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[M] Use equation (6) to solve the problem in Exercise 13. Set \({{\bf{x}}^{\left( {\bf{0}} \right)}} = {\bf{d}}\), and for \[k = {\bf{1}},\,{\bf{2}},\,....\] compute \({{\bf{x}}^{\left( k \right)}} = {\bf{d}} + C{{\bf{x}}^{\left( {k - {\bf{1}}} \right)}}\). How many steps are needed to obtain the answer in Exercise 13 to four significant figures?

Short Answer

Expert verified

12 steps

Step by step solution

01

Write the augmented matrix

The augmented matrix is shown below:

\(A = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{.8412}&{ - 0.0064}&{ - 0.0025}&{ - 0.0304}&{ - 0.0014}&{ - 0.0083}&{ - 0.1594}\\{ - .0057}&{0.7355}&{ - 0.0436}&{ - 0.0099}&{ - 0.0083}&{ - 0.0201}&{ - 0.3413}\\{ - .0264}&{ - 0.1506}&{0.6443}&{ - 0.0139}&{ - 0.0142}&{ - 0.0070}&{ - 0.0236}\\{ - .3299}&{ - 0.0565}&{ - 0.0495}&{0.6364}&{ - 0.0204}&{ - 0.0483}&{ - 0.0649}\\{ - .0089}&{ - 0.0081}&{ - 0.0333}&{ - 0.0295}&{0.6588}&{ - 0.0237}&{ - 0.0020}\\{ - 0.1190}&{ - 0.0901}&{ - 0.0996}&{ - 0.1260}&{ - 0.1722}&{0.7632}&{ - 0.3369}\\{ - 0.0063}&{ - 0.0126}&{ - 0.0196}&{ - 0.0098}&{ - 0.0064}&{ - 0.0132}&{0.9988}\end{array}}&{\begin{array}{*{20}{c}}{74000}\\{56000}\\{10500}\\{25000}\\{17500}\\{196000}\\{5000}\end{array}}\end{array}\,} \right]\)

02

Write the iterations

\({{\bf{x}}^{\left( 0 \right)}} = \left( {74000.0,\,56000.0,\,10500.0,\,25000.0,\,17500.0,\,196000.0,\,5000.0} \right)\)

\({{\bf{x}}^{\left( 1 \right)}} = \left( {89344.2,\,77730.5,\,26708.1,\,72334.7,\,30325.6,\,265158.2,\,9327.8} \right)\)

\({{\bf{x}}^{\left( 2 \right)}} = \left( {94681.2,\;87714.5,\;37577.3,\;100520.5,\;38598.0,\;296563.8,\;11480.0} \right)\)

\({{\bf{x}}^{\left( 3 \right)}} = \left( {97091.9,\;92573.1,\;43867.8,\;115457.0,\;43491.0,\;312319.0,\;12598.8} \right)\)

\({{\bf{x}}^{\left( 4 \right)}} = \left( {98291.6,\;95033.2,\;47314.5,\;123202.5,\;46247.0,\;320502.4,\;13185.5} \right)\)

\({{\bf{x}}^{\left( 5 \right)}} = \left( {98907.2,\;96305.3,\;49160.6,\;127213.7,\;47756.4,\;324796.1,\;13655.9} \right)\)

\({{\bf{x}}^{\left( 6 \right)}} = \left( {99226.6,\;96969.6,\;50139.6,\;129296.7,\;48569.3,\;327053.8,\;13655.9} \right)\)

\({{\bf{x}}^{\left( 7 \right)}} = \left( {99393.1,\;97317.8,\;50928.7,\;130948.0,\;49002.8,\;328240.9,\;13741.1} \right)\)

\({{\bf{x}}^{\left( 8 \right)}} = \left( {99480.0,\;97500.7,\;50928.7,\;130948.0,\;49232.5,\;328864.7,\;13785.9} \right)\)

\({{\bf{x}}^{\left( 9 \right)}} = \left( {99525.5,\;97647.2,\;51147.2,\;131399.2,\;49417.7,\;329364.4,\;13821.7} \right)\)

\({{\bf{x}}^{\left( {10} \right)}} = \left( {99561.9,\;97673.7,\;51186.8,\;131480.4,\;49451.3,\;329454.7,\;13828.2} \right)\)

\({{\bf{x}}^{\left( {11} \right)}} = \left( {99561.9,\;97687.6,\;51186.8,\;131480.4,\;49451.3,\;329502.1,\;13831.6} \right)\)

\({{\bf{x}}^{\left( {12} \right)}} = \left( {99568.4,\;97687.6,\;51207.5,\;131523.0,\;49469.0,\;329502.1,\;13831.6} \right)\)

So, the answer to Exercise 13 is obtained in 12 steps.

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Most popular questions from this chapter

When a deep space probe launched, corrections may be necessary to place the probe on a precisely calculated trajectory. Radio elementary provides a stream of vectors, \({{\bf{x}}_{\bf{1}}},....,{{\bf{x}}_k}\), giving information at different times about how the probe’s position compares with its planned trajectory. Let \({X_k}\) be the matrix \(\left[ {{x_{\bf{1}}}.....{x_k}} \right]\). The matrix \({G_k} = {X_k}X_k^T\) is computed as the radar data are analyzed. When \({x_{k + {\bf{1}}}}\) arrives, a new \({G_{k + {\bf{1}}}}\) must be computed. Since the data vector arrive at high speed, the computational burden could be serve. But partitioned matrix multiplication helps tremendously. Compute the column-row expansions of \({G_k}\) and \({G_{k + {\bf{1}}}}\) and describe what must be computed in order to update \({G_k}\) to \({G_{k + {\bf{1}}}}\).

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.

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

The inverse of \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\C&I&{\bf{0}}\\A&B&I\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\Z&I&{\bf{0}}\\X&Y&I\end{array}} \right]\). Find X, Y, and Z.

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 \(k = {\bf{2}},...,{\bf{6}}\).

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