Difference between revisions of "031 Review Part 1, Problem 3"

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!Solution:    
 
!Solution:    
 
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|-
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|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
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|The eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are &nbsp;<math style="vertical-align: -4px"> 0, 1, -1, -e.</math>
 
|-
 
|-
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|So, we have
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|Hence, the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math> are distinct.
 
|-
 
|-
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|
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|Therefore, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable and the statement is true.
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
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\displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
 
−
&&\\
 
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& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
 
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&&\\
 
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& = & \displaystyle{-\frac{2}{5}}.
 
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\end{array}</math>
 
 
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|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|-
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|&nbsp;&nbsp; &nbsp; &nbsp; False
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|&nbsp;&nbsp; &nbsp; &nbsp; TRUE
 
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|}
 
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]
 
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:38, 9 October 2017

True or false: If    is a    matrix with characteristic equation    then    is diagonalizable.

Solution:  
The eigenvalues of    are  
Hence, the eigenvalues of   are distinct.
Therefore,    is diagonalizable and the statement is true.
Final Answer:  
       TRUE

Return to Sample Exam

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