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

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In order to find other eigenvalues we need to solve the characteristic equation. <math display="block">\operatorname{det}(A-\lambda I)=0.</math>  
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In order to find other eigenvalues, we solve the characteristic equation <math display="block">\operatorname{det}(A-\lambda I)=0,</math>  
  
  
i.e., <math display="block">\operatorname{det}\begin{pmatrix}4-\lambda&-1&7\\0&3-\lambda&0\\1&2&-2-\lambda \end{pmatrix}=(4-\lambda)\cdot\operatorname{det}\begin{pmatrix}3-\lambda&0\\2&-2-\lambda \end{pmatrix}+1\cdot\operatorname{det}\begin{pmatrix}-1&7\\3-\lambda&0 \end{pmatrix}=0.</math>  
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i.e., <math display="block">\operatorname{det}\begin{bmatrix}4-\lambda&-1&7\\0&3-\lambda&0\\1&2&-2-\lambda \end{bmatrix}=(3-\lambda)\cdot\operatorname{det}\begin{bmatrix}4 -
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\lambda & 7 \\ 1 & -2-\lambda\end{bmatrix}=0.</math>  
  
Then we have
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Thus we have
 
<math display="block">
 
<math display="block">
(4-\lambda)(3-\lambda)(-2-\lambda)-7(3-\lambda)=0.
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(3-\lambda)[(4-\lambda)(-2-\lambda)-7]=0.
 
</math>  
 
</math>  
  
After simplification we obtain <math display="inline">(3-\lambda)(\lambda+3)(\lambda-5)=0</math>, for other eigenvalues expect <math display="inline">\lambda_1=3</math>, we also have <math display="inline">\color{blue}\lambda_2=-3, \lambda_3=5</math>.
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After simplification, we obtain <math display="inline">(3-\lambda)(\lambda+3)(\lambda-5)=0</math>; we know of the eigenvalue <math display="inline">\lambda_1=3</math> from part (a), and the others are <math display="inline">\color{blue}\lambda_2=-3, \lambda_3=5.</math>

Latest revision as of 20:45, 8 March 2018

In order to find other eigenvalues, we solve the characteristic equation


i.e.,

Thus we have

After simplification, we obtain ; we know of the eigenvalue from part (a), and the others are