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Q17E

Expert-verified
Found in: Page 400

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Sketch the curves defined in Exercises 15 through 20. In each case, draw and label the principal axes, label the intercepts of the curve with the principal axes, and give the formula of the curve in the coordinate system defined by the principal axes.17. ${\mathbf{3}}{{\mathbf{x}}}_{{\mathbf{1}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{4}}{{\mathbf{x}}}_{{\mathbf{1}}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}$

${\stackrel{\to }{v}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]and{\stackrel{\to }{v}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]$

See the step by step solution

## Step 1: Given Information:

$3{x}_{1}^{2}+4{x}_{1}{x}_{2}=1$

## Step 2: To Find the Eigen values:

$q\left({x}_{1},{x}_{2}\right)=3{x}_{1}^{2}+4{x}_{1}{x}_{2}=1\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]\left[\begin{array}{c}3{x}_{1}+2{x}_{2}\\ 2{x}_{1}\end{array}\right]$

Split the term $4{x}_{1}{x}_{2}$ equally between the two components. Therefore

$q\left(\stackrel{\to }{x}\right)=\stackrel{\to }{x}×A\stackrel{\to }{x},whereA=\left[\begin{array}{cc}3& 2\\ 2& 0\end{array}\right]$

To determine the eigen values of the matrix A

$det\left(A-\lambda l\right)=0\left|\begin{array}{cc}3-\lambda & 2\\ 2& 0-\lambda \end{array}\right|=0\phantom{\rule{0ex}{0ex}}\left(3-\lambda \right)\left(0-\lambda \right)-\left(2\right)\left(2\right)=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-3\lambda -4=0\phantom{\rule{0ex}{0ex}}\left(\lambda -4\right)\left(\lambda +1\right)=0$

The eigen values are ${\lambda }_{1}=4and{\lambda }_{2}=-1$

To find the eigen vectors are ${\lambda }_{1}=4$

$\left(A-4I\right)\stackrel{~}{x}=0\left[\begin{array}{cc}3-4& 2\\ 2& 0-4\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

Apply Row operation ${R}_{2}\to {R}_{2}+2{R}_{1}$

$\left[\begin{array}{cc}-1& 2\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

The first row implies ${x}_{1}=2{x}_{2},{x}_{2}=1,{x}_{1}=2,{\lambda }_{1}=4$

${\stackrel{\to }{u}}_{1}=\left[\begin{array}{c}2\\ 1\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

${\stackrel{\to }{v}}_{1}=\frac{1}{||{\stackrel{\to }{u}}_{1}||}{\stackrel{\to }{u}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]$

When ${\lambda }_{2}=-1$

$\left(A+l\right)\stackrel{\to }{x}=0\left[\begin{array}{cc}3+1& 2\\ 2& 0+1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}4& 2\\ 2& 1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

Apply row operation ${R}_{2}\to 2{R}_{2}-{R}_{1}$

$\left[\begin{array}{cc}4& 2\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

First row implies that ${x}_{1}=-\frac{1}{2}{x}_{2},{x}_{2}=2,{x}_{1}=-1,{\lambda }_{2}=-1$

${\stackrel{\to }{u}}_{2}=\left[\begin{array}{c}-1\\ 2\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

${\stackrel{\to }{v}}_{2}=\frac{1}{||{\stackrel{\to }{u}}_{2}||}{\stackrel{\to }{u}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}4{c}_{1}^{2}-{c}_{1}^{2}=1$

${\stackrel{\to }{v}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]and{\stackrel{\to }{v}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]$

## Step 3: To Find the graph of the axes:

The graph of the coordinate axes is
### Content Structure
The passage contains a mathematical problem from a linear algebra textbook. It involves sketching a curve defined by a quadratic equation and finding the principal axes, intercepts, and the formula of the curve in a new coordinate system defined by the principal axes. The solution includes detailed steps for finding eigenvalues and eigenvectors, and then using them to transform the original equation.

### Analysis
single_question_with_answer

- the text contains a single mathematical question (q17e) that asks to sketch a curve and provide specific details about it.
- the solution to this question is provided in the text, including the steps to find eigenvalues, eigenvectors, and the transformation of the curve.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
### Select your language

Suggested languages for you:

Americas

Europe

Q17E

Expert-verified
Found in: Page 400

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Sketch the curves defined in Exercises 15 through 20. In each case, draw and label the principal axes, label the intercepts of the curve with the principal axes, and give the formula of the curve in the coordinate system defined by the principal axes.17. ${\mathbf{3}}{{\mathbf{x}}}_{{\mathbf{1}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{4}}{{\mathbf{x}}}_{{\mathbf{1}}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}$

${\stackrel{\to }{v}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]and{\stackrel{\to }{v}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]$

See the step by step solution

## Step 1: Given Information:

$3{x}_{1}^{2}+4{x}_{1}{x}_{2}=1$

## Step 2: To Find the Eigen values:

$q\left({x}_{1},{x}_{2}\right)=3{x}_{1}^{2}+4{x}_{1}{x}_{2}=1\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]\left[\begin{array}{c}3{x}_{1}+2{x}_{2}\\ 2{x}_{1}\end{array}\right]$

Split the term $4{x}_{1}{x}_{2}$ equally between the two components. Therefore

$q\left(\stackrel{\to }{x}\right)=\stackrel{\to }{x}×A\stackrel{\to }{x},whereA=\left[\begin{array}{cc}3& 2\\ 2& 0\end{array}\right]$

To determine the eigen values of the matrix A

$det\left(A-\lambda l\right)=0\left|\begin{array}{cc}3-\lambda & 2\\ 2& 0-\lambda \end{array}\right|=0\phantom{\rule{0ex}{0ex}}\left(3-\lambda \right)\left(0-\lambda \right)-\left(2\right)\left(2\right)=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-3\lambda -4=0\phantom{\rule{0ex}{0ex}}\left(\lambda -4\right)\left(\lambda +1\right)=0$

The eigen values are ${\lambda }_{1}=4and{\lambda }_{2}=-1$

To find the eigen vectors are ${\lambda }_{1}=4$

$\left(A-4I\right)\stackrel{~}{x}=0\left[\begin{array}{cc}3-4& 2\\ 2& 0-4\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

Apply Row operation ${R}_{2}\to {R}_{2}+2{R}_{1}$

$\left[\begin{array}{cc}-1& 2\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

The first row implies ${x}_{1}=2{x}_{2},{x}_{2}=1,{x}_{1}=2,{\lambda }_{1}=4$

${\stackrel{\to }{u}}_{1}=\left[\begin{array}{c}2\\ 1\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

${\stackrel{\to }{v}}_{1}=\frac{1}{||{\stackrel{\to }{u}}_{1}||}{\stackrel{\to }{u}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]$

When ${\lambda }_{2}=-1$

$\left(A+l\right)\stackrel{\to }{x}=0\left[\begin{array}{cc}3+1& 2\\ 2& 0+1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}4& 2\\ 2& 1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

Apply row operation ${R}_{2}\to 2{R}_{2}-{R}_{1}$

$\left[\begin{array}{cc}4& 2\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

First row implies that ${x}_{1}=-\frac{1}{2}{x}_{2},{x}_{2}=2,{x}_{1}=-1,{\lambda }_{2}=-1$

${\stackrel{\to }{u}}_{2}=\left[\begin{array}{c}-1\\ 2\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

${\stackrel{\to }{v}}_{2}=\frac{1}{||{\stackrel{\to }{u}}_{2}||}{\stackrel{\to }{u}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}4{c}_{1}^{2}-{c}_{1}^{2}=1$

${\stackrel{\to }{v}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]and{\stackrel{\to }{v}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]$

## Step 3: To Find the graph of the axes:

The graph of the coordinate axes is
user
### Content Structure
The passage contains a mathematical problem from a linear algebra textbook. It involves sketching a curve defined by a quadratic equation and finding the principal axes, intercepts, and the formula of the curve in a new coordinate system defined by the principal axes. The solution includes detailed steps for finding eigenvalues and eigenvectors, and then using them to transform the original equation.

### Analysis
TYPE: SINGLE_QUESTION_WITH_ANSWER

- The text contains a single mathematical question (Q17E) that asks to sketch a curve and provide specific details about it.
- The solution to this question is provided in the text, including the steps to find eigenvalues, eigenvectors, and the transformation of the curve.