Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Find the least square solutions of the system $A \overset{⇀}{x} = \overset{⇀}{b}$ where
$A = [\begin{matrix} 1 & 1 \\ 10^{- 10} & 0 \\ 0 & 10^{- 10} \end{matrix}] and \overset{⇀}{b} = [\begin{matrix} 1 \\ 10^{- 10} \\ 10^{- 10} \end{matrix}]$ .

The solution is $a = (\frac{1 - 10^{- 40}}{10^{- 40} + 10^{- 20} - 1}), a n d b = \frac{10^{- 20} (1 + 10^{- 20})}{(10^{- 40} + 10^{- 20} - 1)}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step:1 Definition of least square

Consider a linear system as $A \overset{⇀}{x} = \overset{⇀}{b}$ .

Here A is an matrix, a vector $\overset{⇀}{x}$ in $R^{n}$ called a least square solution of this system if role="math" localid="1659687531916" $||\overset{⇀}{b} - A \overset{⇀}{x^{k}}|| \leq ||\overset{⇀}{b} - A \overset{⇀}{x^{k}}|| for all \overset{⇀}{x} in R^{m}$ .

Step:2 Explanation of the solution

Consider a linear system as $A \overset{⇀}{x} = \overset{⇀}{b}$ where $A = [\begin{matrix} 1 & 1 \\ 10^{- 10} & 0 \\ 0 & 10^{- 10} \end{matrix}] and \overset{⇀}{b} = [\begin{matrix} 1 \\ 10^{- 10} \\ 10^{- 10} \end{matrix}]$ and A also A is A is an matrix, a vector $\overset{⇀}{x}$ in $R^{n}$ called a least square solution of this system if

localid="1659688022040" $||\overset{⇀}{b} - A {\overset{⇀}{x}}^{k}|| \leq ||\overset{⇀}{b} - A {\overset{⇀}{x}}^{k}|| for all \overset{⇀}{x} in R^{m}$

The term least square solution reflects the fact that minimizing the sum of the squares of the component of the vector $(\overset{⇀}{b} - A \overset{⇀}{x})$ .

To show ${\overset{⇀}{x}}^{k}$ of a linear system $A \overset{⇀}{x} = \overset{⇀}{b}$ .

Consider the following string of equivalent statements.

The vector ${\overset{⇀}{x}}^{k}$ is a least square solution of the system $A \overset{⇀}{x} = \overset{⇀}{b}$ .

Then by the definition as follows.

$\Leftrightarrow ||\overset{⇀}{b} - A {\overset{⇀}{x}}^{k}|| \leq ||\overset{⇀}{b} - A {\overset{⇀}{x}}^{k}|| for all \overset{⇀}{x} in R^{m}$

Also by the theorem 5.4.3 as follows.

$\Leftrightarrow A \overset{⇀}{x} = proj (\overset{⇀}{b}) where V = im (A)$

Similarly by the theorem 5.1.4 and 5.4.1 as follows.

$\Leftrightarrow \overset{⇀}{b} - A \overset{⇀}{x} is in V - = {(imA)}^{⊥} = \ker (A^{T}) ⇕ A^{T} (\overset{⇀}{b} - A \overset{⇀}{x}) = \overset{⇀}{0} ⇕ A^{T} A \overset{⇀}{x} = A^{T} \overset{⇀}{b}$ .

Therefore, the least square of the system $A \overset{⇀}{x} = \overset{⇀}{b}$ are the exact solution of the system .

$A^{T} A \overset{⇀}{x} = A^{T} \overset{⇀}{b}$

The system $A^{T} A \overset{⇀}{x} = A^{T} \overset{⇀}{b}$ is called the normal equation of .

Now, simplify as follows.

$[\begin{matrix} 1 & 10^{- 10} & 0 \\ 1 & 0 & 10^{- 10} \end{matrix}] [\begin{matrix} 1 & 1 \\ 10^{- 10} & 0 \\ 0 & 10^{- 10} \end{matrix}] = [\begin{matrix} 1 & 10^{- 10} & 0 \\ 1 & 0 & 10^{- 10} \end{matrix}] [\begin{matrix} 1 \\ 10^{- 10} \\ 10^{- 10} \end{matrix}] [\begin{matrix} 1 + 10^{10} & 1 \\ 1 & 10^{- 20} \end{matrix}] \overset{⇀}{x} = [\begin{matrix} 1 + 10^{- 20} \\ 1 + 10^{- 20} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} a \\ b \end{matrix}] where x_{1} = a (\frac{1 + 10^{- 40}}{10^{- 40} + 10^{- 20} - 1}) and x_{2} = b = \frac{10^{- 20} (1 + 10^{- 20})}{(10^{- 40} + 10^{- 20} - 1)} Hence, the solution is a = (\frac{1 + 10^{- 40}}{10^{- 40} + 10^{- 20} - 1}) and b = \frac{10^{- 20} (1 + 10^{- 20})}{(10^{- 40} + 10^{- 20} - 1)} .$

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Linear Algebra With Applications

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Short Answer

Find the least square solutions of the system $A \overset{⇀}{x} = \overset{⇀}{b}$ where
$A = [\begin{matrix} 1 & 1 \\ 10^{- 10} & 0 \\ 0 & 10^{- 10} \end{matrix}] and \overset{⇀}{b} = [\begin{matrix} 1 \\ 10^{- 10} \\ 10^{- 10} \end{matrix}]$ .

Step by Step Solution

Step:1 Definition of least square

Step:2 Explanation of the solution

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Find the least square solutions of the system Ax⇀=b⇀ where A=[1110-100010-10] and b⇀=[110-1010-10].

Step by Step Solution

Step:1 Definition of least square

Step:2 Explanation of the solution

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Find the least square solutions of the system $A \overset{⇀}{x} = \overset{⇀}{b}$ where
$A = [\begin{matrix} 1 & 1 \\ 10^{- 10} & 0 \\ 0 & 10^{- 10} \end{matrix}] and \overset{⇀}{b} = [\begin{matrix} 1 \\ 10^{- 10} \\ 10^{- 10} \end{matrix}]$ .