrix for the system of linear equations
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.
Example 2

18. Key Concept 3

19. Key Concept 3

20. A. Determine whether is in row-echelon form.
Identify an Augmented Matrix in Row-Echelon Form
A. Determine whether is 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.
Answer: yes
Example 3

21. B. Determine whether is in row-echelon form.
Identify an Augmented Matrix in Row-Echelon Form
B. Determine whether is 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.
Answer: no
Example 3

22. C. Determine whether is in row-echelon form.
Identify an Augmented Matrix in Row-Echelon Form
C. Determine whether is 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.
Answer: yes
Example 3

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

24. 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.
Example 4

25. 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.
Example 4

26. Augmented Matrix Switch R3 with R1. 3R2 – R3 
Gaussian Elimination with a Matrix
Augmented Matrix
Switch R3 with R1.
3R2 – R3 
Example 4

27. Gaussian Elimination with a Matrix
R3 – 6R1
13R2 + R3
Example 4

28. Answer: hamburgers: $5, French fries: $3, drinks: $2
Gaussian Elimination with a Matrix
R3 
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
Example 4

29. A. senior citizen: $5, adult: $8, children: $2
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
Example 4

30. Use Gauss-Jordan Elimination
Solve the system of equations. x – y + z = 3 –x + 2y – z = 2 2x – 3y + 3z = 8
Augmented Matrix
R1 + R2
Example 5

31. Use Gauss-Jordan Elimination
R1 + R2
R3 – 2R1
3R2 + R3
Example 5

32. Use Gauss-Jordan Elimination
R1 – R3
The solution of the system is x = 1, y = 5, and z = 7. Check this solution in the original system of equations.
Answer: x = 1, y = 5, z = 7
Example 5

33. 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
Example 5

34. No Solution and Infinitely Many Solutions
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.
Augmented Matrix
Example 6

35. R2 – 2R3 R3 – R1 R3 + R2 No Solution and Infinitely Many Solutions
Example 6

36. No Solution and Infinitely Many Solutions
Write the corresponding system of linear equations for the reduced row-echelon form of the augmented matrix.
x + 2y + z = 8 y + 3z = 3
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.
Answer: (5z + 2, –3z + 3, z)
Example 6

37. No Solution and Infinitely Many Solutions
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.
Augmented Matrix
Example 6

38. Because 0 ≠ –12, the system has no solution.
No Solution and Infinitely Many Solutions
R1 – R3 
R1 – R2 
Because 0 ≠ –12, the system has no solution.
Answer: no solution
Example 6

39. Solve the system of equations
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)
Example 6

40. 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
Example 7

41. Interchange R1 with R3. R2 – 3R1  R2  Infinitely Many Solutions
Example 7

42. Infinitely Many Solutions
R3 – 4R1
R3 + 11R2
R3 
Example 7

43. Infinitely Many Solutions
R2 – 2R3 
R1 – 3R2
R1 – 4R3 
Example 7

44. 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.
Example 7

45. Infinitely Many Solutions
So, a solution of the system can be expressed as (2z + 1, –3z – 2, –z + 4, z), where z is any real number.
Answer: (2z + 1, –3z – 2, –z + 4, z)
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 
Example 7

46. Solve the system of equations
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
Example 7

47. Multivariable Linear Systems and Row Operations
LESSON 6–1
Multivariable Linear Systems and Row Operations