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Some matrix programs, such as MATLAB, have a command to create Hilbert matrices of various sizes. If possible, use an inverse command to compute the inverse of a twelfth-order or larger Hilbert matrix, A. Compute \(A{A^{ - 1}}\). Report what you find.

Short Answer

Expert verified

The output matrix is an identity matrix.

Step by step solution

01

Create a Hilbert matrix of large size

Use the MATLAB command to create theHilbert matrix of size\(12 \times 12\).

\( > > {\rm{A}} = {\rm{hilb}}\left( {12} \right)\)

\(\left[ {\begin{array}{*{20}{c}}{1.000}&{0.500}&{0.333}&{0.250}&{0.200}&{0.167}&{0.143}&{0.125}&{0.111}&{0.100}&{0.091}&{0.083}\\{0.500}&{0.333}&{0.250}&{0.200}&{0.167}&{0.143}&{0.125}&{0.111}&{0.100}&{0.091}&{0.083}&{0.077}\\{0.333}&{0.250}&{0.200}&{0.167}&{0.143}&{0.125}&{0.111}&{0.100}&{0.091}&{0.083}&{0.077}&{0.071}\\{0.250}&{0.200}&{0.167}&{0.143}&{0.125}&{0.111}&{0.100}&{0.091}&{0.083}&{0.077}&{0.071}&{0.067}\\{0.200}&{0.167}&{0.143}&{0.125}&{0.111}&{0.100}&{0.091}&{0.083}&{0.077}&{0.071}&{0.067}&{0.063}\\{0.167}&{0.143}&{0.125}&{0.111}&{0.100}&{0.091}&{0.083}&{0.077}&{0.071}&{0.067}&{0.063}&{0.059}\\{0.143}&{0.125}&{0.111}&{0.100}&{0.091}&{0.083}&{0.077}&{0.071}&{0.067}&{0.063}&{0.059}&{0.056}\\{0.125}&{0.111}&{0.100}&{0.091}&{0.083}&{0.077}&{0.071}&{0.067}&{0.063}&{0.059}&{0.056}&{0.053}\\{0.111}&{0.100}&{0.091}&{0.083}&{0.077}&{0.071}&{0.067}&{0.063}&{0.059}&{0.056}&{0.053}&{0.050}\\{0.100}&{0.091}&{0.083}&{0.077}&{0.071}&{0.067}&{0.063}&{0.059}&{0.056}&{0.053}&{0.050}&{0.048}\\{0.091}&{0.083}&{0.077}&{0.071}&{0.067}&{0.063}&{0.059}&{0.056}&{0.053}&{0.050}&{0.048}&{0.045}\\{0.083}&{0.077}&{0.071}&{0.067}&{0.063}&{0.059}&{0.056}&{0.053}&{0.050}&{0.048}&{0.045}&{0.043}\end{array}} \right]\)

02

Obtain the inverse of matrix A

Compute theinverse of matrix A by using the MATLAB command shown below:

\( > > B = {\rm{A}}\^ - 1\)

\[\left[ {\begin{array}{*{20}{c}}{1.41e + 02}&{ - 9.93e + 05}&{2.29e + 05}&{ - 2.54e + 06}&{1.61e + 07}&{ - 6.35e + 07}&{1.62e + 08}\\{ - 9.93e + 03}&{9.35e + 05}&{ - 2.43e + 07}&{2.89e + 08}&{ - 1.91e + 09}&{7.76e + 09}&{ - 2.03e + 10}\\{2.29e + 05}&{ - 2.43e + 07}&{6.75e + 08}&{ - 8.38e + 09}&{5.72e + 10}&{ - 2.37e + 11}&{6.29e + 11}\\{ - 2.54e + 06}&{2.89e + 08}&{ - 8.38e + 09}&{1.07e + 11}&{ - 7.48e + 11}&{3.15e + 12}&{ - 8.48e + 12}\\{1.61e + 07}&{ - 1.91e + 09}&{5.72e + 10}&{ - 7.48e + 11}&{5.30e + 12}&{ - 2.27e + 13}&{6.16e + 13}\\{ - 6.35e + 07}&{7.76e + 09}&{ - 2.37e + 11}&{3.15e + 12}&{ - 2.27e + 13}&{6.16e + 13}&{ - 2.69e + 14}\\{1.62e + 08}&{ - 2.03e + 10}&{6.29e + 11}&{ - 8.48e + 12}&{6.16e + 13}&{ - 2.69e + 14}&{7.44e + 14}\\{ - 2.73e + 08}&{3.47e + 10}&{ - 1.09e + 12}&{1.49e + 13}&{ - 1.09e + 14}&{4.80e + 14}&{ - 1.34e + 15}\\{3.02e + 08}&{ - 3.89e + 10}&{1.24e + 12}&{ - 1.70e + 13}&{1.26e + 14}&{ - 5.57e + 14}&{1.56e + 15}\\{ - 2.10e + 08}&{2.74e + 10}&{ - 8.81e + 11}&{1.22e + 13}&{ - 9.08e + 13}&{4.04e + 14}&{ - 1.14e + 15}\\{8.83e + 07}&{ - 1.10e + 10}&{3.57e + 11}&{ - 4.98e + 12}&{3.72e + 13}&{ - 1.66e + 14}&{4.70e + 14}\\{ - 1.45e + 07}&{1.93e + 09}&{ - 6.29e + 10}&{8.82e + 11}&{ - 6.63e + 12}&{2.98e + 13}&{ - 8.44e + 13}\end{array}} \right.\]

\(\left. {\begin{array}{*{20}{c}}{ - 2.73e + 08}&{3.02e + 08}&{ - 2.10e + 08}&{8.38e + 07}&{ - 1.45e + 07}\\{3.47e + 10}&{ - 3.89e + 10}&{ - 2.74e + 10}&{ - 1.10e + 10}&{1.93e + 09}\\{ - 1.09e + 12}&{1.24e + 12}&{ - 8.81e + 11}&{3.57e + 11}&{ - 6.29e + 10}\\{1.49e + 13}&{ - 1.70e + 13}&{1.22e + 13}&{ - 4.89e + 12}&{8.82e + 11}\\{ - 1.09e + 14}&{1.26e + 14}&{ - 9.08e + 13}&{3.72e + 13}&{ - 6.63e + 12}\\{4.80e + 14}&{ - 5.57e + 14}&{4.04e + 14}&{ - 1.66e + 14}&{2.98e + 13}\\{ - 1.34e + 15}&{1.56e + 15}&{ - 1.14e + 15}&{4.70e + 14}&{ - 8.44e + 13}\\{2.42e + 15}&{ - 2.84e + 15}&{2.80e + 15}&{ - 8.62e + 14}&{1.55e + 14}\\{ - 2.84e + 15}&{3.34e + 15}&{ - 2.45e + 15}&{1.02e + 15}&{ - 1.85e + 14}\\{2.08e + 15}&{ - 2.45e + 15}&{1.81e + 15}&{ - 7.56e + 14}&{1.37e + 14}\\{ - 8.62e + 14}&{1.02e + 15}&{ - 7.56e + 14}&{3.17e + 14}&{ - 5.75e + 13}\\{1.55e + 14}&{ - 1.85e + 14}&{1.37e + 14}&{ - 5.75e + 13}&{1.05e + 13}\end{array}} \right]\)

03

Obtain the product of the matrix and its inverse

Compute matrix C by using the MATLAB command shown below:

\( > > C = {\rm{A}}*{\rm{B}}\)

\(\left[ {\begin{array}{*{20}{c}}{1.000}&{0.000}&{0.000}&{ - 0.001}&{0.001}&{0.006}&{ - 0.010}&{ - 0.019}&{ - 0.054}&{0.000}&{0.007}&{ - 0.001}\\{ - 0.005}&{1.000}&{0.000}&{0.000}&{0.002}&{0.000}&{0.020}&{ - 0.052}&{0.025}&{ - 0.019}&{0.003}&{ - 0.004}\\{ - 0.006}&{0.003}&{1.000}&{0.000}&{0.002}&{ - 0.002}&{0.023}&{ - 0.042}&{0.041}&{ - 0.036}&{0.007}&{ - 0.003}\\{ - 0.007}&{0.004}&{ - 0.001}&{1.000}&{0.002}&{ - 0.003}&{0.015}&{ - 0.035}&{0.049}&{ - 0.047}&{0.003}&{ - 0.003}\\{ - 0.007}&{0.006}&{ - 0.001}&{0.000}&{1.003}&{0.000}&{0.001}&{ - 0.029}&{0.027}&{ - 0.021}&{0.006}&{ - 0.002}\\{ - 0.007}&{0.006}&{ - 0.001}&{ - 0.001}&{0.002}&{0.997}&{0.020}&{ - 0.036}&{0.047}&{ - 0.036}&{0.010}&{ - 0.003}\\{ - 0.007}&{0.006}&{ - 0.001}&{ - 0.002}&{0.002}&{ - 0.004}&{1.012}&{ - 0.040}&{0.047}&{ - 0.034}&{0.010}&{ - 0.002}\\{ - 0.006}&{0.007}&{0.000}&{ - 0.003}&{0.002}&{ - 0.006}&{0.016}&{0.948}&{0.050}&{ - 0.040}&{0.015}&{ - 0.003}\\{ - 0.006}&{0.007}&{0.000}&{ - 0.003}&{0.001}&{0.005}&{ - 0.011}&{0.012}&{1.012}&{ - 0.001}&{ - 0.001}&{ - 0.001}\\{ - 0.006}&{0.006}&{0.000}&{ - 0.004}&{0.003}&{ - 0.002}&{0.005}&{ - 0.031}&{0.049}&{0.975}&{0.13}&{ - 0.002}\\{ - 0.006}&{0.006}&{0.001}&{ - 0.004}&{0.000}&{0.009}&{ - 0.022}&{0.009}&{ - 0.033}&{0.025}&{0.989}&{0.002}\\{ - 0.006}&{0.006}&{0.001}&{ - 0.005}&{0.002}&{0.001}&{0.006}&{ - 0.018}&{0.025}&{ - 0.018}&{0.002}&{0.999}\end{array}} \right]\)

Thus, the resultant matrix is an identity matrix.

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Most popular questions from this chapter

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

5. \[\left[ {\begin{array}{*{20}{c}}A&B\\C&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&Y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\Z&{\bf{0}}\end{array}} \right]\]

Suppose Ais a \(3 \times n\) matrix whose columns span \({\mathbb{R}^3}\). Explain how to construct an \(n \times 3\) matrix Dsuch that \(AD = {I_3}\).

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

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