3.2 Pivoting Strategies
1. For each of the following augmented matrices, identify the entry which would
serve as the first pivot element for
(i) Gaussian elimination with no pivoting;
(ii) Gaussian elimination with partial pivoting; and
(iii) Gaussian elimination with scaled partial pivoting.
(a)
0 1 1 1
303−4
1 1 1 2
2 3 1 3
0
7
6
6
(ii) Among the values
(iii) Since
Among the values
2Section 3.2
(b)
3−1 3 1
6 0 9 −2
−12 0 −10 5
72 −8 48 −19
6
13
−17
93
(ii) Among the values
(iii) Since
Among the values
(c)
−1.78 0.56 4.33
2.53 −1.05 3.02
1.47 −0.54 −0.83
7.23
−1.61
−3.38
with partial pivoting will be the 2.53 in the second row of the first column.
(iii) Since
Pivoting Strategies 3
Among the values
(d)
0.25 0.35 0.15
0.20 0.20 0.25
0.15 0.20 0.25
0.60
0.90
0.70
(i) With no pivoting, the first pivot element will be the 0.25 in the first row
(e)
0.2115 2.296 2.715 3.215
0.4371 3.916 1.683 2.852
6.099 4.324 23.20 1.578
4.623 0.8926 15.32 5.305
8.438
8.888
35.20
26.14
4Section 3.2
(ii) Among the values
(iii) Since
Among the values
For the augmented matrices indicated in Exercises 2 – 6, show the contents of
the matrix after one pass of
(i) Gaussian elimination with no pivoting;
(ii) Gaussian elimination with partial pivoting; and
(iii) Gaussian elimination with scaled partial pivoting.
For (ii) and (iii), show the contents of the row vector, and for (iii), show the
contents of the scale vector.
2. The augmented matrix from Exercise 1(a).
Pivoting Strategies 5
After carrying out the first pass of Gaussian elimination, the augment matrix
(iii) Initialize the row vector to r=1234T.From Exercise 1(a), we
know that the scale vector is
After carrying out the first pass of Gaussian elimination, the augment matrix
3. The augmented matrix from Exercise 1(b).
(i) After carrying out the first pass of Gaussian elimination, the augmented matrix
6Section 3.2
After carrying out the first pass of Gaussian elimination, the augment matrix
4. The augmented matrix from Exercise 1(c).
(i) After carrying out the first pass of Gaussian elimination, the augmented matrix
After carrying out the first pass of Gaussian elimination, the augment matrix
Pivoting Strategies 7
5. The augmented matrix from Exercise 1(d).
(i) After carrying out the first pass of Gaussian elimination, the augmented matrix
(iii) Initialize the row vector to r=123T.From Exercise 1(d), we know
that the scale vector is
6. The augmented matrix from Exercise 1(e).
(i) After carrying out the first pass of Gaussian elimination, the augmented matrix
(ii) Initialize the row vector to r=1234T.From Exercise 1(e), we
(iii) Initialize the row vector to r=1234T.From Exercise 1(e), we
know that the scale vector is
In Exercises 7 – 12:
(a) Solve the indicated system using Gaussian elimination with partial pivoting.
Show all intermediate matrices and the row vector at each step.
(b) Repeat part (a) using Gaussian elimination with scaled partial pivoting.
Show the contents of the scale vector.
Pivoting Strategies 9
7.
2x1+ 3x2+x3=−4
4x1+x2+ 4x3= 9
3x1+ 4x2+ 6x3= 0
The initial augmented matrix for the system is
(a) Initialize the row vector to r=123T.Among the values
The larger value corresponds to r3, so we interchange the second and third
entries in the row vector to obtain
(b) Initialize the row vector to r=123T.Since
Among the values
The first pass of Gaussian elimination transforms the augmented matrix to
Back substitution now yields
8.
2x1−x2+x3= 2
4x1+ 2x2+x3= 7
6x1−4x2+ 2x3= 4
Pivoting Strategies 11
The initial augmented matrix for the system is
(a) Initialize the row vector to r=123T.Among the values
The first pass of Gaussian elimination transforms the augmented matrix to
The larger value corresponds to r2, so there is no need to modify the contents of
the row vector. The next pass of Gaussian elimination reduces the augmented
Back substitution now yields
(b) Initialize the row vector to r=123T.Since
12 Section 3.2
have a common value of 1. Following convention, we choose the first occur-
Back substitution now yields
9.
3x2+x3= 1
x1+ 2x2−2x3= 7
2x1+ 5x2+ 4x3=−1
The initial augmented matrix for the system is
Pivoting Strategies 13
The first pass of Gaussian elimination transforms the augmented matrix to
Now,
Back substitution now yields
(b) Initialize the row vector to r=123T.Since
Among the values
the largest corresponds to r2. We therefore interchange the first two entries
in the row vector to obtain
The first pass of Gaussian elimination transforms the augmented matrix to
14 Section 3.2
The larger value corresponds to r2, so no change needs to be made to the row
vector. The next pass of Gaussian elimination reduces the augmented matrix
Back substitution now yields
10.
x1+ 8x2+ 6x3=−1
−3x1−4x2+ 5x3= 6
2x1+ 4x2−6x3=−8
The initial augmented matrix for the system is
(a) Initialize the row vector to r=123T.Among the values
Pivoting Strategies 15
(b) Initialize the row vector to r=123T.Since
Among the values
the largest corresponds to r2. We therefore interchange the first two entries
in the row vector to obtain
The first pass of Gaussian elimination transforms the augmented matrix to
16 Section 3.2
Back substitution now yields
11.
x1−3x2+ 7x3= 2
2x1+ 4x2−3x3=−1
−3x1+ 7x2+ 2x3= 3
The initial augmented matrix for the system is
(a) Initialize the row vector to r=123T.Among the values
The first pass of Gaussian elimination transforms the augmented matrix to
Pivoting Strategies 17
(b) Initialize the row vector to r=123T.Since
the largest corresponds to r2. We therefore interchange the first two entries
in the row vector to obtain
The larger value corresponds to r3, so we need to interchange the second and
third entries in the row vector. This yields