Q42Q Exercises 42鈥�44 show how to us... [FREE SOLUTION]

91影视

Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 483 pages

ISBN: 978-03219822384

Chapter 1: Q42Q (page 1)

Exercises 42鈥�44 show how to use the condition number of a matrix Ato estimate the accuracy of a computed solution of $Ax = b$. If the entries of Aand b are accurate to about rsignificant digits and if the condition number of Ais approximately ${\bf{1}}{{\bf{0}}^k}$ (with ka positive integer), then the computed solution of $Ax = b$ should usually be accurate to at least $r - k$ significant digits.
42. Find the condition number of the matrix A in Exercise 9. Construct a random vector x in ${\mathbb{R}^{\bf{4}}}$ and compute ${\bf{b}} = A{\bf{x}}$. Then use your matrix program to compute the solution ${{\bf{x}}_{\bf{1}}}$ of $Ax = b$. To how many digits do x and ${{\bf{x}}_{\bf{1}}}$ agree? Find out the number of digits your matrix program stores accurately, and report how many digits of accuracy are lost when ${{\bf{x}}_{\bf{1}}}$ is
used in place of the exact solution x.

Short Answer

Expert verified

The solution has approximately 12 decimal places, and the calculated answer

(${{\bf{x}}_1}$) is accurate.

Step by step solution

Obtain the condition number of matrix A

Consider matrix A as shown below:

$A = \left( {\begin{aligned}{*{20}{c}}4&0&{ - 3}&{ - 7}\\{ - 6}&9&9&9\\7&{ - 5}&{10}&{19}\\{ - 1}&2&4&{ - 1}\end{aligned}} \right)$

Obtain thecondition numberof matrix A using the MATLAB command shown below:

$\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{4 0 }} - {\rm{3 }} - {\rm{7; }} - {\rm{6 9 9 9; 7 }} - {\rm{5 10 19; }} - {\rm{1 2 4 }} - {\rm{1}}} \right);\\ > > {\rm{ C}} = {\rm{cond}}\left( {\rm{A}} \right)\end{aligned}$

It gives the output 23683.

Thus, thecondition number of matrix A is 23683.

By comparing with thecondition number of A, that is ${10^k}$, the condition number is approximately ${10^4}$.

Obtain the solution by using the MATLAB command

It is found that x and${{\bf{x}}_1}$agree to at least 12 or 13significant digits if it run multiple experiments with MATLAB, which properly captures 16 digits.

Obtain a random matrix by using the MATLAB command shown below:

$ > > {\bf{x}} = {\rm{rand}}\left( {4,1} \right)$

${\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{0.9501}\\{0.2131}\\{0.6068}\\{0.4860}\end{aligned}} \right)$

Now, compute${\bf{b}} = A{\bf{x}}$by using the MATLAB command shown below:

$\begin{aligned}{l} > > {\rm{ }}A{\rm{ }} = {\rm{ }}\left( {{\rm{4 0 }} - {\rm{3 }} - {\rm{7; }} - {\rm{6 9 9 9; 7 }} - {\rm{5 10 19; }} - {\rm{1 2 4 }} - {\rm{1}}} \right){\rm{;}}\\ > > x = \left( {0.9501{\rm{; 0}}{\rm{.2131; 0}}{\rm{.6068; 0}}{\rm{.4860}}} \right){\rm{;}}\\ > > b = A*x\end{aligned}$

The output is ${\bf{b}} = A{\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{ - 3.8493}\\{5.5795}\\{20.7973}\\{.8467}\end{aligned}} \right)$.

Obtain the MATLAB solution

Compute${{\bf{x}}_1}$of$A{\bf{x}} = {\bf{b}}$by using the MATLAB command shown below:

$\begin{aligned}{l} > > {\rm{ }}A{\rm{ }} = {\rm{ }}\left( {{\rm{4 0 }} - {\rm{3 }} - {\rm{7; }} - {\rm{6 9 9 9; 7 }} - {\rm{5 10 19; }} - {\rm{1 2 4 }} - {\rm{1}}} \right){\rm{;}}\\ > > b = \left( { - {\rm{3}}{\rm{.8493; 5}}{\rm{.5795; 20}}{\rm{.7973; 0}}{\rm{.8467}}} \right){\rm{;}}\\ > > {x_1} = A\backslash b\end{aligned}$

The output is${{\bf{x}}_1} = \left( {\begin{aligned}{*{20}{c}}{.9501}\\{.2311}\\{.6068}\\{.4860}\end{aligned}} \right)$.

Obtain the difference between x and${{\bf{x}}_1}$.

${\bf{x}} - {{\bf{x}}_1} = \left( {\begin{aligned}{*{20}{c}}{.0171}\\{.4858}\\{ - .2360}\\{.2456}\end{aligned}} \right) \times {10^{ - 12}}$

Thus, the solution has approximately 12 decimal places, and the calculated answer (${{\bf{x}}_1}$) is accurate.