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Chapter 1: Q43Q (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.

43. Repeat Exercise 42 for the matrix in Exercise 10.

Short Answer

Expert verified

The solution has approximately 11 decimal places, and the calculated answer

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

Step by step solution

01

Obtain the condition number of matrix A

Consider matrix A as shown below:

\(A = \left( {\begin{aligned}{*{20}{c}}5&3&1&7&9\\6&4&2&8&{ - 8}\\7&5&3&{10}&9\\9&6&4&{ - 9}&{ - 5}\\8&5&2&{11}&4\end{aligned}} \right)\)

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

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{5 3 1 7 9; 6 4 2 8 }} - {\rm{8; 7 5 3 10 9; 9 6 4 }} - {\rm{9 }} - {\rm{5; 8 5 2 11 4}}} \right);\\ > > {\rm{ C}} = {\rm{cond}}\left( {\rm{A}} \right)\end{aligned}\)

It gives the output 68622.

Thus, thecondition number of matrix A is 68622.

By comparing with thecondition number of A, that is \({10^k}\), the required condition number is approximately in between \({10^4}\) and \({10^5}\).

02

Obtain the solution by using the MATLAB command

It is discovered that x and\({{\bf{x}}_1}\)agree to at least 11 or 12 significant digits if it run multiple experiments with MATLAB, which properly captures 16 digits.

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

\( > > {\bf{x}} = {\rm{rand}}\left( {5,1} \right)\)

\({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{.2190}\\{.0470}\\{.6789}\\{.6793}\\{.9347}\end{aligned}} \right)\)

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

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{5 3 1 7 9; 6 4 2 8 }} - {\rm{8; 7 5 3 10 9; 9 6 4 }} - {\rm{9 }} - {\rm{5; 8 5 2 11 4}}} \right);\\ > > x = \left( {.2190{\rm{; }}{\rm{.0470; }}{\rm{.6789; }}{\rm{.6793; }}{\rm{.9347}}} \right){\rm{;}}\\ > > b = A*x\end{aligned}\)

The output is \({\bf{b}} = A{\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{15.0821}\\{.8165}\\{19.0097}\\{ - 5.8188}\\{14.5557}\end{aligned}} \right)\).

03

Obtain the MATLAB solution

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

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{5 3 1 7 9; 6 4 2 8 }} - {\rm{8; 7 5 3 10 9; 9 6 4 }} - {\rm{9 }} - {\rm{5; 8 5 2 11 4}}} \right);\\ > > b = \left( {15.0821{\rm{; }}{\rm{.8165; 19}}{\rm{.0097; }} - {\rm{5}}{\rm{.8188; 14}}{\rm{.5557}}} \right){\rm{;}}\\ > > {x_1} = A\backslash b\end{aligned}\)

The output is\({{\bf{x}}_1} = \left( {\begin{aligned}{*{20}{c}}{.2190}\\{.0470}\\{.6789}\\{.6793}\\{.9347}\end{aligned}} \right)\).

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

\({\bf{x}} - {{\bf{x}}_1} = \left( {\begin{aligned}{*{20}{c}}{.3165}\\{ - .6743}\\{.3343}\\{.0158}\\{ - .0005}\end{aligned}} \right) \times {10^{ - 11}}\)

Thus, the solution has approximately 11 decimal places, and the calculated answer (\({{\bf{x}}_1}\)) is accurate.

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