Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Using paper and pencil, find the QR factorization of the matrices in Exercises 15 through 28. Compare with Exercises 1 through 14.
19. $[\begin{matrix} 2 & 1 \\ 2 & 1 \\ 1 & 5 \end{matrix}]$

The QR factorization of the matrix is $[\begin{matrix} 2 & 1 \\ 2 & 1 \\ 1 & 5 \end{matrix}] = \frac{1}{3} [\begin{matrix} 2 & - 1 / \sqrt{2} \\ 2 & - 1 / \sqrt{2} \\ 1 & 4 / \sqrt{2} \end{matrix}] [\begin{matrix} 3 & 0 \\ 3 & \sqrt[3]{2} \end{matrix}]$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine column u→1 and entries r11 of R

Consider the matrix $M = [\begin{matrix} 2 & 1 \\ 2 & 1 \\ 1 & 5 \end{matrix}]$ where ${\vec{v}}_{1} = [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$ and ${\vec{v}}_{2} = [\begin{matrix} 1 \\ 1 \\ 5 \end{matrix}]$

By the theorem of QR method, the value of ${\vec{u}}_{1}$ and $r_{11}$ is defined as follows:

$r_{11} = ||\vec{v_{1}}|| {\vec{u}}_{1} = \frac{1}{r_{11}} {\vec{v}}_{1}$ .

Simplify the equation $r_{11} = | | {\vec{v}}_{1} | |$ as follows:

$r_{11} = | | {\vec{v}}_{1} | | r_{11} = ||[\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]|| r_{11} = \sqrt{2^{2} + 2^{2} + 1^{2}} r_{11} = 3$

Substitute the values 7 for $r_{11}$ and $[\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$ for ${\vec{v}}_{1}$ in the equation ${\vec{u}}_{1} = \frac{1}{r_{11}} {\vec{v}}_{1}$ as follows:

${\vec{u}}_{1} = \frac{1}{r_{11}} {\vec{v}}_{1} {\vec{u}}_{1} = \frac{1}{3} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}] {\vec{u}}_{1} = [\begin{matrix} 2 / 3 \\ 2 / 3 \\ 1 / 3 \end{matrix}]$

Therefore, the values localid="1659437134701" ${\vec{u}}_{1} = [\begin{matrix} 2 / 3 \\ 2 / 3 \\ 1 / 3 \end{matrix}] and r_{11} = 3$ .

Step 2: Determine column v→2⊥ and entries r12 of R

As $r_{12} = \vec{u_{1}} \cdot \vec{v_{2}}$ , substitute the values $[\begin{matrix} 1 \\ 1 \\ 5 \end{matrix}]$ for $\vec{V_{2}}$ and ${[\begin{matrix} \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{matrix}]}^{T}$ for $\vec{u_{1}}$ in the equation

localid="1659437594586" $r_{11} = \vec{u_{1}} . \vec{V_{2}}$ as follows:

$r_{11} = \vec{u_{1}} . \vec{V_{2}}$

$r_{12} = {[\begin{matrix} \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{matrix}]}^{T} . [\begin{matrix} 1 \\ 1 \\ 5 \end{matrix}]$

$r_{12} = \frac{2}{3} + \frac{2}{3} + \frac{5}{3} r_{12} = 3$

Substitute the values $[\begin{matrix} 1 \\ 1 \\ 5 \end{matrix}]$ for $\vec{V_{2}}$ ,3 for $r_{12}$ and $[\begin{matrix} 2 / 3 \\ 2 / 3 \\ 1 / 3 \end{matrix}]$ for $\vec{u_{1}}$ in the equation ${\vec{v}}_{2}^{⊥} = \vec{V_{2}} - r_{12} {\vec{u}}_{1}$ as follows:

${\vec{v}}_{2}^{⊥} = \vec{V_{2}} - r_{12} {\vec{u}}_{1}$

${\vec{v}}_{2}^{⊥} = [\begin{matrix} 1 \\ 1 \\ 5 \end{matrix}] - 3 [\begin{matrix} 2 / 2 \\ 2 / 3 \\ 1 / 3 \end{matrix}] {\vec{v}}_{2}^{⊥} = [\begin{matrix} 1 \\ 1 \\ 5 \end{matrix}] - [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}] {\vec{v}}_{2}^{⊥} = [\begin{matrix} - 1 \\ - 1 \\ 4 \end{matrix}] Therefore, {\vec{v}}_{2}^{⊥} = [\begin{matrix} - 1 \\ - 1 \\ 4 \end{matrix}] and r_{12} = 3$

Step 3: Determine column u→2 and entries r22 of R

The value of $\vec{u_{2}}$ and $r_{22}$ is defined as follows.

$\begin{array}{l} r_{22} = ||\vec{V_{2}^{⊥}}|| \end{array} {\vec{u}}_{2} = \frac{1}{r_{22}} \vec{V_{2}^{⊥}}$

Simplify the equation localid="1659438712674" $\begin{array}{l} r_{22} = ||\vec{V_{2}^{⊥}}|| \end{array}$ as follows.

$r_{22} = ||\vec{V_{2}^{⊥}}||$

$r_{22} = ||[\begin{matrix} - 1 \\ - 1 \\ 4 \end{matrix}]||$

$r_{22} = \sqrt{{(- 1)}^{2} + {(- 1)}^{2} + {(4)}^{2}} r_{22} = 3 \sqrt{2}$

Substitute the values $3 \sqrt{2}$ for $r_{22}$ and $[\begin{matrix} - 1 \\ - 1 \\ 4 \end{matrix}]$ for $\vec{V_{2}^{⊥}}$ in the equation $\vec{u_{2}} = \frac{1}{r_{22}} \vec{V_{2}^{⊥}}$ as follows.

${\vec{u}}_{2} = \frac{1}{r_{22}} {\vec{v}}_{2}^{⊥} {\vec{u}}_{2} = \frac{1}{3 \sqrt{2}} [\begin{matrix} - 1 \\ - 1 \\ 4 \end{matrix}] {\vec{u}}_{2} = [\begin{matrix} - 1 / 3 \sqrt{2} \\ - 1 / 3 \sqrt{2} \\ 4 / 3 \sqrt{2} \end{matrix}]$

The values localid="1659439047262" ${\vec{u}}_{2} = [\begin{matrix} - 1 / 3 \sqrt{2} \\ - 1 / 3 \sqrt{2} \\ 4 / 3 \sqrt{2} \end{matrix}]$ and $r_{22} = 3 \sqrt{2}$

Therefore, the matrices $Q = [\begin{matrix} 2 / 3 & - 1 / 3 \sqrt{2} \\ 2 / 3 & - 1 / 3 \sqrt{2} \\ 1 / 3 & 4 / 3 \sqrt{2} \end{matrix}]$ and $R = [\begin{matrix} 3 & 3 \\ 0 & 3 \sqrt{2} \end{matrix}]$ .

Hence, the QR factorization of the matrix is localid="1659439350112" $[\begin{matrix} 2 & 1 \\ 2 & 1 \\ 1 & 5 \end{matrix}] = \frac{1}{3} [\begin{matrix} 2 & - 1 / \sqrt{2} \\ 2 & - 1 / \sqrt{2} \\ 1 & 4 / \sqrt{2} \end{matrix}] [\begin{matrix} 3 & 3 \\ 0 & 3 \sqrt{2} \end{matrix}]$ .

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Linear Algebra With Applications

Answers without the blur.

Short Answer

Using paper and pencil, find the QR factorization of the matrices in Exercises 15 through 28. Compare with Exercises 1 through 14.
19. $[\begin{matrix} 2 & 1 \\ 2 & 1 \\ 1 & 5 \end{matrix}]$

Step by Step Solution

Step 1: Determine column u→1 and entries r11 of R

Step 2: Determine column v→2⊥ and entries r12 of R

Step 3: Determine column u→2 and entries r22 of R

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Linear Algebra With Applications

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Short Answer

Using paper and pencil, find the QR factorization of the matrices in Exercises 15 through 28. Compare with Exercises 1 through 14.19.[212115]

Step by Step Solution

Step 1: Determine column u→1 and entries r11 of R

Step 2: Determine column v→2⊥ and entries r12 of R

Step 3: Determine column u→2 and entries r22 of R

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Using paper and pencil, find the QR factorization of the matrices in Exercises 15 through 28. Compare with Exercises 1 through 14.
19. $[\begin{matrix} 2 & 1 \\ 2 & 1 \\ 1 & 5 \end{matrix}]$