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GaussPivotScaleEtcBronson_001

# GaussPivotScaleEtcBronson_001 - 2.9 Solve the following set...

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Unformatted text preview: 2.9 Solve the following set of equations by (a) standard Gaussian elimination and (b) Gaussian elimination with partial pivoting, rounding all computations to four signiﬁcant ﬁgures: 0.00001x1 + x2 = 1.00001 x, + x2 = 2 (a) We write the system in matrix form, rounding 1.00001 to 1.000. Thcn'we transform the augmented , matrix into row-echelon form using the algorithm of Chapter 1, in the following steps: [0.00001 1 i 1.000 ] 1 1 l 2 —> [ 1 100000 I! 100000] 1 1 I 2 [ 1 100000 E 100000 —* 0 —100000 l—100000] [ 1 100000 E 100000] —> 0 1 l 1 (Note that we round to —100000 twice in the next-to-last step.) The resulting augmented matrix shows that the system is consistent. The equations associated with this matrix are xl + 100000x2 = 100000 x2 = 1 which have the solution x1 = 0 and x2 =1. However, substitution into the original equations shows that this is not the solution to the original system. (b) Transforming the augmented matrix into row-echelon form using partial pivoting yields [0.00001 1 11.000] 1 2 2 Rows 1 and 2 are interchanged 1000 because row 2 has the largest element in column 1, the current work column. ' ‘ '1 1 E 2 ' Rounding to four signiﬁcant ——> 0 1 1 ﬁgures. The system of equations associated with the last augmented matrix is consistent and is 1 [ 1 1 —> 0.00001 1 x, + x2 = 2 x2 = 1 Its solution is x, = x2 = 1, which is also the solution to the original set of equations. All computers round to a number of signiﬁcant ﬁgures k that depends on the machine being used. Then an equation of the form 10—(k+1)x1 + x2 =1 + 10*“ will generate results like that of part a unless some pivoting strategy is used. (We had k = 4 in part a.) as a rule, dividing by very small numbers can lead to signiﬁcant roundoff error and should be avoided when possible. 2.11 To use scaled pivoting, we ﬁrst deﬁne, as the scale factorfor each row of the coefﬁcient matrix A, the largest element in absolute value appearing in that row. The scale factors are computed once and only once and, for easy reference, are added onto the augmented matrix [A I B] as another partitioned column. Then Step 1.3 of Chapter 1 is replaced with the following: _7 Divide the absolute value of each nonzero element that is in the work column and on or below the work row by the scale factor for its row. The element yielding the largest quotient is the new pivot; denote its row as row I. If row I is different from the current work row (row R), then interchange rows I and R. Row interchanges are the only elementary row operations that are performed on the scale factors; all other steps in the Gaussian elimination are limited to A and B. Solve Problem 2.10 using scaled pivoting. The scale factors for the system of Problem 2.10 are s1= max{1, 2, 3} = 3 52 = max{2, 1, |—4|} = 4 s3 = max{|—5|,8, 17} = l7 We add a column consisting of these scale factors to the augmented matrix for the system, and then . transforming it to row-echelon form as follows: 1 2 3 E 18 3 The scale-factor quotients for the 2 1 —4 II —30 4 elements in column 1 are 1/3 = *5 8 17 : 96 17 0.333, 2/4 = 0.500, and 5/17 = ' 0.294. —> 2 1 —4 : —30 4 The largest quotient is 0.500, so 1 2 3 E 18 3 the pivot is 2, which appears in —5 8 17! 96 17 row 2. Since [=2 and R=1, the ﬁrst and second rows are interchanged. —> 1 0.5 —2 I —15 4 1 2 3 E 13 3 —5 8 17 1 96 17 1 0.5 —2 5 —15 4 —> 0 1.5 5 g 33 3 —5 8 17 l 96 17 1 0.5 —2 E —15 4 Now work row is 2, and the work 0 1.5 5 : 33 3 column is 2. The quotients are —> 0 10.5 7 : ‘21 17 1.5/3=0.500 and 10.5/17 = 0.618. 1 0.5 —2 E —15 4 The largest quotient is 0.618, so 0 10.5 7 : 21 17 the pivot is 10.5, which appears in —> 0 1.5 5 I 33 3 row 3. The second and third rows are interchanged. 1 0.5 —2 : —15 4 —> 0 1 0.66667 5 2 17 0 1.5 5 : 33 3 1 0.5 -2 ll —15 4 0 1 0.66667 : 2 17 —> 0 0 4 l 30 3 1 0.5 —2 —15 4 0 1 0.66667 : 2 17 —> 0 0 1 l 7.5 3 Writing the set of equations associated with this augmented matrix (ignoring the column of scale factors) and solvmg them by back substitution, we obtain the solution x1 = 1.5, x2 = —3, x3 = 7.5. 2.12 To use complete pivoting, we replace Step 1.3 of Chapter 1 with the following steps, which involve both row and column interchanges: Let the current work row be R, and the current work column C. Scan all the elements of submatrix A of the augmented matrix that are on Or below row R and on or to'the right of column C, to determine which is largest in absolute value. Denote the row and column in which this element appears as row I and column J. If 1% R, interchange rows 1 and R; if J¢C, interchange rows J and C. Because column interchanges change the order of the unknowns, a bookkeeping mechanism for associating columns with unknowns must be implemented. To do so, add a new partitioned row, row 0, above the usual agumented matrix. Its elements, which are initially in the order 1, 2, . . . , n to denote the subscripts on the unknowns, will designate which unknown is assocrated With each column. 2.13 Gauss-Jordan elimination adds a step between Steps 2.3 and 2.4 of the algorithm for Gaussian elimination. Once the augmented matrix has been reduced to row-echelon form, it is then reduced still further. Beginning with the last pivot element and continuing sequentially backward to the ﬁrst, each pivot element is used to transform all other elements in its column to zero. Use Gauss-Jordan elimination to solve Problem 2.8. The ﬁrst two steps of the Gaussian elimination algorithm are used to reduce the augmented matrix to row—echelon form as in Problems 1.13 and 2.8: 12 —1: 6 0161—4 001:—1 Then the matrix is reduced further, as follows: 1 2 —l ; 6 Add —6 times the third row to the —> O l 0 1 2 second row. 0 0 1 i—l —> 1 2 0 1 5 Add the third row to the ﬁrst row. 0 1 0 .' 2 0 0 1 I —1 -—> 1 0 0 '. 1 Add —2 times the second row to 0 1 0 I 2 the ﬁrst row. 0 0 1 I —l The set of equations associated with this augmented matrix is x1 = 1, x2 = 2, and x3 = —1, which is the solution set for the original system (no back substitution is required). ...
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