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The band matrixAshown below can be used to estimate
the unsteady conduction of heat in a rod when the temperatures at points \({p_{\bf{1}}},...,{p_{\bf{5}}}\) on the rod change with time.

The constantCin the matrix depends on the physical nature
of the rod, the distance \(\Delta x\) between the points on the rod,
and the length of time \(\Delta t\) between successive temperature
measurements. Suppose that for \(k = 0,1,2,...,\) a vector \({{\bf{t}}_k}\)
in \({\mathbb{R}^5}\) lists the temperatures at time \(k\Delta t\). If the two ends of the
rod are maintained at \(0^\circ \), then the temperature vectors satisfy
the equation \(A{t_{k + 1}} = {t_k}\)\(\left( {k = 0,1,...} \right)\), where \[A = \left[ {\begin{array}{*{20}{c}}{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}&{}&{}&{}\\{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}&{}&{}\\{}&{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}&{}\\{}&{}&{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}\\{}&{}&{}&{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}\end{array}} \right]\]

  1. Find the LU factorization ofAwhen\[C = 1\]. A matrix
    such as Awith three nonzero diagonals is called a tridiagonal matrix. The Land Ufactors are bidiagonal matrices.
  1. Suppose \[C = 1\] and\({t_0} = \left( {10,12,12,12,10} \right)\). Use the
    LU factorization of Ato find the temperature distributions \({t_1}\), \({t_2}\), \({t_3}\), and \({t_4}\).

Short Answer

Expert verified
  1. The LU factorizations are

\(L = \left[ {\begin{array}{*{20}{c}}1&0&0&0&0\\{ - 1/3}&1&0&0&0\\0&{ - 3/8}&1&0&0\\0&0&{ - 8/21}&1&0\\0&0&0&{ - 21/55}&1\end{array}} \right]\),\(U = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&0&0&0\\0&{8/3}&{ - 1}&0&0\\0&0&{21/8}&{ - 1}&0\\0&0&0&{55/21}&{ - 1}\\0&0&0&0&{144/55}\end{array}} \right]\).

2. The temperature distributions are \({{\bf{t}}_1} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{9.6667}\\{10.4444}\\{9.6667}\\{6.5556}\end{array}} \right]\), \({{\bf{t}}_2} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{7.6667}\\{8.5926}\\{7.6667}\\{4.7407}\end{array}} \right]\), \({{\bf{t}}_3} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{6.0556}\\{6.9012}\\{6.0556}\\{3.5988}\end{array}} \right]\), \({{\bf{t}}_4} = \left[ {\begin{array}{*{20}{c}}{2.7922}\\{4.7778}\\{5.4856}\\{4.7778}\\{2.7922}\end{array}} \right]\).

Step by step solution

01

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

Consider the matrix at\(C = 1\)as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&0&0&0\\{ - 1}&3&{ - 1}&0&0\\0&{ - 1}&3&{ - 1}&0\\0&0&{ - 1}&3&{ - 1}\\0&0&0&{ - 1}&3\end{array}} \right]\)

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:

\(L = \left[ {\begin{array}{*{20}{c}}1&0&0&0&0\\{ - 1/3}&1&0&0&0\\0&{ - 3/8}&1&0&0\\0&0&{ - 8/21}&1&0\\0&0&0&{ - 21/55}&1\end{array}} \right]\)

\(U = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&0&0&0\\0&{8/3}&{ - 1}&0&0\\0&0&{21/8}&{ - 1}&0\\0&0&0&{55/21}&{ - 1}\\0&0&0&0&{144/55}\end{array}} \right]\)

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

\[ > > LU - A\]

The output is shown below:

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

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

02

(b) Step 2: Solve the temperature distribution

Suppose the solution of\(L{{\bf{s}}_{k + 1}} = {{\bf{t}}_k}\)is\({{\bf{s}}_{k + 1}}\), where \(\left( {k = 0,1,...} \right)\). So, the solution of\(U{{\bf{t}}_{k + 1}} = {{\bf{s}}_{k + 1}}\)is\({{\bf{t}}_{k + 1}}\).

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

\[\begin{array}{l} > > {\rm{b}} = \left[ {\begin{array}{*{20}{c}}{10}&{12}&{12}&{12}&{10}\end{array}} \right]\\ > > {\rm{X}} = {\rm{A}}\backslash {\rm{b}}'\end{array}\]

The solutions are shown below:

\({{\bf{s}}_1} = \left[ {\begin{array}{*{20}{c}}{10.0000}\\{15.3333}\\{17.7500}\\{18.7619}\\{17.1636}\end{array}} \right]\),\({{\bf{t}}_1} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{9.6667}\\{10.4444}\\{9.6667}\\{6.5556}\end{array}} \right]\), \({{\bf{s}}_2} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{11.8519}\\{14.8889}\\{15.3386}\\{12.4121}\end{array}} \right]\), \({{\bf{t}}_2} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{7.6667}\\{8.5926}\\{7.6667}\\{4.7407}\end{array}} \right]\), \({{\bf{s}}_3} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{9.2469}\\{12.0602}\\{12.2610}\\{9.4222}\end{array}} \right]\), \({{\bf{t}}_3} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{6.0556}\\{6.9012}\\{6.0556}\\{3.5988}\end{array}} \right]\), \({{\bf{s}}_4} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{7.2551}\\{9.6219}\\{9.7210}\\{7.3104}\end{array}} \right]\), \({{\bf{t}}_4} = \left[ {\begin{array}{*{20}{c}}{2.7922}\\{4.7778}\\{5.4856}\\{4.7778}\\{2.7922}\end{array}} \right]\).

Thus, the temperature distributions are \({{\bf{t}}_1} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{9.6667}\\{10.4444}\\{9.6667}\\{6.5556}\end{array}} \right]\), \({{\bf{t}}_2} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{7.6667}\\{8.5926}\\{7.6667}\\{4.7407}\end{array}} \right]\), \({{\bf{t}}_3} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{6.0556}\\{6.9012}\\{6.0556}\\{3.5988}\end{array}} \right]\), \({{\bf{t}}_4} = \left[ {\begin{array}{*{20}{c}}{2.7922}\\{4.7778}\\{5.4856}\\{4.7778}\\{2.7922}\end{array}} \right]\).

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Most popular questions from this chapter

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

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.]

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

Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

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