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Chapter 2: Q2.5-31Q (page 93)

The solution to the steady-state heat flow problem for
the plate in the figure is approximated by the solution to the
equation\(A{\bf{x}} = {\bf{b}}\);where\(b = \left( {5,15,0,10,0,10,20,30} \right)\)and

\(A = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&{ - 1}&{}&{}&{}&{}&{}\\{ - 1}&4&0&{ - 1}&{}&{}&{}&{}\\{ - 1}&0&4&{ - 1}&{ - 1}&{}&{}&{}\\{}&{ - 1}&{ - 1}&4&0&{ - 1}&{}&{}\\{}&{}&{ - 1}&0&4&{ - 1}&{ - 1}&{}\\{}&{}&{}&{ - 1}&{ - 1}&4&0&{ - 1}\\{}&{}&{}&{}&{ - 1}&0&4&{ - 1}\\{}&{}&{}&{}&{}&{ - 1}&{ - 1}&4\end{array}} \right]\)

(Refer to Exercise 33 of Section 1.1.) The missing entries in Aare zeros. The nonzero entries of A lie within a band along the main diagonal. Such band matricesoccur in a variety of applications and often are extremely large (with thousands of rows and columns but relatively narrow bands).

  1. Use the method of Example 2 to construct an LU factorization of A, and note that both factors are band matrices (with two nonzero diagonals below or above the main diagonal). Compute \(LU - A\) to check your work.
  1. Use the LU factorization to solve\(A{\bf{x}} = {\bf{b}}\).
  1. Obtain \({A^{ - {\bf{1}}}}\) and note that\({A^{ - {\bf{1}}}}\) is a dense matrix with no
    band structure. When Ais large, LandUcan be stored in
    much less space than\({A^{ - {\bf{1}}}}\). This fact is another reason for
    preferring the LU factorization of Ato \({A^{ - {\bf{1}}}}\) itself.

Short Answer

Expert verified

(a) The LU factorization is shown below:

\(L = \left[ {\begin{array}{*{20}{c}}1&0&0&0&0&0&0&0\\{ - .25}&1&0&0&0&0&0&0\\{ - .25}&{ - .0667}&1&0&0&0&0&0\\0&{ - .2667}&{ - .2857}&1&0&0&0&0\\0&0&{ - .2679}&{ - .0833}&1&0&0&0\\0&0&0&{ - .2917}&{.2921}&1&0&0\\0&0&0&0&{ - .2697}&{ - .0861}&1&0\\0&0&0&0&0&{ - .2948}&{ - .2931}&1\end{array}} \right]\)

\(U = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&{ - 1}&0&0&0&0&0\\0&{3.75}&{ - .25}&{ - 1}&0&0&0&0\\0&0&{3.7333}&{ - 1.0667}&{ - 1}&0&0&0\\0&0&0&{3.4286}&{ - .2857}&{ - 1}&0&0\\0&0&0&0&{3.7083}&{ - 1.0833}&{ - 1}&0\\0&0&0&0&0&{3.3919}&{ - .2921}&{ - 1}\\0&0&0&0&0&0&{3.7052}&{ - 1.0861}\\0&0&0&0&0&0&0&{3.3868}\end{array}} \right]\)

(b) The solution is\({\bf{x}} = \left( {3.9569,{\rm{ }}6.5885,{\rm{ }}4.2392,{\rm{ }}7.3971,{\rm{ }}5.6029,{\rm{ }}8.7608,{\rm{ }}9.4115,{\rm{ }}12.0431} \right)\).

(c) The inverse is \({A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{.2953}&{.0866}&{.0945}&{.0509}&{.0318}&{.0227}&{.0010}&{.0082}\\{.0866}&{.2953}&{.0509}&{.0945}&{.0227}&{.0318}&{.0082}&{.0100}\\{.0945}&{.0509}&{.3271}&{.1093}&{.1045}&{.0591}&{.0318}&{.0227}\\{.0509}&{.0945}&{.1093}&{.3271}&{.0591}&{.1045}&{.0227}&{.0318}\\{.0318}&{.0227}&{.1045}&{.0591}&{.3271}&{.1093}&{.0945}&{.0509}\\{.0227}&{.0318}&{.0591}&{.1045}&{.1093}&{.3271}&{.0509}&{.0945}\\{.0010}&{.0082}&{.0318}&{.0227}&{.0945}&{.0509}&{.2953}&{.0866}\\{.0082}&{.0100}&{.0227}&{.0318}&{.0509}&{.0945}&{.0866}&{.2953}\end{array}} \right]\).

Step by step solution

01

(a) Step 1: State the LU factorization of matrix A

Consider the matrix shown below:

\(A = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&{ - 1}&0&0&0&0&0\\{ - 1}&4&0&{ - 1}&0&0&0&0\\{ - 1}&0&4&{ - 1}&{ - 1}&0&0&0\\0&{ - 1}&{ - 1}&4&0&{ - 1}&0&0\\0&0&{ - 1}&0&4&{ - 1}&{ - 1}&0\\0&0&0&{ - 1}&{ - 1}&4&0&{ - 1}\\0&0&0&0&{ - 1}&0&4&{ - 1}\\0&0&0&0&0&{ - 1}&{ - 1}&4\end{array}} \right]\)

Use the following MATLAB command to obtain theLU factorization of the matrix:

\[ > > \left[ {\begin{array}{*{20}{c}}{\rm{L}}&{\rm{U}}\end{array}} \right] = {\rm{lu}}\left( {\rm{A}} \right)\]

The LU factorization is shown below:

\(L = \left[ {\begin{array}{*{20}{c}}1&0&0&0&0&0&0&0\\{ - .25}&1&0&0&0&0&0&0\\{ - .25}&{ - .0667}&1&0&0&0&0&0\\0&{ - .2667}&{ - .2857}&1&0&0&0&0\\0&0&{ - .2679}&{ - .0833}&1&0&0&0\\0&0&0&{ - .2917}&{.2921}&1&0&0\\0&0&0&0&{ - .2697}&{ - .0861}&1&0\\0&0&0&0&0&{ - .2948}&{ - .2931}&1\end{array}} \right]\)

\(U = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&{ - 1}&0&0&0&0&0\\0&{3.75}&{ - .25}&{ - 1}&0&0&0&0\\0&0&{3.7333}&{ - 1.0667}&{ - 1}&0&0&0\\0&0&0&{3.4286}&{ - .2857}&{ - 1}&0&0\\0&0&0&0&{3.7083}&{ - 1.0833}&{ - 1}&0\\0&0&0&0&0&{3.3919}&{ - .2921}&{ - 1}\\0&0&0&0&0&0&{3.7052}&{ - 1.0861}\\0&0&0&0&0&0&0&{3.3868}\end{array}} \right]\)

Use the following MATLAB command to obtain the\(LU - A\)factorization of the matrix:

\[ > > LU - A\]

The output is shown below:

\(\begin{array}{c}LU - A = \left[ {\begin{array}{*{20}{c}}0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\end{array}} \right]\\ = {\bf{0}}\end{array}\)

Thus, it is verified that \(LU = A\).

02

(b) Step 2: Solve the system of equations

Use the following MATLAB command to solve the system by using LU factorization:

\[\begin{array}{l} > > {\rm{b}} = \left[ {\begin{array}{*{20}{c}}5&{15}&0&{10}&0&{10}&{20}&{30}\end{array}} \right]\\ > > {\rm{X}} = {\rm{A}}\backslash {\rm{b}}'\end{array}\]

The solution is shown below:

\(X = \left[ {\begin{array}{*{20}{c}}{3.9569}\\{6.5885}\\{4.2392}\\{7.3971}\\{5.6029}\\{8.7608}\\{9.4115}\\{12.0431}\end{array}} \right]\)

Thus, the solution is

\({\bf{x}} = \left( {3.9569,{\rm{ }}6.5885,{\rm{ }}4.2392,{\rm{ }}7.3971,{\rm{ }}5.6029,{\rm{ }}8.7608,{\rm{ }}9.4115,{\rm{ }}12.0431} \right)\).

03

(c) Step 3: Find the inverse of the matrix

Use the following MATLAB command to obtain the inverse of matrixA:

\[ > > B = A\^ - 1\]

Theinverse obtained is shown below:

\({A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{.2953}&{.0866}&{.0945}&{.0509}&{.0318}&{.0227}&{.0010}&{.0082}\\{.0866}&{.2953}&{.0509}&{.0945}&{.0227}&{.0318}&{.0082}&{.0100}\\{.0945}&{.0509}&{.3271}&{.1093}&{.1045}&{.0591}&{.0318}&{.0227}\\{.0509}&{.0945}&{.1093}&{.3271}&{.0591}&{.1045}&{.0227}&{.0318}\\{.0318}&{.0227}&{.1045}&{.0591}&{.3271}&{.1093}&{.0945}&{.0509}\\{.0227}&{.0318}&{.0591}&{.1045}&{.1093}&{.3271}&{.0509}&{.0945}\\{.0010}&{.0082}&{.0318}&{.0227}&{.0945}&{.0509}&{.2953}&{.0866}\\{.0082}&{.0100}&{.0227}&{.0318}&{.0509}&{.0945}&{.0866}&{.2953}\end{array}} \right]\)

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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.

3. \[\left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\I&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right]\]

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

If Ais an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).
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