Exercise Set 6.2
1.
(c)
3.03 -12.1 14 -119 -3.03 12.1 -7 120 6.11 -14.2 21 -139
GAUSSIAN ELIMINATION WITH SCALED COLUMN PIVOTING
The reduced system - output by rows:
6.110000000000000 -14.199999999999999 21.000000000000000
-0.495908346972177 5.058101472995091 3.414075286415711
0.495908346972177 -1.000000000000000 7.000000000000000
Has solution vector:
0.000000000000000 10.000000000000000 0.142857142857143
with 1 row interchange(s)
The rows have been logically re-ordered to:
3 2 1
Exercise Set 6.3
4.
(a)
1 -1 2 -1 6 1 0 -1 1 4 2 1 3 -4 -2 0 -1 1 -1 5
GAUSSIAN ELIMINATION WITH SCALED COLUMN PIVOTING The reduced system - output by rows: 1.000000000000000 0.000000000000000 -1.000000000000000 1.000000000000000 0.000000000000000 -1.000000000000000 1.000000000000000 -1.000000000000000 2.000000000000000 -1.000000000000000 6.000000000000000 -7.000000000000000 1.000000000000000 1.000000000000000 0.333333333333333 1.333333333333333 Has solution vector: 3.000000000000000 -6.000000000000000 -2.000000000000000 -1.000000000000000 with 2 row interchange(s) The rows have been logically re-ordered to: 2 4 3 1Then we can re-run the program, replacing the augmented entry in each row for the second system. The output will be exactly the same, except for
Has solution vector: 1.000000000000000 1.000000000000000 1.000000000000000 1.000000000000000 with 2 row interchange(s)(b)
To find 
we solve 4 systems 
to obtain
where 
has a 1 in the j-th position,
and 0's in the other 3 positions, for each 
For example, input
1 -1 2 -1 1 1 0 -1 1 0 2 1 3 -4 0 0 -1 1 -1 0gives output
GAUSSIAN ELIMINATION WITH SCALED COLUMN PIVOTING The reduced system - output by rows: 1.000000000000000 0.000000000000000 -1.000000000000000 1.000000000000000 0.000000000000000 -1.000000000000000 1.000000000000000 -1.000000000000000 2.000000000000000 -1.000000000000000 6.000000000000000 -7.000000000000000 1.000000000000000 1.000000000000000 0.333333333333333 1.333333333333333 Has solution vector: 0.125000000000000 0.125000000000000 0.875000000000000 0.750000000000000 with 2 row interchange(s) The rows have been logically re-ordered to: 2 4 3 1
The solutions for the other 3 columns are
0.625000000000000 -0.375000000000000 -0.625000000000000 -0.250000000000000 0.125000000000000 0.125000000000000 -0.125000000000000 -0.250000000000000 0.000000000000000 -1.000000000000000 -1.000000000000000 -1.000000000000000
14.
1 3 2 -2 5 2 4 -1 -3 4 -2 2 1 3 2 1 3 2 4 -1
GAUSSIAN ELIMINATION WITH SCALED COLUMN PIVOTING The reduced system - output by rows: -2.000000000000000 2.000000000000000 1.000000000000000 3.000000000000000 -1.000000000000000 6.000000000000000 0.000000000000000 0.000000000000000 -0.500000000000000 0.666666666666667 2.500000000000000 -0.500000000000000 -0.500000000000000 0.666666666666667 1.000000000000000 6.000000000000000 Has solution vector: -1.200000000000000 1.000000000000000 0.600000000000000 -1.000000000000000 with 1 row interchange(s) The rows have been logically re-ordered to: 3 2 1 4
Exercise Set 6.4
2.
(c)
1 1 -1 1 0 1 2 -4 -2 0 2 1 1 5 0 -1 0 -2 -4 0
System has no unique solution GAUSSIAN ELIMINATION WITH SCALED COLUMN PIVOTING The reduced system - output by rows: 1.000000000000000 1.000000000000000 -1.000000000000000 1.000000000000000 1.000000000000000 1.000000000000000 -3.000000000000000 -3.000000000000000 2.000000000000000 -1.000000000000000 0.000000000000000 0.000000000000000 -1.000000000000000 1.000000000000000 0.000000000000000 0.000000000000000 Has solution vector: 0.000000000000000 0.000000000000000 0.000000000000000 0.000000000000000 with 0 row interchange(s) The rows have been logically re-ordered to: 1 2 3 4The program stopped (OK=.FALSE.) because no unique solution exists. The matrix as output is the matrix
with the zero
entries below the diagonal replaced by the multipliers from the
first two stages of Gaussian elimination.
Notice that the pivot element 
Hence 
Exercise Set 6.6
10.
(a)
Yes, since
(b)
No, consider the matrix
which is strictly diagonally dominant but 
is not.
(c)
No, consider the matrices
which are both strictly diagonally dominant but
is not.
(d)
No, consider the matrix
which is strictly diagonally dominant but
is not.
(e)
No, consider
which are both diagonally dominant but
is not.
10.
(a)
A is singular when 
since 
(b)
A is never strictly diagonally dominant, regardless of
the value of 
(c)
A is symmetric for all 
(d)
After 2 steps of Gaussian elimination, we obtain the upper triangular matrix
Hence, 
is positive definite if and only if 
by Theorem 6.22.
Note that
where 
using the obvious definition for
the square-root of a diagonal matrix, so we also have
and Choleski factorizations of A when 