1. Pre-Calculus Lesson 7-3 Solving Systems of Equations Using Gaussian Elimination

2. What you’ll learn about Triangular Forms for Linear Systems Gaussian Elimination Elementary Row Operations and Row Echelon Form Reduced Row Echelon Form Applications … and why Many applications in business and science are modeled by systems of linear equations in three or more variables.

3. What is a Dimension? Flat: (x, y coordinate plane)  2 Dimensions x y x y z Solid: (x, y, z coordinate space)  3 Dimensions

4. Flat: (x, y coordinate plane)  2 Dimensions x y (3, 0) (0, 2)

5. 1. Draw the 3 dimensional coordinate Plane below on your paper. x z Solid: (x, y, z coordinate space)  3 Dimensions y (0, 3, 0) (2, 0, 0) 2. Plot the following 3 points on your graph. (0, 0, -2) 3. Connect the dots 4. What geometric “thing” do three dots make? Plane: a simple 3-dimensional shape.

6. What shape does an Equation Make? Ax + By = D Ax + By + Cz = D 3x + 2y = 6  line (in “2” space)  plane (in “3” space) 2x + 3y + 4z = 6

7. Categories of solutions in “3 space” Exactly one point, an “ordered triple” (x, y, z)

8. Along a Line: INFINITELY many solutions  (x, y, z) triples Categories of solutions in “3 space”

9. 2 parallel planes intersected by a 3 rd plane  no common solution for all 3 planes

10. Categories of solutions in “3 space” 3 parallel planes  No intersection  No solution

11. Categories of solutions in “3 space” Another possibility: No planes are parallel  NO POINT SATISFIES ALL 3 EQUATIONS  No solution  (too hard to draw)

12. Categories of solutions in “3 space” 1. One ordered triple (x, y, z)  one solution 3. Infinitely many ordered triples 2. No solution

13. Categories of Solutions in “2-space” Ways 2 lines can be graphed: Cross  one solution Parallel  no solutions Same line  infinite number of number of solutions solutions

14. Solving Systems of Linear Equations (30 April 1777 – 23 February 1855) Charles Fredrick Gauss Gaussian Elimination

15. Re-writing a system of equations as a “matrix”. 1x + 4y = 2 2x + 5y = 7 1 4 2 2 5 7

16. Where does a “matrix” come from? 2x – 3y 7x + 2y Expressions Not equations since no equal sign 2x – 3y = 8 7x + 2y = 2 Equations Are equations since there is an equal sign

17. Where does a “matrix” come from? 2x – 3y 7x + 2y 2 –3 7 2 Matrix of coefficients 2x – 3y = 8 7x + 2y = 2 2 –3 8 7 2 2

18. Big Picture We will perform “row operations” to turn the left side matrix into the matrix on the right side. 3 5 3 -1 2 10 1 0 -4 0 1 3 x + 0y = -4 x + 0y = -4 0x + 1y = 3 3x + 5y = 3 -x + 2y = 10 x = -4 x = -4 y = 3 y = 3

19. We call this reduced row eschelon form. 1 0 -4 0 1 3 1’s on the main diagonal 0’s above/below the main diagonal

20. How do I do that? Similar to elimination, we add multiples of one row to another row. BUT, unlike elimination, we only change one row at a time and we end up with the same number of rows that we started with.

21. Some important principles about systems of equations. Are the graphs of these two systems different from each other? Principle 1: you can exchange rows of a matrix.

22. Some important principles about systems of equations. Are the graphs of these two systems different from each other? Principle 2: you multiply (or divide) any row by a number and it won’t change the graph (or the matrix)

23. 1 st step: we want a zero in the bottom left corner. But, you will see later that this will be easier if the top left number is a one or a negative one. 3 5 3 -1 2 10 Swap rows. -1 2 10 3 5 3

24. -1 2 10 3 5 3 1 st step: we still want a zero in the bottom left corner. Forget about all the numbers but the 1 st column (for a minute).

25. -1 2 10 3 5 3 1 st step: we still want a zero in the bottom left corner. Forget about all the numbers but the 1 st column (for a minute). What multiple of the 1 st row should we add or subtract from row 2 to turn the 3 into a zero? # in 2 nd row # in 1 st row This gives us the pattern of what to do to each other number in row 2.

26. 3 3 5 3 -1 2 10 3 5 3 1 st step: we still want a zero in the bottom left corner. New Row 2 + 3(-1 2 10) +3(-1) 0 +3(2) 11 +3(10) 33

27. 3 3 5 3 -1 2 10 3 5 3 1 st step: we still want a zero in the bottom left corner. New Row 2 + 3(-1 2 10) +3(-1) 0 +3(2) 11 +3(10) 33 -1 2 10 0 11 33

28. 2 nd step: we want a one in 2 nd position of the 2 nd row. New Row 2 0 11 33 -1 2 10 0 11 33 11 01 3

29. New Row 2 0 11 33 -1 2 10 0 11 33 11 01 3 -1 2 10 0 1 3 2 nd step: we want a one in 2 nd position of the 2 nd row.

30. -1 2 10 0 1 3 3 rd step: we want a zero in 2 nd position of the 1 st row. Forget about all the numbers but the 2 nd column (for a minute).

31. -1 2 10 0 1 3 3 rd step: we want a zero in 2 nd position of the 1 st row. # in 1 st row # in 2 nd row Forget about all the numbers but the 2 nd column (for a minute). What multiple of the 2 nd row should we add or subtract from the 1 st row to turn the 2 into a zero? This gives us the pattern of what to do to each other number in row 1.

32. -1 2 10 0 1 3 New Row 1 -2(0 1 3) -2(0) -2(1) 0 -2(3) 4 3 rd step: we want a zero in 2 nd position of the 1 st row.

33. -1 2 10 0 1 3 New Row 1 -2(0 1 3) -2(0) -2(1) 0 -2(3) 4 -1 0 4 0 1 3 3 rd step: we want a zero in 2 nd position of the 1 st row.

34. 4 th step: we want a one in the top left corner. -1 0 4 0 1 3

35. New Row 2 1 0 -4 -1 0 4 0 1 3 4 th step: we want a one in the top left corner. 1 0 -4 0 1 3

36. Look at the circular pattern Don’t freak out: this goes faster than you think. -1 2 10 3 5 3

37. Look at the circular pattern Don’t freak out: this goes faster than you think. -1 2 10 3 5 3 -1 2 10 0 11 33

38. -1 2 10 0 11 33 Look at the circular pattern Don’t freak out: this goes faster than you think. -1 2 10 3 5 3 -1 2 10 0 11 33

39. -1 2 10 0 11 33 Look at the circular pattern Don’t freak out: this goes faster than you think. -1 2 10 3 5 3 -1 2 10 0 11 33 -1 2 10 0 1 3

40. -1 2 10 0 11 33 Look at the circular pattern Don’t freak out: this goes faster than you think. -1 2 10 3 5 3 -1 2 10 0 11 33 -1 2 10 0 1 3 -1 2 10 0 1 3

41. -1 2 10 0 11 33 Look at the circular pattern Don’t freak out: this goes faster than you think. -1 2 10 3 5 3 -1 2 10 0 11 33 -1 2 10 0 1 3 -1 2 10 0 1 3 -1 0 4 0 1 3

42. -1 0 4 0 1 3 -1 2 10 0 11 33 Look at the circular pattern Don’t freak out: this goes faster than you think. -1 2 10 3 5 3 -1 2 10 0 11 33 -1 2 10 0 1 3 -1 2 10 0 1 3 -1 0 4 0 1 3

43. -1 0 4 0 1 3 -1 2 10 0 11 33 Look at the circular pattern Don’t freak out: this goes faster than you think. 1 0 -4 0 1 3 -1 2 10 3 5 3 -1 2 10 0 11 33 -1 2 10 0 1 3 -1 2 10 0 1 3 -1 0 4 0 1 3

44. Systems of linear equations in Three Variables 2x + y – z = 5 3x – 2y + z = 16 4x + 3y – 5z = 3

45. Row Echelon Form of a Matrix A matrix is in row echelon form if the following conditions are satisfied. 1. Rows consisting entirely of 0’s (if there are any) occur at the bottom of the matrix. 2. The first entry in any row with nonzero entries is 1. 3. The column subscript of the leading 1 entries increases as the row subscript increases. (this means the ones lie on a diagonal)

46. Row Echelon Form of a Matrix 1 -3 2 -3 0 1 7 2 0 0 1 5 z = 5 z = 5 y + 7z = 2 (1) plug z = 5 into 2 nd equation and solve for ‘y’ x – 3y + 2z = -3 x – 3y + 2z = -3 (2) plug z = 5 and the value of ‘y’ found in step 1 into 1 st equation and solve for ‘x’ equation and solve for ‘x’

47. Reduced ow Echelon Form of a Matrix 1 0 0 -2 0 1 0 4 0 0 1 -5 We are going to perform “row operations” on a matrix to change it into this form. on a matrix to change it into this form. z = -5 z = -5 y = 4 x = -2 x = -2

48. Finding the Row Echelon Form x – y + 2z = -3 x – y + 2z = -3 2x + y – z = 0 -x + 2y – 3z = 7 -x + 2y – 3z = 7 1.Convert to “Augmented” matrix form. matrix form. 1 -1 2 -3 1 -1 2 -3 2 1 -1 0 2 1 -1 0 -1 2 -3 7 2.The right most column are the numbers on the right side of the equal sign. The dotted line is not usually drawn in textbooks, usually drawn in textbooks, but I like to draw it, to show but I like to draw it, to show where the equal sign is. where the equal sign is.

49. Row Operations -- the “big picture” We are going to rewrite the matrix numerous times. Each rewrite will change one row only. Our objective is to eliminate the will change one row only. Our objective is to eliminate the x-variables below the 1 st row like this. x-variables below the 1 st row like this. 1 -1 2 -3 1 -1 2 -3 2 1 -1 0 2 1 -1 0 -1 2 -3 7 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 0 2 -3 7 0 2 -3 7 1 5 0 -2 1 5 0 -2 0 3 0 3 0 3 0 3 0 0 -3 7 0 0 -3 7 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 0 0 -3 7 0 0 -3 7 Next: eliminate the y-variables below the 2 nd row like this. Next: eliminate the Z-variables above the 3 rd row like this. 1 0 0 -3 1 0 0 -3 0 3 0 6 0 3 0 6 0 0 -3 9 0 0 -3 9 Next: eliminate the y-variables above the 2 nd row like this.

50. Row Operations Then: turn the major diagonal into 1’s. 1 0 0 -3 1 0 0 -3 0 1 0 2 0 1 0 2 0 0 1 -3 0 0 1 -3 This final matrix represents the following system of equations: 1 0 0 -3 1 0 0 -3 0 3 0 6 0 3 0 6 0 0 -3 9 0 0 -3 9 1x + 0y + 0z = -3 1x + 0y + 0z = -3 0x + 1y + 0z = 2 0x + 1y + 0z = 2 0x + 0y + 1z = -3 0x + 0y + 1z = -3 x = -3 y = 2 y = 2 z = -3 z = -3 The solution to the system of equations:

51. Solve the following system using Row Operations 1 -1 2 -3 1 -1 2 -3 2 1 -1 0 2 1 -1 0 -1 2 -3 7 x - y + 2z = -3 x - y + 2z = -3 2x + y - z = 0 2x + y - z = 0 -x + 2y - 3z = 7 -x + 2y - 3z = 7 Convert to a matrix

52. Row Operations The new Row #2 = (-2)Row #1 + Old Row #2 1 -1 2 -3 1 -1 2 -3 2 1 -1 0 2 1 -1 0 -1 2 -3 7 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 -1 2 -3 7 A small explanation below the new below the new matrix will help matrix will help to avoid confusion. to avoid confusion. 2 1 -1 0 (-2)(1 -1 2) (-3)(-2) + 0 3 -5 6 (1) Eliminate the x-variable on the second row.

53. Row Operations (2) Eliminate the x-variable in the 3 rd row. 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 -1 2 -3 7 The new Row #3 = Row #1 + Old Row #3 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 0 1 -1 4 0 1 -1 4 1 -1 2 -3 + -1 2 -3 7 0 1 -1 4

54. Row Operations 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 0 1 -1 4 0 1 -1 4 1 -1 2 -3 1 -1 2 -3 0 1 -1 4 0 1 -1 4 (3) Eliminate the y-variable on the third row. Can only use 2 nd row (or you will introduce a number other than “0” in the x-position of the 3 rd row)!!! other than “0” in the x-position of the 3 rd row)!!! For example: New row 3 = row 1 + old row 3 1 -1 2 -3 0 1 -1 4 + 1 0 1 1 Can only use rows 2 and 3 at this point!!!

55. Row Operations 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 0 1 -1 4 0 1 -1 4 1 -1 2 -3 1 -1 2 -3 0 1 -1 4 0 1 -1 4 0 3 -5 6 0 3 -5 6 (3) Eliminate the y-variable on the third row. Yuck: fractions!! 0 1 -1 4 (-1/3)(0 3 -5) (6)(-1/3) + Wouldn’t it be nice if the 2 nd and 3 rd rows where switched? Then we could just multiply row two by -3 and then add to row 3. Let’s do it!!! Short way of saying this saying this

56. Row Operations 1 -1 2 -3 1 -1 2 -3 0 3 -5 6 0 3 -5 6 0 1 -1 4 0 1 -1 4 1 -1 2 -3 1 -1 2 -3 0 1 -1 4 0 1 -1 4 0 3 -5 6 0 3 -5 6 (3) Eliminate the y-variable on the 3 rd row. 0 3 -5 6 (-3)(0 1 -1) (4)(-3) (-3)(0 1 -1) (4)(-3) + 0 0 -2 -6 1 -1 2 -3 1 -1 2 -3 0 1 -1 4 0 1 -1 4 0 0 -2 -6 0 0 -2 -6

57. Row Operations (4) Eliminate the z-variable on the 2 nd row. 0 1 -1 4 (-0.5)(0 0 -2) (-6)(-0.5) (-0.5)(0 0 -2) (-6)(-0.5) + 0 1 0 7 1 -1 2 -3 1 -1 2 -3 0 1 -1 4 0 1 -1 4 0 0 -2 -6 0 0 -2 -6 Can only use 3 rd row (or you will introduce a number other than “0” in the x-position of the 2 nd row)!!! other than “0” in the x-position of the 2 nd row)!!! 1 -1 2 -3 1 -1 2 -3 0 1 0 7 0 1 0 7 0 0 -2 -6 0 0 -2 -6

58. Row Operations (5) Eliminate the z-variable on the 1st row. 1 -1 2 -3 1 -1 2 -3 0 0 -2 -6 + 1 -1 0 -9 1 -1 2 -3 1 -1 2 -3 0 1 0 7 0 1 0 7 0 0 -2 -6 0 0 -2 -6 1 -1 0 -9 1 -1 0 -9 0 1 0 7 0 1 0 7 0 0 -2 -6 0 0 -2 -6

59. Row Operations (6) Eliminate the y-variable on the 1st row. 0 1 0 7 0 1 0 7 1 -1 0 -9 1 -1 0 -9 + 1 0 0 -2 1 -1 0 -9 1 -1 0 -9 0 1 0 7 0 1 0 7 0 0 -2 -6 0 0 -2 -6 Can only use 2 nd row (or you will introduce a number other than “0” in the z-position of the 1 st row)!!! other than “0” in the z-position of the 1 st row)!!! 1 0 0 -2 1 0 0 -2 0 1 0 7 0 1 0 7 0 0 -2 -6 0 0 -2 -6

60. Row Operations (7) Get all “1’s” on the main diagonal. 1 0 0 -2 1 0 0 -2 0 1 0 7 0 1 0 7 0 0 -2 -6 0 0 -2 -6 1 0 0 -2 1 0 0 -2 0 1 0 7 0 1 0 7 0 0 1 3 0 0 1 3 1x + 0y + 0z = -2 1x + 0y + 0z = -2 0x + 1y + 0z = 7 0x + 1y + 0z = 7 0x + 0y + 1z = -3 0x + 0y + 1z = -3 x = -2 y = 7 y = 7 z = 3 z = 3 The solution to the system of equations:

61. Classes of Solutions for 3 Equations with 3 unknowns Gaussian Elimination results in: - a unique solution (the planes intersect at a point) -something silly like: 3 = 3 (Infinitely many solutions) -Again something silly like: 5 = -9 (there are no solutions)

62. Your turn: x - y + z = 0 2x – 3z = -1 -x - y + 2z = -1 1. Solve using Gaussian Elimination

63. HOMEWORK 7-3