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Chapter 2: Q2.5-32Q (page 93)

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

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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

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 \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

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 \(D\left( {ABC} \right) = I\).

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

Suppose AB = AC, where Band Care \(n \times p\) matrices and A is invertible. Show that B = C. Is this true, in general, when A is not invertible.

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.)
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