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Chapter 2: Q2.3-45Q (page 93)

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

Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If Ais larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the “input” to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the “output” and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the “state” vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\( - 2A\), \(B - 2A\), \(AC\), \(CD\).

Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).
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