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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

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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

Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).

Show that if ABis invertible, so is B.

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 \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

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

Assume \(A - s{I_n}\) is invertible and view (8) as a system of two matrix equations. Solve the top equation for \({\bf{x}}\) and substitute into the bottom equation. The result is an equation of the form \(W\left( s \right){\bf{u}} = {\bf{y}}\), where \(W\left( s \right)\) is a matrix that depends upon \(s\). \(W\left( s \right)\) is called the transfer function of the system because it transforms the input \({\bf{u}}\) into the output \({\bf{y}}\). Find \(W\left( s \right)\) and describe how it is related to the partitioned system matrix on the left side of (8). See Exercise 15.

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]
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