Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (a)/Solution 1"

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Since the eigenvector <math>x</math> of <math>A</math> for eigenvalue <math>\lambda_1</math> is the solution to <math display="inline">Ax=\lambda_1x</math>, i.e., <math display="inline">(A-\lambda_1I)x=0</math>, all we need to do is to solve following system. <math display="block">(A-\lambda_1I)x=
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Since an eigenvector <math>\mathbf{x}</math> of <math>A</math> of eigenvalue <math>\lambda_1</math> satisfies <math display="inline">A\mathbf{x}=\lambda_1\mathbf{x}</math>, i.e., <math display="inline">(A-\lambda_1I)\mathbf{x}=\mathbf{0}</math>, all we need to do is solve the following system.  
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<math display="block">(A-\lambda_1I)\mathbf{x}=
 
\left(\begin{bmatrix}
 
\left(\begin{bmatrix}
 
4 & -1 & 7 \\
 
4 & -1 & 7 \\
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0 & 1 & 0\\
 
0 & 1 & 0\\
 
0 & 0 & 1
 
0 & 0 & 1
\end{bmatrix}\right)\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}1&-1&7\\0&0&0\\1&2&-5 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=0.</math> By applying Gaussian elimination
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\end{bmatrix}\right)\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}1&-1&7\\0&0&0\\1&2&-5 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\mathbf{0}.</math>  
  
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By applying Gaussian elimination,
  
 
<math display="block">
 
<math display="block">
\begin{bmatrix}1&-1&7\\0&0&0\\1&2&-5 \end{bmatrix}\xrightarrow{R_2\leftrightarrow R_3, R_2-R_1} \begin{bmatrix}1&-1&7\\0&3&-12\\0&0&0 \end{bmatrix}\xrightarrow{R_2/3}\begin{bmatrix}1&-1&7\\0&1&-4\\0&0&0 \end{bmatrix}\xrightarrow{R_1-R_2}\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0 \end{bmatrix}
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\begin{bmatrix}1&-1&7\\0&0&0\\1&2&-5 \end{bmatrix}\xrightarrow{R_2\leftrightarrow R_3, R_2-R_1} \begin{bmatrix}1&-1&7\\0&3&-12\\0&0&0 \end{bmatrix}\xrightarrow{R_2/3}\begin{bmatrix}1&-1&7\\0&1&-4\\0&0&0 \end{bmatrix}\xrightarrow{R_1-R_2}\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0 \end{bmatrix},
 
</math>
 
</math>
  
  
we also have <math display="block">\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=0.</math> Thus <math display="inline">x_1+3x_3=0</math> and <math display="inline">x_2-4x_3=0</math>, in vector form we get <math display="block">\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}-3x_3\\4x_3\\x_3 \end{bmatrix}=x_3\begin{bmatrix}-3\\4\\1 \end{bmatrix}.</math> So the eigenvector corresponding to <math display="inline">\lambda_1=3</math> is <math display="block">\color{blue}\begin{bmatrix}-3\\4\\1 \end{bmatrix}.</math>
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we obtain <math display="block">\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=0.</math>  
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Thus <math display="inline">x_1+3x_3=0</math> and <math display="inline">x_2-4x_3=0</math>; in vector form we get <math display="block">\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}-3x_3\\4x_3\\x_3 \end{bmatrix}=x_3\begin{bmatrix}-3\\4\\1 \end{bmatrix},</math> where <math>x_3 \in \mathbb{R}</math> is free. So an eigenvector corresponding to <math display="inline">\lambda_1=3</math> is <math display="block">\color{blue}\begin{bmatrix}-3\\4\\1 \end{bmatrix}.</math>

Revision as of 20:36, 8 March 2018

Since an eigenvector of of eigenvalue satisfies , i.e., , all we need to do is solve the following system.

By applying Gaussian elimination,


we obtain

Thus and ; in vector form we get where is free. So an eigenvector corresponding to is