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

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(Created page with "As in hint (2), following similar steps in (a) gives other two eigenvectors<math display="block">v_2=\begin{bmatrix}-1\\0\\1\end{bmatrix}, \ v_3=\begin{bmatrix}1\\0\\1\end{b...")
 
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As in hint (2),  following similar steps in (a) gives other two eigenvectors<math display="block">v_2=\begin{bmatrix}-1\\0\\1\end{bmatrix}, \  v_3=\begin{bmatrix}1\\0\\1\end{bmatrix} .</math>
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According to the definition, if we want to prove three eigenvectors are linearly independent,  we have to check the solution to following equation
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<math display="block">a\begin{bmatrix}-3\\4\\1\end{bmatrix}+b\begin{bmatrix}-1\\0\\1\end{bmatrix}+c\begin{bmatrix}1\\0\\1\end{bmatrix}=0 </math>
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is <math display="inline">a=b=c=0</math>.
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It is equivalent to
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<math display="block">\begin{bmatrix}-3&-1&1\\4&0&0\\1&1&1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix}=0 .</math>
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Applying row reductions as follows
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<math display="block">\begin{bmatrix}-3&-1&1\\4&0&0\\1&1&1\end{bmatrix}\xrightarrow{R2/4, R_1\leftrightarrow R_1}\begin{bmatrix}1&0&0\\-3&-1&1\\1&1&1\end{bmatrix}\xrightarrow {R_2+3R_1, R_3-R_1}\begin{bmatrix}1&0&0\\0&-1&1\\0&1&1\end{bmatrix} \xrightarrow{R_3+R_2}\begin{bmatrix}1&0&0\\0&-1&1\\0&0&2\end{bmatrix}\xrightarrow{R_2-1/2R_3}\begin{bmatrix}1&0&0\\0&-1&0\\0&0&2\end{bmatrix}.</math>
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Thus we get
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<math display="block">\begin{bmatrix}1&0&0\\0&-1&0\\0&0&2\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix}=0 .</math>
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That is <math display="inline">a=b=c=0</math>.
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