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. Compute the products shown in Exercises 1–4.

2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

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

How many rows does \(B\) have if \(BC\) is a \({\bf{3}} \times {\bf{4}}\) matrix?

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

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