Q43E For聽A=[211111121111112]聽, fin... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 8: Q43E (page 393)

For $A = [\begin{matrix} 2 & 11 & 11 \\ 11 & 2 & 11 \\ 11 & 11 & 2 \end{matrix}]$ , find a nonzero vector $\overset{鈫�}{v}$ in ${鈩�}^{3}$ such that $A \overset{鈫�}{v}$ is orthogonal to $\overset{鈫�}{v}$ .

Short Answer

Expert verified

A nonzero vector $\overset{鈫�}{v} = [\begin{matrix} 0 \\ 2 \\ - 11 + 3 \sqrt{13} \end{matrix}]$

Step by step solution

Symmetric matrix:

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

Find the dot product, which must equal zero:

For $A \overset{鈫�}{v}$ to be orthogonal $\overset{鈫�}{v}$ , their dot product needs to equal zero. Let

$\overset{鈫�}{v} = [\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix}]$

So,

$A \overset{鈫�}{v} . \overset{鈫�}{v} = 0 鈫� [\begin{matrix} 2 & 11 & 11 \\ 11 & 2 & 11 \\ 11 & 11 & 2 \end{matrix}] [\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix}] . [\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix}] = 0 [\begin{matrix} 2 v_{1} & 11 v_{2} & 11 v_{3} \\ 11 v_{1} & 2 v_{2} & 11 v_{3} \\ 11 v_{1} & 11 v_{2} & 2 v_{3} \end{matrix}] . [\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix}] = 0 v_{1} (\begin{matrix} 2 v_{1} & 11 v_{2} & 11 v_{3} \end{matrix}) + v_{2} (\begin{matrix} 11 v_{1} & 2 v_{2} & 11 v_{3} \end{matrix}) + v_{3} (\begin{matrix} 11 v_{1} & 11 v_{2} & 2 v_{3} \end{matrix}) = 0 2 v_{1}^{2} + 22 v_{1} v_{2} + 22 v_{1} v_{3} + 2 v_{2}^{2} + 22 v_{2} v_{3} + 2 v_{3}^{2} = 0 v_{1}^{2} + 11 v_{1} v_{2} + 11 v_{1} v_{3} + v_{2}^{2} + v_{2} {\overset{鈫�}{v}}_{3} + v_{3}^{2} = 0$

Find a nonzero vector v→ in ℝ3 that Av→ is orthogonal to v→ :

choosing $v_{1} = 0$ yields

$v_{2}^{2} + 11 v_{2} v_{3} + v_{3}^{2} = 0$

choosing $v_{2} = 2$ yields

$4 + 22 v_{3} + v_{3}^{2} = 0 v_{3}^{2} + 22 v_{3} + 4 = 0 v_{3} = \frac{- 22 卤 \sqrt{484 - 16}}{2} = - 11 卤 3 \sqrt{13}$

we can choose either $v_{3} = - 11 + 3 \sqrt{13}$ or $v_{3} = - 11 - 3 \sqrt{13}$ , therefore,

$\overset{鈫�}{v} = [\begin{matrix} 0 \\ 2 \\ - 11 + 3 \sqrt{13} \end{matrix}]$

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91影视

For $A = [\begin{matrix} 2 & 11 & 11 \\ 11 & 2 & 11 \\ 11 & 11 & 2 \end{matrix}]$ , find a nonzero vector $\overset{鈫�}{v}$ in ${鈩�}^{3}$ such that $A \overset{鈫�}{v}$ is orthogonal to $\overset{鈫�}{v}$ .

Short Answer

Step by step solution

Symmetric matrix:

Find the dot product, which must equal zero:

Find a nonzero vector v→ in ℝ3 that Av→ is orthogonal to v→ :

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91影视

ForA=[211111121111112], find a nonzero vectorv鈫� in鈩�3such thatAv鈫�is orthogonal tov鈫�.

Short Answer

Step by step solution

Symmetric matrix:

Find the dot product, which must equal zero:

Find a nonzero vector v→ in ℝ3 that Av→ is orthogonal to v→ :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Calculus

Discrete Mathematics

Statistics

Geometry

Mechanics Maths

Study anywhere. Anytime. Across all devices.

For $A = [\begin{matrix} 2 & 11 & 11 \\ 11 & 2 & 11 \\ 11 & 11 & 2 \end{matrix}]$ , find a nonzero vector $\overset{鈫�}{v}$ in ${鈩�}^{3}$ such that $A \overset{鈫�}{v}$ is orthogonal to $\overset{鈫�}{v}$ .