Question

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.

44. Solve an equation \(Ax = b\) for a suitable b to find the last column of the inverse of the fifth-order Hilbert matrix

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

How many digits in each entry of x do you expect to be correct? Explain. [Note:The exact solution is \(\left( {630, - 12600,56700, - 88200,44100} \right)\).]

Short answer

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

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

Step-by-step solution

Obtain the column matrix b

Consider the matrix\(A = \left[ {\begin{array}{*{20}{c}}1&{1/2}&{1/3}&{1/4}&{1/5}\\{1/2}&{1/3}&{1/4}&{1/5}&{1/6}\\{1/3}&{1/4}&{1/5}&{1/6}&{1/7}\\{1/4}&{1/5}&{1/6}&{1/7}&{1/8}\\{1/5}&{1/6}&{1/7}&{1/8}&{1/9}\end{array}} \right]\).

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

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

\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\\1\end{array}} \right]\)

Obtain the MATLAB solution

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ \begin{array}{l}1{\rm{ 1/2 1/3 1/4 1/5; 1/2 1/3 1/4 1/5 1/6; 1/3 1/4 1/5 1/6 1/7; }}\\{\rm{1/4 1/5 1/6 1/7 1/8; 1/5 1/6 1/7 1/8 1/9}}\end{array} \right]{\rm{;}}\\ > > b = \left[ {0{\rm{; 0; 0; 0; 1}}} \right]{\rm{;}}\\ > > x = A\backslash b\end{array}\)

The output is \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{630.00}\\{ - 12600.00}\\{56700.00}\\{ - 88200.00}\\{44100.00}\end{array}} \right]\).

Obtain the condition number of matrix A

Consider matrix A as shown below:

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

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

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

It gives the output 476607.25.

Thus, thecondition number of matrix A is 476608 or\(4.8 \times {10^5}\).

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

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