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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 rowechelon 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 nexttolast 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 rowechelon 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 rowechelon form as follows: 1 2 3 E 18 3 The scalefactor 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 GaussJordan 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 rowechelon 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 GaussJordan 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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 Winter '10
 Taylor
 augmented matrix, Row

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