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Chapter 6: Q-17-SE (page 331)

In Exercises 17–20, solve \(Ax = b\) and \(A\left( {\Delta x} \right) = \Delta b\) and show that the inequality (2) holds in each case. (See the discussion of ill-conditioned matrices in Exercises 41–43 in Section 2.3.)

\(17.\,\,A = \left[ {\begin{array}{*{20}{c}}{4.5}&{3.1}\\{1.6}&{1.1}\end{array}} \right],\,\,b = \left[ {\begin{array}{*{20}{c}}{19.249}\\{6.843}\end{array}} \right],\,\,\,\Delta b = \left[ {\begin{array}{*{20}{c}}{0.001}\\{ - 0.003}\end{array}} \right]\)

Short Answer

Expert verified

Thus, the required condition \[\frac{{\parallel \Delta x\parallel }}{{\parallel x\parallel }} \le cond(A) \cdot \frac{{\parallel \Delta b\parallel }}{{\parallel b\parallel }}\] is satisfied.

Step by step solution

01

The condition for solving matrix

Enter the matrices in MATLAB

\(\begin{array}{c}A = \left[ {4.5{\rm{ }}3.1{\rm{ }};{\rm{ }}1.6{\rm{ }}1.1} \right]\\b = \left[ {19.249;{\rm{ }}6.843} \right]\\delta\,\,b = \left[ {0.001; - 0.003} \right]\end{array}\)

02

Solve the equation matrix

Solve the system of equation

\(\)

\(\begin{array}{c}Ax = b\\x = {A^{ - 1}}b\\x{\rm{ }} = \left[ {3.94,\,\,0.49} \right]\end{array}\)

Solve the system of equation

\(\begin{array}{c}A\left( {delta\,\,x} \right) = delta\,\,b\\delta\,\,x = {A^{ - 1}}\left( {delta\,\,b} \right)\\delta\,\,x = \left[ { - 1.04,\,\,1.51} \right]\end{array}\)

03

Find norm and condition number

\(\begin{array}{l}norm\left( {delta\,\,x} \right)/norm\left( x \right)\\ans{\rm{ }} = 0.4618\\norm\left( {delta\,\,b} \right)/norm\left( b \right)\\ans{\rm{ }} = 1.5479e - 04\end{array}\)

\(\begin{array}{l}ans{\rm{ }} = \\1.5479e - 04\\Condition{\rm{ }}Number{\rm{ }}of{\rm{ }}A:\\cond\left( A \right)\\\end{array}\)

Condition Number of A:

\(\begin{array}{l}Cond\left( A \right)\\ans{\rm{ }} = 3.3630e + 03\end{array}\)

04

Find ratio condition 

Compute the ratio condition

\(\begin{array}{l}(A) \cdot \frac{{\parallel \Delta b\parallel }}{{\parallel b\parallel }}\\cond\left( A \right)*{\rm{ }}norm\left( {deltab} \right)/norm\left( b \right)\\ans{\rm{ }} = 0.5206\end{array}\)

So, the condition satisfied,\(\frac{{\parallel \Delta x\parallel }}{{\parallel x\parallel }} \le cond(A) \cdot \frac{{\parallel \Delta b\parallel }}{{\parallel b\parallel }}\)

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Most popular questions from this chapter

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

In Exercises 9-12, find a unit vector in the direction of the given vector.

10. \(\left( {\begin{aligned}{*{20}{c}}{ - 6}\\4\\{ - 3}\end{aligned}} \right)\)

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1, and let \({\bf{x}} = \left( {{\bf{1}},{\bf{1}}} \right)\) and \({\bf{y}} = \left( {{\bf{5}}, - {\bf{1}}} \right)\).

a. Find\(\left\| {\bf{x}} \right\|\),\(\left\| {\bf{y}} \right\|\), and\({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).

b. Describe all vectors\(\left( {{z_{\bf{1}}},{z_{\bf{2}}}} \right)\), that are orthogonal to y.
See all solutions

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