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.

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.

Consider matrix A as shown below:

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

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

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

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 \({10^4}\).

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:

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

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{0.9501}\\{0.2131}\\{0.6068}\\{0.4860}\end{array}} \right]\)

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

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

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

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

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

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

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

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

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