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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

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

Show that if ABis invertible, so is B.

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

15. a. If A and B are \({\bf{2}} \times {\bf{2}}\) with columns \({{\bf{a}}_1},{{\bf{a}}_2}\) and \({{\bf{b}}_1},{{\bf{b}}_2}\) respectively, then \(AB = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}{{\bf{b}}_1}}&{{{\bf{a}}_2}{{\bf{b}}_2}}\end{aligned}} \right)\).

b. Each column of ABis a linear combination of the columns of Busing weights from the corresponding column of A.

c. \(AB + AC = A\left( {B + C} \right)\)

d. \({A^T} + {B^T} = {\left( {A + B} \right)^T}\)

e. The transpose of a product of matrices equals the product of their transposes in the same order.
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