text: ### Select your language
  
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  ### 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
explanation: The provided text is a detailed solution to a linear algebra problem. It contains a math question and a step-by-step solution to that question. The question asks to sketch the curve defined by the equation \(3x_1^2 + 4x_1x_2 = 1\) and to label the principal axes, intercepts, and the formula of the curve in the coordinate system defined by the principal axes. The solution includes finding the eigenvalues and eigenvectors of the matrix associated with the quadratic form, and then using these to sketch the curve.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ### 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed solution to a linear algebra problem. It contains a math question and a step-by-step solution to that question. The question asks to sketch the curve defined by the equation \(3x_1^2 + 4x_1x_2 = 1\) and to label the principal axes, intercepts, and the formula of the curve in the coordinate system defined by the principal axes. The solution includes finding the eigenvalues and eigenvectors of the matrix associated with the quadratic form, and then using these to sketch the curve.
  
  Label: question-and-answer