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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

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Show that if the columns of Bare linearly dependent, then so are the columns of AB.

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.

Suppose the last column of ABis entirely zero but Bitself has no column of zeros. What can you sayaboutthe columns of A?

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 \({G_{k + {\bf{1}}}}\).

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