Week4

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Week4

  1. 1. A row of a matrix is a zero row if it consists entirely of zeros. A row which is not a zero row is a nonzero row. A leading entry of a row refers to the leftmost nonzero entry in a nonzero row A unit column of a matrix is a column with one entry a 1 and all other entries 0. non-zero row 0 2 1 1 0 3 0 0 0 0 0 0 Leading entry 1 6 1 0 2 0 zero row 0 0 0 0 0 0 unit column
  2. 2. A matrix is in row-echelon form if and only if: 1. All zero rows occur below all nonzero rows. 2. The leading entry of a nonzero row lies strictly to the right of the leading entry of any other preceding row. 3. All entries in a column below a leading entry are zeros. 3 0 6 3 3. 2. 0 2 -1 4 0 0 1 2 1. 0 0 0 0
  3. 3. A matrix A is in or if it is in echelon form and satisfies the following additional conditions: 1- All the leading entries are 1. 2- Every column containing a leading one is a .
  4. 4. unit columns leading 1s
  5. 5. The following matrices are in echelon form. 2 3 2 1 1 2 1 0 0 1 4 8 0 1 4 4 0 0 0 5 2 0 0 1 3
  6. 6. And the following matrices 1 0 0 29 1 0 0 0 0 1 0 16 0 1 0 4 0 0 1 3 0 0 0 0 are in reduced echelon form.
  7. 7. Example 3: not a unit column 1 2 3 A is in row-echelon form. A fails to be in reduced row-echelon form A 0 1 7 because the second column contains a 0 0 0 leading 1, but is not a unit column. leading 1 C fails to be in row-echelon form because 1 2 1 5 the leading 1 in the second row is to the C 0 0 0 1 right of the leading 1 in the third row. Since 0 0 1 0 C is not in row-echelon form, it cannot be in reduced row-echelon form. leading 1s
  8. 8. The following matrices are in echelon form. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 where 0 0 0 0 0 0 any real number
  9. 9. And the following matrices are in reduced echelon form. 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0
  10. 10. A is one that is in echelon form. A is one that is in reduced echelon form.
  11. 11. Each matrix is row equivalent to one and only one reduced echelon matrix.
  12. 12. The rank of the matrix is the number of nonzero rows of its row echelon form
  13. 13. Example 4 . 1 3 2 R1 R 2 1 3 A 2 6 0 0 Thus r (A ) 1 1 3 2R1 R 2 1 3 R2 1 3 3R 2 R1 1 0 B 2 5 0 1 0 1 0 1 Thus, r(B) = 2
  14. 14. Step 1: Begin with the leftmost nonzero column say Cr . If a1r =0 and asr 0 interchange R1 by a row Rs. Step 2: Use elementary row operations to create zeros in all positions below a1r -(asr /a1r) R1+Rs s= 2,3, ,m
  15. 15. Step 3: go to the next leftmost nonzero column and repeat steps 1and 2 to the remaining submatrix. Repeat the process until there are no more nonzero rows to modify.
  16. 16. Row reduce the matrix below to echelon form and find the rank of . 0 3 6 4 9 1 2 1 3 1 A 2 3 0 3 1 1 4 5 9 7
  17. 17. Step 1: R1 R4 1 4 5 9 7 1 2 1 3 1 2 3 0 3 1 0 3 6 4 9
  18. 18. R1 R 2 2R 1 R 3 1 4 5 9 7 0 2 4 6 6 0 5 10 15 15 0 3 6 4 9
  19. 19. R3 5 2 R2 R4 32R2 1 4 5 9 7 0 2 4 6 6 0 0 0 0 0 0 0 0 5 0
  20. 20. 3 4 1 4 5 9 7 0 2 4 6 6 0 0 0 5 0 0 0 0 0 0 r(A) = 3
  21. 21. Row reduce the matrix below to echelon form and find the rank of . 0 3 6 6 4 5 3 7 8 5 8 9 3 9 12 9 6 15
  22. 22. 3 9 12 9 6 15 R1 R3 3 7 8 5 8 9 0 3 6 6 4 5 3 9 12 9 6 15 3R1 R2 0 2 4 4 2 6 0 3 6 6 4 5
  23. 23. 3 9 12 9 6 15 3 R2 R3 0 2 4 4 2 6 2 0 0 0 0 1 4 Row Echelon Form r(A) = 3
  24. 24. Assume that A is an invertible matrix of order n Step 1: Construct the augmented matrix B=[A In] Step 2: Reduce B to its (unique) reduced echelon form The resulting equivalent matrix is -1 n
  25. 25. Use the Gauss-Jordon elimination to find the inverse of 2
  26. 26. 5 1 1 1 0 R1 2 2 2 1 3 0 1 5 1 1 0 R1 R 2 2 2 1 1 0 - 1 2 2
  27. 27. 5 1 1 0 2R 2 2 2 0 1 -1 2 5 1 0 3 -5 R2 R1 2 0 1 -1 2 1 3 5 A 1 2
  28. 28. Use the Gauss-Jordon elimination to find the inverse of 2 1 0 A 1 2 1 0 1 2
  29. 29. 2 1 0 1 0 0 A I 1 2 1 0 1 0 0 1 2 0 0 1 1 1 1 - 0 0 0 2 2 1 R 1 1 2 1 0 1 0 2 0 1 2 0 0 1
  30. 30. 1 1 1 - 0 0 0 2 2 3 1 R1 R 2 0 1 1 0 2 2 0 1 2 0 0 1 1 1 1 - 0 0 0 2 2 2 2 1 2 R 2 0 1 0 3 3 3 3 0 1 2 0 0 1
  31. 31. 1 2 1 1 0 0 3 3 3 1 2 1 2 R2 R1 0 1 0 2 3 3 3 0 1 2 0 0 1 1 2 1 1 0 0 3 3 3 2 1 2 R2 R3 0 1 0 3 3 3 4 1 2 0 0 1 3 3 3
  32. 32. 1 2 1 1 0 0 3 3 3 3 2 1 2 R3 0 1 0 4 3 3 3 0 0 1 1 1 1 4 2 2 1 1 0 1 0 3 3 3 2 1 2 R3 R2 0 1 0 1 3 2 3 0 0 1 1 1 1 4 2
  33. 33. 3 1 1 1 0 0 4 2 3 1 R3 R1 0 1 0 1 2 1 2 3 3 0 0 1 1 1 1 4 2 3 1 1 4 2 3 1 1 2 A 1 2 3 1 1 1 4 2
  34. 34. Solve x1 x2 7 2 x1 3 x2 6 Solution x1 3, x2 4
  35. 35. Solve x1 3 x2 2 x3 2 x5 0 2 x1 6 x2 5 x3 2 x 4 4 x5 3 x6 1 5 x3 10 x 4 15 x6 5 2 x1 6 x2 8 x 4 4 x5 18 x6 6 Solution x1 ? , x2 ? , x3 ? , x4 ? , x5 ? , x6 ?
  36. 36. A linear equation in n variables x 1, x 2 , , x n is defined to be in the form a1x 1 a2x 2 an x n b where a1, a2 , , an ,b R . The variables in a linear equation are called the unknowns.
  37. 37. Linear N o t lin e a r 2 x 3y 7 x 3y 7 x1 2 x2 3 x3 x4 7 3x 2 z xz 4 1 y 2 x 3z 1 y sin x 0 x1 x2 xn 1 x1 2x 2 x3 1
  38. 38. A finite set of linear equations in the variables x 1 , x 2 , , x n is called a system of linear equations . a11x 1 a12x 2 a1n x n b1 a21x 1 a22x 2 a2n x n b2 am1x 1 am 2x 2 amn x n bm A sequence of numbers s1, s2 , , sn is called a solution of the system if x1 s1, x2 s2 , , xn sn is a solution of every equation in the system.
  39. 39. Matrix form of a system of linear equations: a11 x1 a12 x2 a1n xn b1 a21 x1 a22 x2 a2 n xn b2 m linear equations am 1 x1 am 2 x2 amn xn bm a11 a12 a1n x1 b1 a21 a22 a2 n x2 b2 Single matrix equation Ax b m n n 1 m 1 am 1 am 2 amn xn bm A x b
  40. 40. A system of equations that has no solution is said to be inconsistent; if there is at least one solution of the system, it is said to be consistent. The system x y 4 2x 2 y 6 has no solutions since the equivalent system x y 4 x y 3 has contradictory equations.
  41. 41. Every system of linear equations has either solutions, exactly solution, or solutions.
  42. 42. If m < n, the system has more than one of solutions. If m > n, the system can be reduced to equivalent system where m n, or it is inconsistant. If m = n , the system has either no solution or one solution.
  43. 43. Solution possibilities for two lines Consider the solutions to the general system of two linear equations in the two unknowns x and y: a1 x b1 y c1 (a1 , b1 not both zero) a2 x b2 y c2 (a2 , b2 not both zero) The graphs of these equations are lines; call them l1 and l2 . Then the solutions of the system correspond to the intersections of the lines.
  44. 44. We will study the solution of linear systems under the following assumptions: m = n (#of equations = # of unknowns) b is a non-zero matrix non- The system is consistent . Under these assumptions the system has either no solution or one solution.
  45. 45. For the linear system AX=b , if A is invertible then the system is consistent and it has one solution X = A-1 b
  46. 46. Solvin g a syst e m 1-GAUSS ELIMINATION Reduce the augmented matrix to the echelon form and use back substitution. substitution. 2-GAUSS-JORDON ELIMINATION GAUSS- Reduce the augmented matrix to the reduced [A b] [In X]
  47. 47. To find the solution and the inverse of the matrix at the same time reduce the matrix [A In b] to the reduced echelon form to get [I A-1 X] -1 n
  48. 48. x y 2z 9 2x 4y 3z 1 3x 6y 5z 0 1 1 2 9 1 1 2 9 2 R1 R 2 2 4 3 1 0 2 7 17 3 6 5 0 3 6 5 0
  49. 49. 1 1 2 9 1 1 2 9 3 R1 R 3 0.5 R 2 0 2 7 17 0 1 3.5 8.5 0 3 11 27 0 3 11 27 1 1 2 9 3R2 R3 0 1 3.5 8.5 0 0 0.5 1.5 1 1 2 9 2 R3 0 1 3.5 8.5 0 0 1 3
  50. 50. You may stop at this step and use back substitution. z=3 y-3.5 z=-8.5 y=-8.5+10.5=2 x+y+2z=9 x=9-2-6=1
  51. 51. Or we may continue to find the reduced echelon form and the solution
  52. 52. 1 0 5.5 17.5 1 0 5.5 17.5 R 2 R1 3.5 R3 R 2 0 1 3.5 8.5 0 1 0 2 0 0 1 3 0 0 1 3 1 1 0 1 5.5 R3 R1 0 1 0 2 0 0 1 3
  53. 53. x 2y 4z 12 2x y 5z 18 x 3y 3z 8 1 2 4 12 1 2 4 12 0 3 3 6 2 1 5 18 R2 ( 2)R1 0 1 1 4 1 3 3 8 R3 R1 1 2 4 12 1 2 4 12 1 0 1 1 2 0 1 1 2 R2 0 1 1 4 R3 ( 1)R2 0 0 2 6 3 1 2 4 12 x 2y 4z 12 x 2 1 0 1 1 2 y z 2 solution y 1. R3 2 0 0 1 3 z 3 z 3
  54. 54. 4 x 8 y 12 z 44 Solve the system 3x 6 y 8 z 32 2x y 7 4 8 12 44 1 2 3 11 1 2 3 11 3 6 8 32 1 3 6 8 32 R2 ( 3)R1 0 0 1 1 R1 2 1 0 7 4 2 1 0 7 R3 2R1 0 3 6 15 1 2 3 11 1 2 3 11 R2 R3 0 3 6 15 1 0 1 2 5 R2 0 0 1 1 3 0 0 1 1 The solution is x 2, y 3, z 1.
  55. 55. 3 x1 3 x2 3 x3 9 Solve, if possible, the system of equations 2 x1 x2 4 x3 7 3 x1 5 x2 x3 7 3 3 3 9 1 1 1 3 1 1 1 3 2 1 4 7 1 2 1 4 7 R2 ( 2)R1 0 1 2 1 R1 3 5 1 7 3 3 5 1 7 R3 ( 3)R1 0 2 4 2 1 1 1 3 R3 2R2 0 1 2 1 x1 x2 x3 4 x1 x2 x3 4 0 0 0 0 x2 2x3 1 x2 2x3 1 The general solution to the system is x1 3r 4 x2 2r 1 x3 r , where r is real number (called a parameter).
  56. 56. Example 6 2 x1 + x2 + x3 =7 3 x1 +2 x2 +x3 =-3 x2 + x3 =5
  57. 57. 2 1 1 7 1 0 .5 0 .5 3 .5 1 3 2 1 -3 R1 3 2 1 -3 2 0 1 1 5 0 1 1 5 1 0 .5 0 .5 3 .5 3R 1 R2 0 0 .5 -0 .5 -1 3 .5 0 1 1 5
  58. 58. 1 0 .5 0 .5 3 .5 2 R 2 0 1 -1 -2 7 0 1 1 5 1 0 1 1 7 0 .5 R 2 R 1 0 1 -1 -2 7 0 1 1 5 1 0 1 1 7 R 2 R 3 0 1 -1 -2 7 0 0 2 3 2
  59. 59. 1 0 1 17 0 .5 R 3 0 1 -1 -2 7 0 0 1 16 x 3 16 x 2 x 3 27 x 2 27 16 11 x 1 x 3 17 x 1 17 16 1
  60. 60. 6 7
  61. 61. 12- 12- SOLVE THE FOLLOWING SYSTEMS OF LINEAR EQUATIONS
  62. 62. 13- 13- SOLVE THE FOLLOWING SYSTEMS OF LINEAR EQUATIONS

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