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Week4

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Week4 Week4 Presentation Transcript

  • 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
  • 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
  • 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 .
  • unit columns leading 1s
  • 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
  • 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.
  • 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
  • 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
  • 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
  • A is one that is in echelon form. A is one that is in reduced echelon form.
  • Each matrix is row equivalent to one and only one reduced echelon matrix.
  • The rank of the matrix is the number of nonzero rows of its row echelon form
  • 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
  • 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
  • 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.
  • 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
  • Step 1: R1 R4 1 4 5 9 7 1 2 1 3 1 2 3 0 3 1 0 3 6 4 9
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • Use the Gauss-Jordon elimination to find the inverse of 2
  • 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
  • 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
  • Use the Gauss-Jordon elimination to find the inverse of 2 1 0 A 1 2 1 0 1 2
  • 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
  • 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
  • 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
  • 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
  • 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
  • Solve x1 x2 7 2 x1 3 x2 6 Solution x1 3, x2 4
  • 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 ?
  • 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.
  • 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
  • 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.
  • 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
  • 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.
  • Every system of linear equations has either solutions, exactly solution, or solutions.
  • 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.
  • 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.
  • 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.
  • For the linear system AX=b , if A is invertible then the system is consistent and it has one solution X = A-1 b
  • 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]
  • 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
  • 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
  • 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
  • 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
  • Or we may continue to find the reduced echelon form and the solution
  • 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
  • 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
  • 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.
  • 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).
  • Example 6 2 x1 + x2 + x3 =7 3 x1 +2 x2 +x3 =-3 x2 + x3 =5
  • 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
  • 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
  • 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
  • 6 7
  • 12- 12- SOLVE THE FOLLOWING SYSTEMS OF LINEAR EQUATIONS
  • 13- 13- SOLVE THE FOLLOWING SYSTEMS OF LINEAR EQUATIONS