Page 39 - Matrix Analysis & Applied Linear Algebra

1.5 Making Gaussian Elimination Work 31

1.5.4. Consider the following system in which the coefficient matrix is the
Hilbert matrix:
1 1 1
x + y + z = ,
2 3 3
1 1 1 1
x + y + z = ,
2 3 4 3
1 1 1 1
x + y + z = .
3 4 5 5
(a) First convert the coefficients to 3-digit floating-point numbers,
and then use 3-digit arithmetic with partial pivoting but with
no scaling to compute the solution.
(b) Again use 3-digit arithmetic, but row scale the coefficients (after
converting them to floating-point numbers), and then use partial
pivoting to compute the solution.
(c) Proceed as in part (b), but this time row scale the coefficients
before each elimination step.
(d) Now use exact arithmetic on the original system to determine
the exact solution, and compare the result with those of parts
(a), (b), and (c).

1.5.5. To see that changing units can affect a floating-point solution, consider
a mining operation that extracts silica, iron, and gold from the earth.
Capital (measured in dollars), operating time (in hours), and labor (in
man-hours) are needed to operate the mine. To extract a pound of silica
requires $.0055, .0011 hours of operating time, and .0093 man-hours of
labor. For each pound of iron extracted, $.095, .01 operating hours, and
.025 man-hours are required. For each pound of gold extracted, $960,
112 operating hours, and 560 man-hours are required.
(a) Suppose that during 600 hours of operation, exactly $5000 and
3000 man-hours are used. Let x, y, and z denote the number
of pounds of silica, iron, and gold, respectively, that are recov-
ered during this period. Set up the linear system whose solution
will yield the values for x, y, and z.
(b) With no scaling, use 3-digit arithmetic and partial pivoting to
compute a solution (˜x, ˜y, ˜) of the system of part (a). Then
z
approximate the exact solution (x, y, z) by using your machine’s
(or calculator’s) full precision with partial pivoting to solve the
system in part (a), and compare this with your 3-digit solution
by computing the relative error defined by

2 2 2
z
y
(x − ˜x) +(y − ˜) +(z − ˜)
e r = .
2
x + y + z 2
2