Exercises 42–44 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.

Consider matrix A as shown below:

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

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

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

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 \({10^5}\).

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:

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

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

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

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

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

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

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

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

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

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

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