Homework #2, due Feb. 19, 2009. You must show ALL steps to get credits.
1: Solve the following linear system and compare your solution with the actual solution.
0.03x1 + 58.9x2 = 59.2
5.31x1 ! 6.10x2 = 47.0
(a)
(b)
(c)
(d)
the actual solution (10,1).
Use Gaussian elimination and three-digit chopping arithmetic.
Use Gaussian elimination and three-digit rounding arithmetic.
Use Gaussian elimination with partial pivoting and three-digit chopping.
Use Gaussian elimination with scaled partial pivoting and three-digit chopping.
The key idea to solve this problem is (1) understand 3-digit chopping/rounding. (2) why
you need to do pivoting. You need to work exactly like a computer with 3-digit limited
memory. For example, in part a, you first compute the Gaussian multiplier,
m=5.31/0.03=177 (3-digit only, not 177.0). For us, operation (E2-m*E1) becomes
!10431.4x2 = !10431.4, and give us x2 = 1 , back-substitution yields, x1 = 10 . However,
for a computer, operation (E2-m*E1) means to compute (m*E1) first. This yields
5.31x1 + 1.04e4 x2 = 1.04e4 with 3-digit chopping. Then operation (E2-m*E1) gives
(!1.04e4 ! 6.1) x2 = (!1.04e4 + 47.0) . With 3-digit chopping, the equation is equivalent
to !1.04e4 x2 = !1.03e4 , this gives you x2 = 0.990 , back substitution using E1 yields
0.03x1 + 58.9 ! 0.990 = 59.2 , which is equivalent to 0.03x1 + 58.3 = 59.2 with 3-digit
chopping, solve for x1 = 30.0 . This problem is a nice example demonstrating how a
BIG number eats small ones.
2: Consider the linear system
3x1 + 4x2 + 3x3 = 16
x1 + 5x2 ! x3 = !12
6x1 + 3x2 + 7x3 = 102
(a) Find matrix L and U using four significant digits.
(b) Solve the linear system using L and U found in part a.
If you review the lecture notes and understand the meaning of 4 significant digits, it is
straightforward to form matrix L using Gaussian multiplier and compute matrix U using
the algorithm discussed in class. Note that if you use Matlab to solve this problem, it will
give you a permutation matrix P. Once you have L and U ready, then Ax = b is
equivalent to LUx = b . Let Ux = y , you have Ly = b and you can solve for y using
forward substitution. Once have y ready you can solve triangular system Ux = y for x
using back substitution.
" 2 ! -1%
3: Consider matrix A = $ ! 2
1'
$
'
# -1 1 4 &
(a) Find ! so that matrix is positive definite.
(b) The matrix A is symmetric. It is known that A can be factored as
A = LDLT ,
where L is the lower triangular matrix with 1’s on its diagonal and D is the
diagonal matrix with
find L and D.
(2)
(n)
on its diagonal. For the case with ! = 1 ,
a11(1) , a22
,..., ann
For part a, you have to make sure, det(A1)>0, det(A2)>0 and det(A)>0, where A1=(2),
" 2 !%
, to solve for ! yields !2 < " < 1.5 . For part b, you need to form diagonal
A2 = $
# ! 2 '&
(2)
(3)
matrix D using a11(1) , a22
from usual Gaussian elimination process. Then form L
, a33
using Gaussian multipliers.
4: Compute the maximum-norm condition number
!"
of the
3 ! 3 Hilbert matrix.
You need to compute the infinity norm for the Hilbert matrix we discussed in class.
! " =748.
! 1 1 0$
5: Compute the spectral radius, ! (A) of matrix A = # 1 2 1 & and its L2 norm, i.e.
#
&
" -1 1 2 %
|| A ||2 = !(AT A)
Review the lecture notes for the definition of spectral radius,
!(A) = max{| "1 |, | "2 |, | "3 |} , that is you need to solve eigenvalues of matrix A,
!1 = 2, !2 = 0, !3 = 3 . So, the answer !(A) =3. Similarly the spectral radius
!(AT A) =9.6458, which yields the L2 norm || A ||2 = 3.1058